TFS-education website

P a g e - 1

KoAunglwinoo( T.F.S – EDUCATION WEBSITE ) Page 1

P a g e - 2


Chapter ( 2 ) The Remainder Theorem and The Factor Theorem

အခကးပစကာလမာ Remainder theorem is divided by ( ၄ငးစကာလေတျ႕ရငး Remainder theorem သ ) Factor theorem is divisible , is exactly divisible , is the factor of , is the root of …. ( ၄ငးစကာလမာေတျ႕ရငး Factor theorem သ ) The Remainder Theorem ဒဂရအမငးဆေတျ႕ေအာငးရြာ f ( x ) မြတးပါ ငယးတကနးမြာ “0” ကပး “x” တနးဖေလထျကး အဒ “x” တနးဖေလဟာ f ( x ) မြာ သျငးရမြာ ေနာကးဆရတ ရလာဒးေလ The remainder လ႔ေရ။ ( Polynomial တစးခ၏ ဒဂရ = အဆပါ Polynomial ရြ အမငးဆထပးညႊနး )

ဥပမာ - x3 + x2 – 6x ဒဂရ 3 ရြ Polynomial

X3 + x2 – x7 + x ဒဂရ 7 ရြ Polynomial မြတးခကး - တညးကနး = f ( x ) အခကးပစကာလ the expression / the polynomial စသညး စကာလမာ ေနာကးရြ ကနးတနး ( သ႔ ) is divided by / is dividible by / is exactiy dividible by / leave the remainder of ေရြ႕ရြ ကနးတနး စာကနး အခကးပစကာလ is divided by / is dividible by / is exactiy dividible by ေနာကးရြ ကနးတနး ( သ႔ ) is a factor of , is the root of ေရြ႕ရြ ကးနးတနး Remainder Theorem အတျကး ေရရမညးစာ ( 1 ) တညးကနးတစးခ စာကနးတစးခ ပစာၦမာ When f ( x ) is* divided by ( x – a ) , the remainder is* f ( a ). ( ေဖားမလာအရ တနးဖ ) f ( a ) = ေပခကးအရတနးဖ ( 2 ) တညးကနးတစးခ စာကနးႏြစးခ ပစာၦမာ When f ( x ) is* divided by ( x – a ) and ( x – b ) , the remainders* are* f ( a) and* f ( b ). ( ေဖားမလာအရ တနးဖ ) f ( a ) = ေပခကးအရတနးဖ , f ( b ) = ေပခကးအရတနးဖ

P a g e - 3


( 3 ) တညးကနးႏြစးခ စာကနးတစးခ ပစာၦမာ When f ( x ) and* g ( x ) are* divided by ( x – a ) , the remainders* are* f ( a) and* g ( a ). ( ေဖားမလာအရ တနးဖ ) f ( a ) = ေပခကးအရတနးဖ g ( a ) = ေပခကးအရတနးဖ The Factor Theorem Factor ဆရငး ေပာငးပနးပါ f ( x ) အတျကး “0” ရြာ ရြာလ႔ရတ ရလာဒးေလ လကၡဏာေပာငးပ x ေနာကးေရ Factor Theorem အတျကး ေရရမညးစာ ( 1 ) တညးကနးတစးခ ဆချကနးတစးခ ပစာၦမာ ( x – a ) is* a factor of f ( x ). f ( a ) = 0 ( 2 ) တညးကနးတစးခ ဆချကနးႏြစးခ ( သ႔ ) ႏြစးထပးဆချကးနးပစာၦမာ ( x – a ) and* ( x – b ) are* the factors* of f ( x ). f ( a ) = 0 and f ( b ) = 0 ( 2 ) တညးကနးႏြစးခ ဆချကနးတစးခ ပစာၦမာ ( x – a ) is* the factor of f ( x) and* g ( x ). f ( a ) = 0 g ( a ) = 0

Note f ( x ) တနးဖတျငး ထပးညႊနးအမငးဆ 3 ရြပါက factor တစးကျငး အရငးရြာ long division နညးဖငးကနးႏြြစးကျငးရြာ f ( x ) အာ သကျငး ချေပရမညး။ ထပးညႊနးအမငးဆ 4 ရြပါက factor ႏြစးကျငးအရငးရြာ အဆပါႏြစးကျငးအခငးခငးေမြာကးပ long division နညးဖငး ကနးႏြြစးကျငး ရြာ f ( x ) အာ ေလကျငးချေပရမညး။ ပစာၦတျငး ( 1 ) Factorize f ( x ) completely ေမပါက f ( x ) = ( x – a ) ( x – b ) ( x – c ) အထရြာ ( 2 ) Find the factors of f ( x ) ေမပါက f ( x ) = ( x – a ) ( x – b ) ( x – c ) အထရြာပ The factors are ( x – a ) , ( x – b ) and ( x – c ) ဟ ဆကးေရေပ။

P a g e - 4


( ၃ ) Solve f ( x ) = 0 / Find the roots of f ( x ) / Solve ေပထာေသာ ညမ ြခငး = 0 ေမပါက f ( x ) = ( x – a ) ( x – b ) ( x – c ) အထရြာပ The factors are ( x – a ) , ( x – b ) and ( x – c ) အစာ f ( x ) = 0 ( x – a ) ( x – b ) ( x – c ) = 0 x – a = 0 ( or ) x – b = 0 ( or ) x – c = 0 x = a ( or ) x = b ( or ) x = c အထ ဆကးရြာေပရမညး။ သတပရနး** ( 1 ) f ( x ) တနးဖသညး ( x – a ),( x – b ) ႏြငး ( x – c ) သကျငးရပါက ၄ငး သကျငးသညး f ( x ) ၏ factor မာ ဖစးသညး။ If f ( x ) = ( x – a ) ( x – b ) ( x – c ) ( x – a ),( x – b ) and ( x – c ) are the factors of f ( x ) ( သ႔ ) The factors of f ( x ) are ( x – a ),( x – b ) and ( x – c ) ( သ႔ ) The factors are ( x – a ),( x – b ) and ( x – c ) ၾကကးသလ ေရႏငးသညး။ ( 2 ) f ( x ) က 0 ႏြငး ညေပပရလာေသာ တနးဖသခ a , b , c သခသညး f ( x ) = 0 ၏ root မာ ဖစးသညး။ If f ( x ) = 0 , ( x – a ) ( x – b ) ( x – c ) = 0 and x = a or x = b or x = c a , b and c are the roots of f ( x ). ( သ႔ ) The roots of f ( x ) are a , b and c. ( သ႔ ) The three roots are a , b and c. ၾကကးသလ ေရႏငးသညး။ ( 3 ) အပနးအလြနးအာဖငး ပစာၦတျငး root သခ ေပလာပါက factor သခအဖစး ပငးေရတတးရမညး။ ဥပမာ - – 1 , 2 and 4 are the roots of f ( x ) ဟေပလာလ ြငး ( x + 1 ) , ( x – 2 ) and ( x – 4 ) are the factors of f ( x ) and f ( x ) = ( x + 1 )( x – 2 )( x – 4 ) ဟပငးေရတတးရမညး။ **မြာတတးေသာ အမြာမာ** ( 1 )f ( x ) = ( x – a ) ( x – b ) ( x – c ) အစာ f ( x ) = ( x – a ) , ( x – b ) , ( x – c ) ဟ မြာတတးသညး။ ( 2 ) The factors are ( x – a ) , ( x – b ) and ( x – c ). ဟ ေရရမညးအစာ The factors are ( x – a ) ( x – b ) ( x – c ) ဟ မြာတတးသညး။ ( 3 ) The factors are အစာ The factorize are ဟ မြာတတးသညး။ ( 4 ) f ( x ) leave the remainder of 13 when divided by ( x – 2 ) ဟေပထာေသာပစာၦတျငး When f ( x ) is divided by ( x – 2 ) , the remainder is f ( 2 ) ဟ ေရရမညးအစာ When f ( x ) is divided by ( x – 2 ) , the remainder is 13 ဟ မြာတတးသညး။

P a g e - 5


( 5 ) f ( x ) is divisible by x – 2 ဟေပထာေသာပစာၦတျငး x – 2 is a factor of f ( x ) . f ( 2 ) = 0 ဟ ေရရမညးအစာ When f ( x ) is divisible by ( x – 2 ) , the remainder , f ( 2 ) = 0 ဟ မြာတတးသညး။ ( ဤသ႔ေရလ ြငး သေဘာတရာအရ ရေသားလညးfactor theorem သရမညးဖစးသဖငး တတးႏငးသမ ြ remainder စကာလ မသေစခငးပါ၊ ) ပစာၦပစ ( 3 ) မ ( 1 ) Remainder သသနး႔ ပစာၦမာ ( 2 ) Factor သသနး႔ ပစာၦမာ ( 3 ) Remainder / Factor ေရာရာ ပစာၦမာ

P a g e - 6


( 1 ) Remainder သသနး႔ ပစာၦမာ

P a g e - 7


( 1 ) Remainder သသနး႔ ပစာၦမာ ပစ ( 1 ) တညးကနးတစးခ စာကနးတစးခ ပစာၦမာ ( a ) တညးကနးမသကနးပစာၦမာ ( ဤပစမ 5 marks Question တျငး မေမဘေသပါ။ )

------------------------------------------------------------------------ ( b ) စာကနးမသကနးပစာၦမာ ( ဤပစမ 5 marks Question တျငး မေမဘေသပါ။ )

------------------------------------------------------------------------ ( iii ) တညးကနးေရာ စာကနးပါ မသကနးပစာၦမာ ( ဤပစမ 5 marks Question တျငး မေမဘေသပါ။ )

------------------------------------------------------------------------ ပစ ( 2 ) တညးကနးတစးခ စာကနးႏြစးခ ပစာၦမာ ( a ) တညးကနးမသကနးပစာၦမာ Type ( 1 )

1. The remainder when 2x3 + kx2 + 7 is divided by x – 2, is half the remainder when the same expression is divided by 2x – 1. Find the value of k. Ans; k = – 5 2007 စစးကငး၊ခငး ( No 67 ) / 2002 ႏငးငခာ ( No 118 ) ( ဤပစာၦအာ သမြတးတနးအဖစးလညး ေမဖပါသညး။ )

2. The remainder when ax3 + bx2 + 2x + 3 is divided by x – 1, is twice that when it is divided by x + 1, show that b = 3a + 3. 2002 မႏ ေလ ( No 111 )

3. The expression x3 + ax2 + b has the same remainder of when divided by x + 1 and x – 2. Given that the remainder of when the expression is divided by x – 4 is 10, find the values a and b 2002 ရနးကနး ( No 110 ) Ans; a = – 3 , b = – 6 ------------------------------------------------------------------------------------------------------------------------ Type ( 2 )

4. The remainder when f ( x ) = ax2 + bx + c is divided by x – 1 , x + 1 , x – 2 are 1 , 25 , 1 respectively. Show that f ( x ) is a perfect square.

Ans; f ( x ) = 4x2 – 12x + 9 = ( 2x – 3 )2 , f ( x ) is a perfect square. 2012 ကခငး ( No 20 ) / 2003 မႏ ေလ ( No 100 ) ------------------------------------------------------------------------------------------------------------------------ Type ( 3 )

5. The expression ax3 – 6ax + b leaves the remainders of 1 and 2 when divided by x – 1 and x – 2 respectively. Find the values a and b and find the remainder when the expression is divided by x + 4. 2014 ကခငး ( No 9 ) Ans; a = 1 , b = 6 , the remainder = – 34

P a g e - 8


ပစတ ပစာၦမာ 2013 စစးကငး၊ခငး ( No 4 ) / 2011 ပခ၊ကခငး ( No 27 )/ 2004 ပခ ( No 95 )

The expression ax3 – x2 + bx – 1 leaves the remainders of – 33 and 77 when divided by x + 2 and x – 3 respectively. Find the values of a and b and the remainder when divided by x – 2. Ans; a = 3 , b = 2 , The remainder = 23 2009 မျနး၊ကရငး၊တနသၤာရ ( No 49 ) / 2006 စစးကငး၊ခငး ( No 77 )

The polynomial x3 + ax2 + bx – 3 leaves a remainders of 27 when divided by x – 2 and a remainders of 3 when divided by x + 1. Calculate the remainder when the polynomial is divided by x – 1. Ans; a = 6 , b = – 1 , The remainder = 3 2007 မေကျ ( No 66 )

The expression ax3 + x2 + bx has the remainder 3 when it is divided by x + 1 . When the expression is divided by x– 2 the remainder is 6. Find the remainder when the expression divided by x – 3. Ans; a = 1 , b = – 3 , The remainder = 27 2004 ကခငး ( No 96 ) ( ဤပစာၦမမြာ 3 marks Question အတျကးသာ တနးပါသညး။ )

The expression x2 + ax + b leaves a remainder 3 when it is divided by x – 1 and leaves a remainder 1 when it is divided by x – 3. Find the value of a. Ans; a = – 5

------------------------------------------------------------------------

( b ) စာကနးမသကနးပစာၦမာ

6. Given that the polynomial x2 – 10x + 14 leaves the same remainder when divided by x + 2b or x + 2c where 3b – 2c = 0 and b ≠ c. Find the value of b and c. Ans; b = – 2 , c = – 3 2014 ပခ ( No 5 )

7. Given that the expression x2 – 10x + 14 leaves the same remainder when divided by x + 2b or x + 2c, Where b ≠ c , show that b + c + 5 = 0. 2003 ဧရာတ ( No 105 )

8. Given that the expression x2 – 10x + 14 leaves the same remainder when divided by x + 2b or x + 2c, where b – c = 1 . Find the value of b and c. Ans; b = – 2 , c = – 3 2010 ကခငး ( No 40 ) ပစတပစာၦမာ 2007 ရနးကနး ( No 64 )

The expression x2 – 10x + 4 has the same remainder when divided by x – 2p or x + q where 2p ≠ – q. Find the value of 2p – q. Ans; 2p – q = 2

P a g e - 9


စပးဆကးေလကငးရနး ဖညးစျကးပစာၦမာ

*** The remainders when the expression 2x2 + 4x + 8 is divided by x + a or x – 2b are the same , where a + b = 5 and a ≠ – 2b . Find the value of a and b , and show that a – 4b = 0 . Ans; a = 4 , b = – 1 , a – 4b = 0

----------------------------------------------------------------------------------- *** Given that the expression x2 + 4x + k leaves the same remainder when divided by x + b or x + 3c,

where b – 3c = 2 . Find the value of b and c , and show that bc = 1. Ans; b = 2 , c = 1

3 , bc = 1

----------------------------------------------------------------------------------- *** The remainder when the expression x2 – x + 8 is divided by x – 2b is four times the remainder when the expression is divided by x – c , where b + c = 1 . Find the value of b and c.

-----------------------------------------------------------------------------------

Ans; b = 12 , c = – 11 *** The remainder when the expression 2x2 – 4x + 8 is divided by x – a is four times the remainder when the expression is divided by 2x – b , where a – b = 2. Find the value of a and b. Ans; a = 4 , b = 2

( *** ပထာေသာ ပစာၦမာသညး ဂဏးထမြနးေသာ ကေလမာ ေလကငးရနး ဖညးစျကးေပထာခငး ဖစးသညး။ ဤပစာၦမာက ေနာကးပငးတျငး မမတ႔တျကးခကးေလကငးမႈႏြငး တကးဆငးစစးေဆႏငးေစရနး တျကးပထာပါသညး။ ) ( c ) တညးကနးေရာ စာကနးပါ မသကနးပစာၦမာ

9. Given that the remainder when f ( x ) = x3 – x2 + ax is divided by x + a where a > 0 , is twice the remainder when f ( x ) is divided by x – 2a , find the value of a. Find also the remainder when f ( x ) is

divided by x – 2. Ans; a = 2

17 , the remainder = 4

4

17

2010 ရနးကနး ( No 33 ) စပးဆကးေလကငးရနး ဖညးစျကးပစာၦမာ

Given that the remainder when f ( x ) = x3 + ax2 – 12ax is divided by x – 3a where a > 0 , is three times the remainder when f ( x ) is divided by x + a , find the value of a. Find also the remainder when f ( x ) is divided by x – 2a. Ans; a = 2 , the remainder = 0

-----------------------------------------------------------------------------------

P a g e - 10


Given that the remainder when f ( x ) = 2x3 – ax2 + 22a2 is divided by x + 2a , is two times the remainder when f ( x ) is divided by x – a , find the value of a and the remainder when f ( x ) is divided by x – 2. Ans; a = 0 , the remainder = 16 ( or ) a = – 1 , the remainder = 21

------------------------------------------------------------------------

ပစ ( 3 ) တညးကနးႏြစးခ စာကနးတစးခ ပစာၦမာ ( a ) တညးကနးမသကနးပစာၦမာ Type ( 1 )

10. When the expressions ax3 + 5x2 + bx + 4 and bx3 + 9x2 + ax – 6 are divided by x + 3 , the remainders are – 14 and – 12. Find the value of a and b. Ans; a = 2 , b = 3 2004 ရနးကနး ( No 93 ) Type ( 2 )

11. The expressions x3 – ax + a2 and ax3 + x2 – 17 have the same remainder when divided by x – 2. Find the possible values of a and the remainder. Ans; a = 3 or a = 7 , The remainder = 11 or 43 2003 မေကျ ( No 101 )

------------------------------------------------------------------------

( b ) စာကနးမသကနးပစာၦမာ Type ( 1 )

12. The expression x3 – x2 – 4x + 24 and x3 – 7x + 6 have the same remainder when divided by x + k. Find the possible values of k. Ans; k = – 6 or k = 3 2010 ရခငး ( No 39 ) Type ( 2 )

13. Given that x3 – 2x2 – 3x – 11 and x3 – x2 – 9 have the same remainder when divided by x + a , determine the values of a and the corresponding remainders. 2006 ဧရာတ ( No 79 ) Ans; a = 1 and the remainder = – 11 or a = 2 and the remainder = – 21 Type ( 3 )

14.The remainder when x4 + 3x2 – 2x + 2 is divided by x + a is the square of the remainder when x2 – 3 is

divided by x + a . Find a. Ans ; a = 7

9 or a = – 1

2014 ႏငးငခာ ( No 11 ) / 2006 ႏငးငခာ ( No 84 )

P a g e - 11


စပးဆကးေလကငးရနး ပစာၦမာ

1.The remainder when 9x4 – 5x2 – 4x – 7 is divided by x – a is the square of the remainder when 3x2 – 2 is divided by x – a . Find a. Ans ; a = 1 or a = 3

----------------------------------------------------------------------------------- 2.The remainder when 4x4 – 10x2 + 10x + 21 is divided by x + a is the square of the remainder when

3 – 2x2 is divided by x + a . Find a. Ans ; a = 2 or a = 3

----------------------------------------------------------------------------------- *** 3.The remainder when x4 + 4x3 + x2 + 48 is divided by x + 2a is the square of the remainder when

x2 + 2x is divided by x + 2a ,where a >0 .Find a and the remainder when x2 + 2x is divided by x – a. Ans ; a = 2 , The remainder = 8

------------------------------------------------------------------------ *** 4.The remainder when x3 + x2 + 5x + 11 is divided by x – a is three times the remainder when 4 + 2x – x2

– x3 is divided by x – a, determine the values of a.

( *** ပထာေသာ ပစာၦမာသညး ဂဏးထမြနးေသာ ကေလမာ ေလကငးရနး ဖညးစျကးေပထာခငး ဖစးသညး။ ဤပစာၦမာက ေနာကးပငးတျငး မမတ႔တျကးခကးေလကငးမႈႏြငး တကးဆငးစစးေဆႏငးေစရနး တျကးပထာပါသညး။ )

------------------------------------------------------------------------

( c ) တညးကနးေရာ စာကနးပါ မသကနးပစာၦမာ ( ဤပစမ 5 marks Question တျငး မေမဘေသပါ။ )

------------------------------------------------------------------------

ပစ ( 4 ) တညးကနးႏြစးခ စာကနးႏြစးခ ပစာၦမာ ( ဤပစမ 5 marks Question တျငး မေမဘေသပါ။ )

------------------------------------------------------------------------ ( a ) တညးကနးမသကနးပစာၦမာ ( ဤပစမ 5 marks Question တျငး မေမဘေသပါ။ )

------------------------------------------------------------------------

( b ) စာကနးမသကနးပစာၦမာ ( ဤပစမ 5 marks Question တျငး မေမဘေသပါ။ )

P a g e - 12


( c ) တညးကနးေရာ စာကနးပါ မသကနးပစာၦမာ

15. The remainder when 3x3 – 3x2 – 2ax +5 is divided by x – a is three times the remainder when

19 + ax – 5x2 – x3 is divided by x + a , find the positive value of a. Ans; a = 2 2008 ကခငး ( No 61 ) စပးဆကးေလကငးရနး ပစာၦမာ

1.The remainder when x3 + ( a – 4 )x2 – 3a is divided by x – a is two times the remainder when

x3 + 2x is divided by x + a where a ≠ 0, find the remainder when x3 + ( a – 4 )x2 – 3a is divided by x – 2a. Ans; a = 2 , the remainder = – 4

----------------------------------------------------------------------------------- 2.The remainder when ( a – 3 )x3 + ( a – 2 )x2 + ( a – 1 ) x is divided by x – a is half of the remainder

when 2x4 + x3 – 2ax2 – ax is divided by x + a where a ≠ 0, find the possible values of a. Ans; a = – 2 or a = – 1

------------------------------------------------------------------------

P a g e - 13


( 2 ) Factor သသနး႔ ပစာၦမာ

P a g e - 14


( 2 ) Factor သသနး႔ ပစာၦမာ ( 1 ) ဆချကနးချခငးေသာ ပစာၦမာ

16. Find the factor of x4– 14x3 + 71x2 – 154x + 120. Ans; The factors are ( x – 5 ), ( x – 4 ), ( x – 3 ) and ( x – 2 ) 2011 မျနး၊ကရငး၊တနသၤာရ ( No 29 )

----------------------------------------------------------------------------------- 17. Show that 3x – 2 is a factor of 6x3 – x2 – 20x + 12 , ( factorize the polynomial completely ) and find

the other factors. Hence solve the equation 6x3 – x2 – 20x + 12 = 0.

Ans; the other factors are 2x – 3 and x+2 , x = 2

3 or x =

3

2 or x = – 2

2012 ဧရာတ ( No 17 ) / 2007 ကခငး ( No 72 ) ပစတပစာၦမာ 2011 ဧရာတ ( No 28 )

Show that 2x + 1 is a factor of 6x4– 5x3 – 10x2+ 5x + 4 and find the other factors. Ans; the other factors are x – 1 , 3x – 4 and x + 1 2006 ရခငး ( No 81 )

Prove that x + 2 is a factor of 6x3 + 5x2 – 17x – 6. Solve the equation 6x3 + 5x2 – 17x = 6

Ans; x = – 2 or x = 1

3 or x =

3

2

2005 ရခငး ( No 90 )

Prove that 2x – 3 is a factor of 2x3 – 13x2+ 23x – 12, and find the other factors. Hence solve

2x3 – 13x2+ 23x – 12 = 0 . Ans; The other factors are x – 1 and x – 4 , x = 1 or x = 4 or x = 3

2

2005 ရြမး။ကယာ ( No 91 )

Let f ( x ) = x4 – 9x2 – 4x + 12. Prove that x + 2 is a factor of f ( x ). Hence factorixe f ( x ) completely.

Ans; f ( x ) = ( x – 1 )( x – 3 )( x + 2 )2

------------------------------------------------------------------------

( 2 ) Solve the Equation / Find the factors. ပစာၦမာ

18. Solve the equation x4– 9x2 = 4x – 12. Ans; x = – 2 or x = 1 or x = 3 2014 ရြမး၊ကယာ ( No 10 ) ပစတ ပစာၦမာ 2013 မျနး၊ကရငး၊တနသၤာရ ( No 7 )

Solve x4 + 5x3 + 5x2 – 5x – 6 = 0. Ans; x = – 3 or x = – 2 or x = – 1 or x = 1

P a g e - 15


2011 ရခငး ( No 30 )

Solve the equation 2x3+ 3x2 = 5x + 6. Ans; x = – 2 or x = – 1 or x = 1 1

2

2009 ပခ ( No 47 )

Solve the equation 2x3 – 13x2 = 12 – 3x. Ans; x = 1 or x = 4 or x = 1 1

2

2008 မႏ ေလ ( No 55 )

Solve x4 – x3 – 16x2 + 4x + 48 = 0. Ans; x = – 3 or x = – 2 or x = 2 or x = 4 2008 မေကျ ၊မျနး၊ကရငး၊တနသၤာရ ( No 56 )

Solve 8x3 – 2x2 – 5x = 1. Ans; x = 1 or x = – 1

2 or x = –

1

4

2008 ပခ ( No 58 )

Solve the equation x3 + 15 = 17 – x2. Ans; x = – 5 or x = 1 or x = 3 2002 ပခ ( No 58 )

Solve the equation 2x3 + 5x2 = 2 – x. Ans; x = – 2 or x = – 1 or x = 1

2

----------------------------------------------------------------------------------- ( 3 ) တညးကနးတစးခ စာကနးတစးခ ပစာၦမာ ( a ) တညးကနးမသကနးပစာၦမာ Type ( 1 )

19. Given that f ( x ) = x3 + px2 – 2x + 4 3 has a factor x + 2 , find the value of p. Show that x – 2 3 is also a factor and solve the equation f ( x ) = 0. Ans; p = – 2 3 , x = 2 3 or x =– 2 or x = 2 2015 ႏငးငခာ ( No 2 ) / 2002 ရြမး၊ကယာ ( No 117 ) ပစတ ပစာၦမာ

Given that f ( x ) = x3 + px2 – 2x + 4 3 has a factor x – 2 3 , find the value of p. Show that x + 2 is also a factor and solve the equation f ( x ) = 0. Ans; p = – 2 3 , x = 2 3 or x =– 2 or x = 2 2018 ႏငးငခာ ( No 8 ) ( အထကးပါ နပါတး 19 ပစာၦအာ factor ႏြစးခ ေနရာ change ထာခငးသာဖစးပါသညး။ )

Given that f ( x ) = 2x3 + x2 – 24x + q has a factor x + 2 3 , find the value of q. Show that x – 2 3 is also a factor of f ( x ) and find the other factors. Ans; q = – 12 , the other factor = ( 2x + 1 ) 2012 ႏငးငခာ ( No 22 )

P a g e - 16


စပးဆကးေလကငးရနး ပစာၦမာ

The expression x3 + kx2 – 2x – 6 2 has a factor x – 3 , find the value of k. Show that x + 2 2 is also a factor of f ( x ) and factorize f ( x ) completely. Ans; k = 2 2 , f ( x ) = ( x – 3 )( x + 3 )( x + 2 2 )

If f ( x ) = x3 – 2x2 – ( a – 1 )x + 4 has a factor x – 2 , find the value of a and solve the equation f ( x ) = 0. Ans; a = 3 , x = – 2 or x = 2 or x = 2

------------------------------------------------------------------------------------------------ Type ( 2 )

20. The expression x2n – k has x + 3 as a factor and leaves a remainder of – 80 when divided by x + 1.

Calculate the values of n and k where both are integers. With these values n and k , factorise x2n – k

completely. Ans; n = 2 , k = 81 , f ( x ) = ( x – 3 )( x + 3 )( x2 + 9 ) 2014 ဧရာတ ( No 6 )

------------------------------------------------------------------------------------------------ Type ( 3 )

21. Given that f ( x ) = x2n – ( p + 1 ) x2 + p , where n and p are positive integers. Show that x – 1 is a factor of f ( x ) , for all values of p. When p = 4, find the value of n for which x – 2 is a factor of f ( x ) and factorize f ( x ) completely. Ans; n = 2 , f ( x ) = ( x – 1 )( x + 1 )( x – 2 )( x + 2 ) 2014 ရနးကနး ( No 1 ) / 2013 ရခငး ( No 8 ) စပးဆကးေလကငးရနး ပစာၦမာ

*** Given that f ( x ) = 2q( x2n + 1) – ( 4q – 3 )x2 – 3 , where n and p are positive integers. Show that x + 1 is a factor of f ( x ) , for all values of q and if q = 2 find the value of n for which 2x – 1 is a factor of f ( x ) and factorize f ( x ) completely. Ans; n = 2 , f ( x ) = ( 2x – 1 )( 2x + 1 )( x – 1 )( x + 1 )

*** Given that f ( x ) = x2n + 4kx2n–1 – x2 – ( 4k + 12 ) x – 12 ,where n and q are positive integers. Show that x + 1 is a factor of f ( x ) , for all values of k and when k = 2 find the value of n for which x – 2 is a factor of f ( x ) and show also that ( x + 2 ) is a factor of f ( x ). Ans; n = 2

*** Given that f ( x ) = p( x – 1 )2n–1 – 6x2 + 16x – p – 8 ,where n and p are positive integers. Show that x – 2 is a factor of f ( x ) , for all values of p and when p = 2 find the value of n for which x – 3 is a factor of f ( x ) and factorize f ( x ) completely. Ans; n = 2 , f ( x ) = ( x – 1 )( x – 2 )( x – 3 )

------------------------------------------------------------------------

P a g e - 17


Type ( 4 )

22. Show that the expression x3 + ( k + 1 )x2 + ( k – 6 )x – 6 has a factor ( x + 1 ) for all values of k. If the expression also has a factor of x + 3 , find k and the third factor. Ans; k = 1 , the third factor = x – 2 2013 ရနးကနး ( No 1 ) ပစတပစာၦမာ

Show that the expression 3x3 + ( k – 6 )x2 + ( k – 5 )x + 4 has a factor x + 1 for all values of k. If the expression also has a factor of x – 2 , find k and the third factor. Ans; k = 1 , the third factor = 3x – 2 2010 မႏ ေလ ( No 34 )

Show that the expression f ( x ) = x3 + ( k – 2 )x2 + ( k – 7 )x – 4 has a factor x + 1, for all values of k. If f ( x ) also has a factor of x – 2 , find the value of k and hence solve f ( x ) = 0. 2008 ဧရာတ ( No 59 ) Ans; k = 3 , x = – 2 or x = – 1 or x = 2

Show that the expression f ( x ) = x3 + ( k – 2 )x2 + ( k – 7 )x – 4 has a factor x + 1 for all values of k. If f ( x ) also has a factor of x + 2 , find the value of k and the third factor. 2002 မႏ ေလ ( No 112 ) Ans; k = 3 , the third factor = x – 2

------------------------------------------------------------------------

23. If x + 2 is a factor of the expression f ( x ) = ax3 + x2 – 19x + 6 . Find the value of a and then solve

the equation f ( x ) = 0. Ans; a = 6 , x = –2 or x = 1

3 or x =

3

2

2012 ရခငး ( No 19 )( တျကးရနးလျယး ) ပစတပစာၦမာ 2008 ရခငး ( No 60 )

Let f ( x ) = x4 + px2 – 4x + 12 where p is a constant. If x + 2 is a factor of f ( x ), find the value of p. With this value of p, prove that f ( x ) has a factor x – 3.Hence factorize f ( x ) completely.

Ans; p = – 9 , f ( x ) = ( x – 3 )( x – 1 )( x + 2 )2

------------------------------------------------------------------------


------------------------------------------------------------------------

P a g e - 18


( c ) တညးကနးေရာ စာကနးပါ မသကနးပစာၦမာ Type ( 1 )

24. Given that 4x4 – 9a2x2 + 2( a2 – 7 )x – 18 is exactly divisible by 2x – 3a , show that a3 – 7a – 6 = 0 and hence find the possible values of a. Ans; a = – 1 or a = – 2 or a = 3 2017 ႏငးငခာ ( No 6 )/ 2007 မျနး၊ကရငး၊တနသၤာရ ( No 70 )/ 2003 မႏေလ ပစတပစာၦမာ 2014 စစးကငး၊ခငး ( No 4 ) / 2003 မျနး၊ကရငး၊တနသၤာရ ( No 107 )( a အစာ b န႔တျကး )

Given that 16x4 – 4x3 – 4a2x2 + 7ax + 18 is divisible by 2x + a , show that a3 – 7a2 + 36 = 0 and find the possible values of a. Ans; a = – 2 or a = 3 or a = 6 စပးဆကးးေလကငးရနး ပစာၦမာ

***

Given that 27 x4 – 3 a2x2 – 3 ( 2a3 – 3a2 + 9a ) x + 24 is exactly divisible by 3x – 2a , show that

a3 – 3a2 + 4 = 0 and find the possible values of a. Ans; a = – 1 or a = 2

***

The expression ( 4𝑏

𝑎2 ) x4 + (

𝑎

2 ) x2 – ( 3a + 12ab )x – 40a2b + 8 is exactly divisible by x – 2a , find

the possible value of a. Ans; a = – 1 or a = 2 Type ( 2 ) *****

25. Find the value of k for which a – 3b is a factor of a4 – 7a2b2 + kb4. Hence for this value of k, factorize

a4 – 7a2b2 + kb4 completely. Ans; k = – 18 , f ( a ) = ( a2 + 2b2 )( a + 3b )( a – 3b ) 2012 စစးကငး၊ခငး ( No 15 ) စပးဆကးးေလကငးရနး ဖညးစျကးပစာၦမာ

Find the value of k for which b + a is a factor of b4 + 3ab3 + a2b2 + ka3b – 2a4. Hence for this value of k,

factorize b4 + 3ab3 + a2b2 + ka3b – 2a4 completely.

Ans; k = – 3 , f ( b ) = ( b – a )( b + a )2( b + 2a )

Find the value of k for which b + a is a factor of kb4 – 5ka2b2 + 8a4. Hence for this value of k, factorize

kb4 – 5ka2b2 + 8a4. completely. Ans; k = 2 , f ( b ) = ( b – a )( b + a )( b – 2a )( b + 2a )

The expression 6xy3 – 4x2y2 – p2yx4 has a factor of y – 2x , and find the values of p.With this value of p2 , when y – 2 is a actor of the expression , find the possible values of x.

Ans p = ± 4 , x = 0 or x = – 3

2 or x = 1

P a g e - 19


If the polynomial 4y4 – ( 4k – 1 )x2y2 – kx4 has a factor of y + 2x , obtain k and factorize the polynomial

completely. Ans; k = 4 , f ( y ) = ( y – x )( y + x )( 4y2 + x2 )

The expression 2b4 – ( k – 2 ) a2b2 + ka4 has a = 𝑏

2 as a root. Find the value of k and factorize the

expression completely. Ans; k = 2 , f ( b ) = ( b – a )( b + a )( b – 2a )( b + 2a )

Find the remainder when 2y2x3 + yx4– 2x5 is divided by ( y – x ). ( for 3 marks )

------------------------------------------------------------------------

Type ( 3 )

26. If x – k is a factor of the expression kx3 + 5x2 – 7kx – 8 , where k is a positive integer, find the numerical value of k. Hence find the other factors of the expression. 2013 မေကျ ( No 3 ) Ans; k = 2 , the other factors are x+4 and 2x+1 ပစတပစာၦမာ 2004 ရခငး / 2006 မျနး၊ကရငး၊တနသၤာရ ( No 80 )

If x – k is a factor of the expression kx3 – 3x2 – 4kx + 12 , where k is a positive integer, find the value of k. Hence factorize the expression completely. Ans; k = 2 , f ( x ) = ( x – 2 )( x + 2 )( 2x – 3 )

------------------------------------------------------------------------

( 4 ) တညးကနးတစးခ စာကနးႏြစးခ ပစာၦမာ ( a ) တညးကနးမသကနးပစာၦမာ Type ( 1 )

27. If f ( x ) = 2x4 + x3 – ax2 + bx + a + b – 1 has factors x – 2 and x + 3 , find the constants a and b. Hence factorize f ( x ) completely. Ans; a = 16 , b = 3 , f ( x ) = ( x – 2 )( 2x – 3 )( x + 1 )( x + 3 ) 2013 ပခ ( No 5 ) ( or )

Given that x2 + x – 6 is a factors of 2x4 + x3 – ax2 + bx + a + b – 1. Find the values of a and b. 2003 ပခ ( No 103 ) (ဤပစာၦသညး အထကးပါပစာၦႏြငး ေမပသာကျာပါသညး။ ) Ans; a = 16 , b = 3

------------------------------------------------------------------------------------------------ Type ( 2 )

28. The equation 6x3 + px2 = qx – 10 has roots x = 2 and x = 1

2 .Find the values of the constants p and q

and the third root. Ans; p = – 5 , q = 19 , The third root = – 5

3

2009 မေကျ ( No 45 )

P a g e - 20


------------------------------------------------------------------------------------------------ ပစတပစာၦမာ 2013 ရြမး၊ကယာ ( No 10 )

Given that the equation 2x3 + ax2 + bx – 12 = 0 has roots x = 1 and x = 4 . Find the values of a , b and

the third root. Ans; a = – 13 , b = 23 , x = 3

2

------------------------------------------------------------------------------------------------------------------------ 29. If x4 + px3 + qx2 – 16x – 12 has factors x – 2 and x + 1 , find the constants p and q and the remaining factors. Ans; p = 4 , q = – 1 , The remaining factors are x + 2 and x + 3 2006 မႏေလ - 2004 မႏ ေလ ( No 75 ) Type ( 3 )

30. Given that the expression ( b – a )x2 – x – ab is divisible by x + 2 and x – 3, find the values of a and b. Ans; a = – 3 and b = – 2 or a = 2 and b = 3 2006 ရနးကနး ( No 74 ) ပစတပစာၦမာ

Given that the expression ( a + 4b )x2 + 2x + b – a is divisible by x – 2 and x + 3, find the value of a and b. Ans; a = 10 , b = – 2 2003 ဧရာတ ( No 104 ) / 2002 ပခ ( No 114 )

31. If x2 – 5x + 6 is a factor of the polynomial f ( x ) = x3 + ax2 + ( a2 – 25 )x – 6 , find a . 2009 မႏ ေလ ( No 44 ) Ans; a = – 6 ပစတပစာၦမာ 2009 ရြမး၊ကယာ ( No 52 )

If x2 – 5x + 6 is a factor of f ( x ) = x4 – 5x3 + ( a – 1 )x2 + bx – 24 , find a and b, hence factorize f ( x )

completely. Ans; a = 3 , b = 20 , f ( x ) = ( x – 3 )( x – 2 )2( x + 2 ) 2004 ႏငးငခာ ( No 98 )

Given that 2x2 – x – 1 is a factor of ax4 + x3 – bx2 + 5x + 6 , find the values of a and b. Ans; a = 2 , q = 14 2005 ရနးကနး၊မျနး၊ကရငး၊တနသၤာရ ( No 85 )

If x2 + 2x – 3 is a factor of f ( x ) = x4 + 2x3 – 7x2 + ax + b , find a and b, hence factorize f ( x ) completely. Ans; a = – 8 , b = 12 , f ( x ) = ( x – 2 )( x – 1 ) ( x + 2 ) ( x + 3 ) 2009 ကခငး ( No 51 )

Given that x2 – x – 6 is a factor of 2x4 – 7x3 – 4x2 + px + q , find the values of p and q. Find the other factors. Ans; p = 27 , q = – 18 , The other factors are x – 1 and 2x – 3 2013 ႏငးငခာ ( No 11 )

P a g e - 21


Given that 2x2 – x – 1 is a factor of ax4 + x3 – bx2 + 5x + 6 , find the value of a and b. Ans; a = 2 , b = 14 2008 ရြမး၊ကယာ ( No 62 )

The expression ax3 + bx2 – 3x – 2 is exactly divisible by x2 + x – 2. Find the value of a and b. Ans; a = 2 , b = 3 2008 ရခငး ( No 71 )

Given that the expression 2x3 + px2 – 8x + q is exactly divisible by 2x2 – 7x + 6. Evaluate p and q and factorize the expression completely. Ans; p = – 3 , q = 12 , f ( x ) = ( x – 2 )( 2x – 3 )( x + 2 ) 2011 ႏငးငခာ ( No 32 )

Find the values of p and q for which the expression 12x4 + 16x3 + px2 + qx – 1 is divisible by 4x2 – 1. Hence, find the other factors of the expression. Ans; p = 1 , q = – 4 , The other factors are x+1 and 3x+1 2009 ရနးကနး ( No 43 )

Find the values of p and q if 6x3 + 13x2 + px + q is exactly divisible by 2x2 + 7x – 4. Show that 3x – 4 is a factors of the given polynomial. Ans; p = – 40 , q = 16 2010 ဧရာတ၊မျနး၊ကရငး၊တနသၤာရ ( No 38 )

The polynomial x3 – 6x2 + px + q is divisible by x2 – 5x – 6. Find the value of p and the value of q. Ans; p = 11 , q = – 6 2012 ရနးကနး ( No 12 )

The polynomial px3 + qx2 – 5x – 6 is exactly divisible by 2x2 + x – 6. Calculate the value of p and q , and factorize the polynomial completely. Ans; p = 2 , q = 3 , f ( x ) = ( x + 2 )( 2x – 3 )( x + 1 )

------------------------------------------------------------------------


------------------------------------------------------------------------

( c ) တညးကနးေရာ စာကနးပါ မသကနးပစာၦမာ

32. Given f ( x ) = 2x3 + ax2 – 7a2x – 6a3 , determine whether or not x – a and x + a are factors of f ( x ). Hence , find the roots of f ( x ) = 0 in term of a.

Ans; f ( a ) = 3a3 – 13a3 ≠ 0 , x – a is not a factor of f ( x ).

f (– a ) = 8a3 – 8a3 = 0 , x + a is a factor of f ( x ).

x = – a , x = 3𝑎

2 and x = 2a , the three roots are – a ,

3𝑎

2 and 2a.

2013 ဧရာတ ( No 6 ) / 2005 ပခ ( No 89 )

P a g e - 22


စပးဆကးးေလကငးရနး ဖညးစျကးပစာၦမာ

Given f ( x ) = x3 + ax2 – 4a2x – 4a3 , determine whether or not x – a and x + 2a are factors of f ( x ). Hence , find the roots of f ( x ) = 0 in term of a.

Ans; f ( a ) = 2a3 – 8a3 ≠ 0 , x – a is not a factor of f ( x ).

f (– a ) = 12a3 – 12a3 = 0 , x + 2a is a factor of f ( x ). x = – 2a , x = – a and x = 2a , the three roots are – 2a , – a and 2a ------------------------------------------------------------------------------------------------------------------------ ( 5 ) တညးကနးႏြစးခ စာကနးတစးခ ပစာၦမာ ( a ) တညးကနးမသကနးပစာၦမာ

**** 33. Let F ( x ) = f ( x ) + g ( x ). If ( x – 2 ) is a common factor of both f ( x ) and g ( x ) then show that ( x – 2 ) is also a factor of F (x ). 2006 ကခငး ( No 82 ) ------------------------------------------------------------------------------------------------------------------------

**** 34. f ( x ) = 2x3 + px2 + qx – 20 , where p and q are constants . Given that x + 2 is a factor of f ( x ) and

that x + 2 is also a factor of f ’ ( x ) , find the values of p and q. Ans; p = 3 , q = – 12 2012 မေကျ ( No 14 ) / 2002 စစးကငး၊ခငး ( No 113 ) စပးဆကးးေလကငးရနး ဖညးစျကးပစာၦမာ

f ( x ) = px3 + qx2 + 24x – 10p , where p and q are constants . Given that x + 2 is a common factor of

f ( x ) and f ’ ( x ) , find the values of p and q. Ans; p = 3 , q = – 12 ------------------------------------------------------------------------------------------------------------------------ 35. Given that x + 5 is a common factor of x3 + px2 – qx + 15 and x3 – x2 – ( q + 5 )x + 40. Find thre

values of p and q . Hence factorize x3 + px2 – qx + 15 completely. 2014 မေကျ ( No 3 ) Ans; p = 1 , q = 17 , f ( x ) = ( x + 5 )( x – 1 )( x – 3 ) ပစတပစာၦမာ 2011 မေကျ ( No 25 )

Given that x + 2 is a common factor of ax2 + ( a + k )x + 6 and ( k – a )x2 + 4x – a. Then find the values of a and k. Ans; a = 4 , k = 7 2003 ရနးကနး ( No 99 )

The expression px4 – 5x + q and x4 – 2x3– px2 – qx – 8 = 0 have a common factor x – 2. Tind the value of p and q. Ans; p = 1 , q = – 6

P a g e - 23


2014 ရခငး ( No 8 ) / 2013 ကခငး ( No 9 ) / 2011 ရြမး၊ကယာ ( No 31 )

Given that ax3 + 4x2 – 5x – 10 = 0 and ax3– 9x – 2 = 0 have a common root ( or factor ). What are the possible values of a ? Ans; a = 11 or 2 2005 ႏငးငခာ ( No 92 ) အထကးပါပစာၦႏြငး ေပခကးေမခကးအတတသာဖစးသညး။

If the equation ax3 + 4x2 – 5x – 10 = 0 and ax3– 9x – 2 = 0 have a common root ( or factor ).Then show that a = 2 or a = 11. 2004 ရြမး၊ကယာ ( No 97 )

If the expression ax4 + x3 – x2 + 3x + 2 = 0 and ax4– x2 + 3x+ 1 = 0 have a common root ( or factor ), then find the value of a. Ans; a = 3 2005 မႏ ေလ ( No 86 )

Given that kx3 + 2x2 + 2x + 3 and kx3– 2x + 9 have a common root ( or factor ),what are the possible

values of k ? Ans; k = – 7 or 5

9

------------------------------------------------------------------------

( b ) စာကနးမသကနးပစာၦမာ

36. Given that x – is a common factor of x2 – 5x + k and x2 – 6x + 3k where k ≠ 0, Find the numerical

values of . Ans; = 9

2

2005 မေကျ ( No 87 )

------------------------------------------------------------------------

( c ) တညးကနးေရာ စာကနးပါ မသကနးပစာၦမာ ( ဤပစမ 5 marks Question တျငး မေမဘေသပါ။

------------------------------------------------------------------------

( 6 ) တညးကနးႏြစးခ စာကနးႏြစးခ ပစာၦမာ ( a ) တညးကနးမသကနးပစာၦမာ ( ဤပစမ 5 marks Question တျငး မေမဘေသပါ။ )

------------------------------------------------------------------------


------------------------------------------------------------------------

( c ) တညးကနးေရာ စာကနးပါ မသကနးပစာၦမာ ( ဤပစမ 5 marks Question တျငး မေမဘေသပါ။ )

P a g e - 24


( 3 ) Remainder / Factor ေရာရာ

ပစာၦမာ

P a g e - 25


( 3 ) Remainder / Factor ေရာရာ ပစာၦမာ ( 1 ) တညးကနးတစးခ စာကနးတစးခ ပစာၦမာ ( i ) တညးကနး မသကနး ပစာၦမာ

37. The cubic polynomial f ( x ) is such that the coefficient of x3 is – 1 and the roots of the equation f ( x ) = 0are 1,2 and k . Given that f ( x ) has a remainder of 8 when divided by x – 3 , find the value of k and the remainder when f ( x ) is divided by x + 3. Ans; k = 7 , The remainder = 200 2016 ပညးတျငး ( No 3 ) စပးဆကးေလကငးရနး ပစာၦမာ ( 1 ) The three roots of the equation f ( x ) = 0 are 1,3 and p. If f ( x ) is a polynomial of degree 3 and the

coefficient of x3 is – 2. Given that f ( x ) has a remainder of 210 when divided by x + 2 , find the value of p and the remainder when f ( x ) is divided by x – 2 . Ans; p = 5 , The remainder = – 6 ( 2 ) The three roots of the equation f ( x ) = 0 are 1, a +1 and – 1. If f ( x ) is a polynomial of degree 3

and the coefficient of x3 is – 1. Given that f ( x ) has a remainder of – 12 when divided by x – 2 , find the value of a and show that x – 1 is a factor of f ( x ). Ans; a = 3

( 3 ) The cubic polynomial f ( x ) is such that the coefficient of x3 is – 3 and the roots of the equation f ( x ) = 0 are 1, –1 and a . Given that f ( x ) has a remainder of ( x – 20 ) when divided by x – 2 , find the value of a and the remainder when f ( x ) is divided by x + 2. Ans; a = 0 , The remainder = – 18

( 4 ) The cubic polynomial f ( x ) is such that the coefficient of x3 is – 4 and the roots of the equation f ( x ) = 0 are 1, 2 and k – 1 . Given that f ( x ) has a remainder of 12x when divided by x – 4 , find the value of k and the remainder when f ( x ) is divided by x – 3. Ans; k = 7 , The remainder = 24

( 5 ) The cubic polynomial f ( x ) is such that the coefficient of x3 is 2 and the roots of the equation f ( x ) = 0 are 1, – 1 and k – 4 . Given that f ( x ) has a remainder of 6k – 12 when divided by x – 2 , find the value of k and the remainder when f ( x ) is divided by x – 3. Ans; k = 4 , The remainder = 24

( 6 ) The cubic polynomial f ( x ) is such that the coefficient of x3 is 3 and the roots of the equation f ( x ) = 0 are – 1, 0 and k – 1 . Given that f ( x ) has a remainder of 4k + 10 when divided by x + 2 , find the value of k and the remainder when f ( x ) is divided by x + 2. Ans; k = 2 , The remainder = – 6

( 7 ) The cubic polynomial f ( x ) is such that the coefficient of x3 is – 2 and the roots of the equation f ( x ) = 0 are – 1, 0 and k + 2 . Given that f ( x ) has a remainder of 4k + 16 when divided by x – 2 ,

find the value of k and express f ( x ) in the form of ax3 + bx , stating the values of a and b. Ans; k = – 1 , a = – 2 and b = 2

P a g e - 26


------------------------------------------------------------------------

( ii ) စာကနးမသကနးပစာၦမာ ( ဤပစမ 5 marks Question တျငး မေမဘေသပါ။ )

------------------------------------------------------------------------

( iii ) တညးကနးေရာ စာကနးပါ မသကနးပစာၦမာ ( ဤပစမ 5 marks Question တျငး မေမဘေသပါ။ )

------------------------------------------------------------------------

( 2 )တညးကနးတစးခ စာကနးႏြစးခ ပစာၦမာ ( i ) တညးကနး မသကနး ပစာၦမာ

38. The expression ax2 + bx – 1 leaves remainder of R when divided by x + 2 and a remainder of 3R + 5 when divided by x – 3. Show that a = 3b – 1 . Given also that the expression is exactly divisible by 2x – 1, evaluate a and b. Ans; a = 2 , b = 1 2013 မႏ ေလ ( No 2 ) / 2007 ႏငးငခာ ( No 73 ) ( or ) 2003 စစးကငး၊ခငး ( No 102 )

The expression ax2 + bx – 1 leaves remainder of R when divided by x + 2 and a remainder of 3R + 5 when divided by x – 3. Show that a – 3b + 1 = 0 ဤပစာၦအာ R မာေနရာတျငး ဆကးသျယးခကးပစဖငးေရႏငးပါေသသညး။ ဥပမာ -

The remainder when the expression ax2 + bx – 1 is divided by x + 2 is 5 more than three time of the remainder when it is diided by x – 3. Show that a = 3b – 1 . Given also that the expression is exactly divisible by 2x – 1, evaluate a and b. Ans; a = 2 , b = 1 ( အထကးပါ ပစာၦႏြစးပဒး အေရအသာကျာေသားလညး သေဘာတရာ အတတသာဖစးသညး . ) ပစတပစာၦမာ 2003 ရခငး ( No 108 )

The expression x2 + ax + b leaves remainder of p when divided by x – 1 and leaves a remainder of p + 6 when divided by x – 2. Find the value of a. Ans; a = 3 ------------------------------------------------------------------------------------------------------------------------

39. x + 4 is a factor of f ( x ) = a ( x + 1 )2 + b ( x + 1 ) + 9. The remainder when f ( x ) is divided by x + 3 is – 11. Find the value of a and b. Find also the solution set of the equation f ( x ) = 0.

Ans; a = 7 , b = 24 , The solution set = { – 4, – 10

7 }

2014 မျနး၊ကရငး၊တနသၤာရ ( No 7 ) 2008 ႏငးငခာ ( No 63 )

P a g e - 27


ပစတပစာၦမာ

f ( x ) = 8x3 – 12x2 + ax + 3 has a factor of 2x – 3. Find a and the solution set of f ( x ) = 0.

2007 ပခ၊ရြမး၊ကယာ ( No 68 ) Ans; a = – 2 , The solution set = { 3

2 ,

1

2 , –

1

2 }

40. If x + 2 is a factor of x3 – ax – 6 , then find the remainder when 2x3 + ax2 – 6x + 9 is dividedby ( x + 1 ). 2017 ပညးတျငး ( No 5 ) Ans; a = 7 , The remainder = 20 ------------------------------------------------------------------------------------------------------------------------ 41. The polynomial 2ax3 + 2bx2 – 5x + 4a is exactly divisible by x2 – 3x – 4. Calculate the values of a

and b. What is the remainder when it is divided by x – 5 ? Ans; a = 1 , b = – 7

2 , The remainder = 54

2008 ရနးကနး ( No 54 ) ------------------------------------------------------------------------------------------------------------------------ 42. The expression x3 + 3ax2 + 5x + 4b has a factor x – 4 but leaves a remainder of 12 when divided by x – 5. Find the values of a and b. Factorize the expression compltely. Ans; a = – 2 , b = 3 , f ( x ) = ( x – 4 )( x – 3 )( x + 1 ) 2009 စစးကငး၊ခငး ( No 46 ) ပစတပစာၦမာ 2009 ဧရာတ ( No 48 )

Given that f ( x ) = 2x3 + px2 + qx – 9 has a factor x – 3 but leaves a remainder of 8 when it is divided by x+1.Find the values of p and q. Show that 2x + 1 is a factor of f ( x ). Ans; p = 1 , q = – 18 2003 မျနး၊ကရငး၊တနသၤာရ ( No 106 ) ( အထကးပါပစာၦႏြငး p,q အစာ a,b ဖစးသညးပကျာသညး၊အာလတသညး။ )

Given that f ( x ) = 2x3 + ax2 + bx – 9 , find the values of a and b if f ( x ) has a factor x – 3 but leaves a remainder of 8 when divided by x+1. Ans; a = 1 , b = – 18 2009 ရခငး ( No 50 )

The expression x3 + 2x2 + px + q leaves a remainder of –18 when divided by x – 2 but has a factor x – 3. Find the values of p and q and factorize the expression compltely. Ans; p = – 11 , b = – 12 , f ( x ) = ( x – 3 ) ( x + 1 ) ( x + 4 ) ------------------------------------------------------------------------------------------------------------------------ 2018 ပညးတျငး ( No 7 ) 2012 မႏ ေလ ( No 13 ) / 2006 မေကျ ( No 76 )

The expression px3 – 5x2 + qx + 10 has factor 2x – 1 but leaves a remainder of – 20 when divided by x + 2. Fnd the values of p and q and factorize the expression completely. Ans; p = 6 , q = – 19 , f ( x ) = ( 2x – 1 )( 3x + 5 )( x – 2 ) 2006 ရြမး၊ကယာ ( No 83 ) အထကးပါ ပစာၦႏြငး ေပခကးတ ေမခကးသာကျာပါသညး။

P a g e - 28


The expression px3 – 5x2 + qx + 10 has a factor 2x – 1 but leaves a remainder of – 20 when it is divided by x + 2. Fnd the remainder when it is divided by x + 3. Ans; p = 6 , q = – 19 , The remainder = – 140 2012 ပခ ( No 16 )

The expression ax3 + bx2 – 2x + 3 has a factor of x + 3 but leave a remainder of 91 when divided by x – 4. Find the value of and b. What is the remainder when it is divided by x + 2. Ans; a = 1 , b = 2 , the remainder = 7 2012 ရြမးကယာ ( No 21 )

The expression 2x4 + ax3 + bx2 – 3x – 4 is exactly divisible by x – 4 but leave a remainder of – 9 when divided by x – 1. What is the remainder when it is divided by x – 2. Ans; a = – 9 , b = 5 , the remainder = – 30 2011 ရနးကနး ( No 23 )

The expression ax3 + bx2 – 13x – 6 has a factor of x – 2 but leave a remainder of – 4 when divided by x – 1. Find the value of and b. What is the remainder when it is divided by x + 2. Ans; a = – 7 , b = – 22 , the remainder = 164 2011 မႏ ေလ ( No 24 )

Let f ( x ) = x3 + ax2 + bx – 22. When f ( x ) is divided by x + 3 , the remainder is – 28. If x + 2 is a factor of f ( x ) , find the value of and b. Ans; a = – 8 , b = – 31 2014 မႏ ေလ ( No 2 ) 2009 ႏငးငခာ ( No 53 )

The expression 6x3 + ax2 + bx + 10 has a factor of 2x – 1 but leave a remainder of – 20 when divided by x + 2. Find the value of and b and factorize the expression completely. Ans; a = – 5 , b = – 19 , f ( x ) = ( 2x – 1 )( x – 2 )( 3x + 5 ) 2008 စစးကငး၊ခငး ( No 57 )

The expression x3 + 2ax2 + bx + 3 has a factor of x + 3 but leave a remainder of 91 when divided by x – 4. What is the remainder when it is divided by x – 2 ? Ans; a = 1 , b = – 2 , The remainder = 15 2004 မေကျ ၊စစးကငး၊ခငး ( No 94 )

The expression x3 + ax2 + bx + 3 has a factor of x + 3, but leave a remainder of 91 when divided by x – 4. What is the remainder when it is divided by 2 – x. Ans; a = 2 , b = – 2 , The remainder = 15 2015 ပညးတျငး ( No 1 ) / 2008 ဧရာတ ( No 69 )

The expression x3 + ax2 + bx + 3 is exactly divisible by x + 3 but leave a remainder of 91 when divided by x – 4. What is the remainder when it is divided by x + 2. Ans; a = 2 , b = – 2 , The remainder = 7 2005 စစးကငး၊ခငး၊ကခငး ( No 88 )

When ( x ) = x3 – 3x2 + ax + b is divided by x + 1 , the remainder is 12. If x – 2 is a factor of f ( x ) , then find a and b. Hence solve f ( x ) = 0. Ans; a = – 4 , b = 12 , x = – 2 or x = 2 or x = 3

P a g e - 29


2003 ႏငးငခာ ( No 109 )

f ( x ) = kx3 + ( 3k – 2 )x2– 4 , where k is a constant. Given that ( x + 2 ) is a factor of f ( x ). Find the value of k. With this value of k, find the remainder when f ( x ) is divided by ( 2x – 1 ).

Ans; k = 3 , The remainder = – 15

8

------------------------------------------------------------------------

မသကနးသလပစာၦမာ

43. The expression 2x3 + bx2 – cx + d leaves the same remainders when divided by x – 1 or x + 2 or 2x – 1. Evaluate b and c. Given also that the expression is exactly divisible by x – 2 , evaluate d. Ans; b = 1 , c = 5 , d = – 10 2012 မျနး၊ကရငး၊တနသၤာရ ( No 18 ) ပစတပစာၦမာ

The expression 2x3 + bx2 – cx + d leaves the same remainders when divided by x + 1 or x –2 or 2x – 1. Evaluate b and c. Given also that the expression is exactly divisible by x + 2 , evaluate d. Ans; b = – 3 , c = 3 , d = 22 2006 ပခ ( No 78 )

44. Given that the expression x3 + ax2 + bx + c leaves the same remainder when divided by x – 1 or x + 2, prove that a = b + 3 . Given also that the remainder is 3 when the expression is divided by x + 1. Calculate the value of c. Ans; c = 1 2007 မႏ ေလ ( No 65 ) ( *** ဤပစာၦသညး Remainder/ Factor ေရာရာပစာၦမဟတးေသားလညး ပစသေဘာတရာတသဖငး ဤ အပးစထတတျငး ထညးသျငးထာရခငး ဖစးသညး။ ဤပစာၦသညး တျကးရမခကးေသားလညး ေကာငးသာအခ႕အတျကး မကးေစ ေမြာကးေစႏငး သဖငး သတထာရမညးဖစးပါသညး။ ) ပစတပစာၦမာ 2010 စစးကငး၊ခငး ( No 36 )

Given that the expression 2x3 + bx2 – cx + d leaves the same remainder when divided by x + 1 or x – 2 or 2x – 1, find the values b and c. Given also that the expression is exactly divisible by x + 2 , evaluate d. Ans; b = – 3 , c = 3 , d = 22 2010 ႏငးငခာ ( No 42 )

The expression x3 + ax2 + bx + c is divisible by both x and x – 3 and leaves a remainder of – 40 when divided by x + 2 . Find the value of a, of b and of c , hence factorize the expression completely. Ans; a = – 5 , b = 6 , c = 0 , f ( x ) = x ( x – 2 ) ( x – 3 ) 2016 ႏငးငခာ ( No 4 )

P a g e - 30


The expression ax3 – (a + 3b)x2 + 2bx + c is divisible by x2 – 2x. When the expression is divided by x – 1, the remainder is 8 more than when it is divided by x + 1. Find the value of a,b and c , hence factorize the expression completely. Ans; a = 2 , b = 1 , c = 0 , f ( x ) = x ( x – 2 ) ( 2x – 1 ) ------------------------------------------------------------------------------------------------------------------------ ( ii ) စာကနးမသကနးပစာၦမာ ( ဤပစမ 5 marks Question တျငး မေမဘေသပါ။ )

------------------------------------------------------------------------

( iii ) တညးကနးေရာ စာကနးပါ မသကနးပစာၦမာ

45. Given that x3 + 4x2 – ax + b is exactly divisible by x + 2 , but leave a remainder a3 when divided by

x – a, calculate the values of a and b. Ans; a = – 4

3 , b = –

16

3 or a = 2 , b = – 12

2002 ရြမး၊ကယာ ( No 116 )

------------------------------------------------------------------------

( 4 )တညးကနးႏြစးခ စာကနးႏြစးခ ပစာၦမာ ( i ) တညးကနးမသကနးပစာၦမာ ( ဤပစမ 5 marks Question တျငး မေမဘေသပါ။ )

------------------------------------------------------------------------

( ii ) စာကနးမသကနးပစာၦမာ ( ဤပစမ 5 marks Question တျငး မေမဘေသပါ။ )

------------------------------------------------------------------------

( iii ) တညးကနးေရာ စာကနးပါ မသကနးပစာၦမာ ( ဤပစမ 5 marks Question တျငး မေမဘေသပါ။ )

------------------------------------------------------------------------

P a g e - 31


Remainder Theorem & Factor Theorem Concept ပငးဆငးရာ ရြငးလငးခကး

ကနးဂဏနးမာစာခငးက သခၤာပစဖငး ေဖားပခငး ဥပမာ။ ။ 20 က 3 ဖငးစာလ ြငး အၾကျငးမညးမ ြရမညးနညး။ အေဇမြာ ရြငးရြငးေလပါ။ သေခာကးလ ဆယးရြစး ။ ႏြစးဆယးထက ဆယးရြစးႏႈတး ႏြစးၾကျငး ။ ဤသညးက သခၤာပစဖငးေရလ ြငး အရြညးစာနညး ( Lond division ) နညးဖငး ေဖားပမညးဆလ ြငး

6 ( စာကနး / Quotient ) 3 ( စာကနး / Divisor )

20 ( တညးကနး / Dividend )

18 ( = 3 x 6 ) ( စာကနး x စာလဒး ) 2 ( အၾကျငး / Remainder ) 18 ႏြငး 2 အာ ပနးေပါငးလ ြငး 20 ပနးရမညးဖစးသဖငး

20 = 18 + 2

20 = ( 3 x 6 ) + 2

တညးကနး = ( စာကနး x စာလဒး ) + အၾကျငး

တညးကနး = ( စာကနး x စာလဒး ) + အၾကျငး

Dividend = ( Divisor x Quotient ) + Remainder

ဆသညး အႏြစးခပး ပစေလ ရလာမညးဖစးပါသညး။ ဆကးပ အထကးပါ ကနးတနးဖမာေနရာမြာ Polynomial ကနးတနးမာ အစာထေလလာၾကညးရေအာငး။

ဥပမာ။ ။ f ( x ) = x2 – 7x + 10 က ( x – 3 ) ဖငး စာလ ြငး အၾကျငး မညးမ ြရမညးနညး။ ( အရြညးစာနညးဖငး စာခငးမြ Remainder Theorem ၏ ပစထတးယမညးဖစးသဖငး long division method ဖငး အရငးစာၾကညးၾကပါစ႔ )

20 – ( 3 x 6 ) = 2

P a g e - 32


x – 4 x – 3 x2 – 7x + 10

x2 – 3x – 4x + 10 – 4x + 12 – 2

( ဤေနရာတျငး တစးခသထာရမညးအခကးမြာ အမငးဆ x ထပးညႊနး 2 ထပးဆလ ြငး long division စာခငးတျငး ႏြစးဆငး ဆငးေပရပါမညး ။ 10 ထပးဆလ ြငး ဆယးထပးစာပဆငးရမညးဖစးသညး။ Long Division စာနညး၏ ဤအာနညးခကး ေၾကာငးပငး Remainder Theorem က Develop လပးယခၾကခငး ဖစးမညးထငးပါသညး။ ) အထကးပါဥပမာတျငး

x2 – 7x + 10 = တညးကနး ( Dividend ) = f ( x ) x – 3 = စာကနး ( Divisor ) x – 4 = စာလဒး ( Quotient ) = Q ( x ) – 2 = အၾကျငး ( Remainder ) ( တနးဖေလ သခလမြာ ကးနးရြငး x အေပၚတျငး မြေနေသာ function ေလမာခညး ဖစးေနသညးက ေတျ႕ရမညး။ ထ႔ေၾကာငး တညးကနးေလအာ f ( x ) ဟေရသလ စာလဒးေလအာလညး Quotient of x / Q ( x ) ဟ ပငးေရပါမညး။ ) Dividend = ( Divisor x Quotient ) + Remainder ဆေသာ ပစေလတျငး အစာသျငးလ ြငး f ( x ) = ( x – 3 ) Q( x ) + ( – 2 ) အကယး၍ x တနးဖ 3 ( x = 3 ) ဖစးမညးဆလ ြငး အထကးပါ ညမ ြခငးေလသညး

f ( 3 ) = ( 3 – 3 ) Q( 3 ) + ( – 2 )

f ( 3 ) = 0 x Q( 3 ) – 2 = 0 – 2

f ( 3 ) = 0 – 2

f ( 3 ) = – 2

ဆသညး ပစေလ ရလာပါမညး။ ဤပစေလမြတစးဆငး အကယး၍ f ( x ) အာ x – 3 ဖငးစာသညးအခါ x – 3 က 0 ဖငးညေပခငးဖငး ဥပမာ x – 3 = 0 x = 3 ရလာသညး တနးဖေလ 3 အာ f ( x ) တနးဖေလတျငး f ( 3 ) ဟ အစာသျငးတျကးခကးမညးဆလ ြငး လအပးေသာ အၾကျငးတနးဖအာ ရႏငးသညး ဆသညး ပစေလ ရလာပါမညး။

Check Point When f ( x ) = x2 – 7x + 10 is divided by x – 3 , the remainder is f ( 3 )

f ( 3 ) = ( 3 )2 – 7( 3 ) + 10 = 9 – 21 + 10 = 19 – 21 = – 2 အကယး၍ f ( x ) အာ x – 3 ဖငး မဟတးပ x – k ဖငး စာမညးဆလ ြငး ၄ငး၏အၾကျငးမြာ f ( k ) ဖစးလာပါမညး။

P a g e - 33


Remainder Theorem Remainder Theorem အတျကး ဤသေဘာတရာေလက နာလညးမညးဆလ ြငး Foctor Theorem အတျကး သေဘာ တရာက နာလညးရနးမခကးေတာပါ။ တစးခသာ ထပးသထာရနးလပါမညး။

Remainder Theorem အရ f ( x ) အာ ( x – k ) ဖငး စာသညးအခါ ၄ငး၏ အၾကျငးတနးဖမြာ f ( k ) ဖစးသညး။

အကယး၍ ( x – k ) သညး f ( x ) ၏ ဆချကနး ( factor ) တစးခ ဖစးသညးအခါ ( x – k ) ဖငး f ( x ) အာ စာသညးအခါ အၾကျငး 0 ရသညး။ ထ႔ေၾကာငး ( x – k ) သညး f ( x ) ၏ factor တစးခ ဖစးေနသညးအခါ ၄ငးဖငး f ( x ) ကစာလ ြငး အၾကျငးတနးဖမြာ f ( k ) = 0 ဖစးသညး။

Factor Theorem

နာလညးတတးကျ မး အသခေအာငးမငးႏငးၾကပါေစ။

ကေအာငးလျငးဥ ( T.F.S – Education Website )

When f ( x ) is divided by ( x – k ) , the remainder is f ( k )

ကနးတစးခအာ ၄ငး၏ဆချကနးဖငးစာလြငး အၾကျငးတနးဖ သည ဖစးသညး။

When ( x – k ) is a factor of f ( x ), the remainder = f ( k ) = 0 ( or )

( x – k ) is a factor of f ( x ). Then, f ( k ) = 0

TFS-education website

Documents