Which of the following polynomials has a double root? a) x 2 -5x+6 b) x 2 -4x+4 c) x 4 -14x 2 +45 d) Both (a) and (b) e) Both (b) and (c)
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Which of the following polynomials has a double root?
a) x2-5x+6
b) x2-4x+4
c) x4-14x2+45
d) Both (a) and (b)
e) Both (b) and (c)
Which of the following polynomials has a double root?
a) x2-5x+6 =(x-2)(x-3)
b) x2-4x+4 =(x-2)(x-2)
c) x4-14x2+45 =(x2-9)(x2-5)=(x-3)(x+3)(x-√5)(x+√5)
d) Both (a) and (b)
e) Both (b) and (c)
B
Which of the following polynomials has a double root?
a) x2-5x+6
b) x2-4x+4
c) x4-14x2+45
d) Both (a) and (b)
e) Both (b) and (c)
B
Polynomial Division
Factoring Polynomials
• Let’s say I have a polynomial x3-6x2+32 and I want to factor it.– Factoring cubics is hard.
• Maybe I graph it and I notice that it looks like I have a root at x=4.– I can guess that my factoring will look something
like• x3-6x2+32=(x-4)(…………….)
Polynomial Division
• x3-6x2+32=(x-4)(…………….)• In order to find the (…………….), I have to divide
both sides by (x-4).• (x3-6x2+32)/(x-4)=(…………….)• Now I need a way to divide polynomials.
Two Methods
• Polynomial Long Division– Long, takes up a lot of space– Easier to read
• Synthetic Division– Short, fast
Polynomial Long Division
(x3-6x2+32)/(x-4)
• Write out the factor, the division sign, and the full polynomial
x-4 |x3-6x2+0x+32
(x3-6x2+32)/(x-4)
• x3/x =x2, put x2 on top
x2 x-4 |x3-6x2+0x+32
(x3-6x2+32)/(x-4)
• x2(x-4)=x3-4x2, put x3-4x2 underneath an line it up.
x2 x-4 |x3-6x2+0x+32
x3-4x2
(x3-6x2+32)/(x-4)
• Subtract down to get a new polynomial
x2 x-4 |x3-6x2+0x+32
x3-4x2
-2x2+0x+32
(x3-6x2+32)/(x-4)
• Repeat steps: divide to the top (-2x2/x), multiply to the bottom (-2x(x-4)), subtract down.
x2-2x x-4 |x3-6x2+0x+32
x3-4x2
-2x2+0x+32 -2x2+8x
-8x+32
(x3-6x2+32)/(x-4)
• Repeat steps: divide to the top (-8x/x), multiply to the bottom (-8(x-4)), subtract down.
x2-2x -8 x-4 |x3-6x2+0x+32
x3-4x2
-2x2+0x+32 -2x2+8x
-8x+32 -8x+32
0
(x3-6x2+32)/(x-4)
• Our remainder is 0, meaning that x-4 really is a factor of x3-6x2+32
x2-2x -8 x-4 |x3-6x2+0x+32
x3-4x2
-2x2+0x+32 -2x2+8x
-8x+32 -8x+32
0
(x3-6x2+32)/(x-4)
• Write down the factorization x3-6x2+0x+32=(x-4)(x2-2x-8)
x2-2x -8 x-4 |x3-6x2+0x+32
x3-4x2
-2x2+0x+32 -2x2+8x
-8x+32 -8x+32
0
Example with a remainder
x2-2x -8 x-4 |x3-6x2
x3-4x2
-2x2
-2x2+8x -8x -8x+32
-32
Example with a remainder
x2-2x -8 x-4 |x3-6x2
x3-4x2
-2x2
-2x2+8x -8x -8x+32
-32 x-4 Is NOT a factor of x3-6x2
Example with a remainder
x2-2x -8 x-4 |x3-6x2
x3-4x2
-2x2
-2x2+8x -8x -8x+32
-32 x-4 Is NOT a factor of x3-6x2
Synthetic Division
Synthetic Division
• Does exactly the same thing as polynomial long division– Faster– Takes up less space– Easier (for me, at least)
Factoring Polynomials
• Let’s say I have a polynomial x3-6x2+32 and I want to factor it.– Factoring cubics is hard.
• Maybe I graph it and I notice that it looks like I have a root at x=4.– I can guess that my factoring will look something
like• x3-6x2+32=(x-4)(…………….)
Polynomial Division
• x3-6x2+32=(x-4)(…………….)• In order to find the (…………….), I have to divide
both sides by (x-4).• (x3-6x2+32)/(x-4)=(…………….)
What is the quotient when the polynomial 3x3 − 18x2 − 27x + 162 is divided by x-3?
a) 3x2+9x-54
b) 3x2+9x+54
c) 3x2-9x+54
d) 3x2-9x-54
e) None of the above is completely
correct
What is the quotient when the polynomial 3x3 − 18x2 − 27x + 162 is divided by x-3?
3x2 -9x -54 x-3 |3x3-18x2-27x+162 3x3-9x2
-9x2-27x+162 -9x2+27x
-54x+162-54x+162
0
3 -9 -54 0 3 |3 -18 -27 162 9 -27 -162
D) 3x2-9x-54
Fun Tricks with Synthetic Division
• If you divide ƒ(x) and (x-c), then the remainder is the value of ƒ(c)
Example: ƒ(x)=3x3-18x2-27x+162
3 -9 -54 0 3 |3 -18 -27 162 9 -27 -162
Remainder is 0, so ƒ(3)=0
Fun Tricks with Synthetic Division
• If you divide ƒ(x) and (x-c), then the remainder is the value of ƒ(c)
Example: ƒ(x)=3x3-18x2-27x+162
3 -9 -54 0 3 |3 -18 -27 162 9 -27 -162
Remainder is 0, so ƒ(3)=0 3 is a root
(x-3) is a factor
Fun Tricks with Synthetic Division
• If you divide ƒ(x) and (x-c), then the remainder is the value of ƒ(c)
Example: ƒ(x)=3x3-18x2-27x+162
3 -15 -42 120 1 |3 -18 -27 162 3 -15 -42
Remainder is 120, so ƒ(1)=120
Fun Tricks with Synthetic Division
• If you divide ƒ(x) and (x-c), then the remainder is the value of ƒ(c)
Example: ƒ(x)=3x3-18x2-27x+162
3 -15 -42 120 1 |3 -18 -27 162 3 -15 -42
Remainder is 120, so ƒ(1)=120 1 is NOT a root
(x-1) is NOT a factor
Final Thought
• Your book (and possibly your recitation instructor) write synthetic division upside down. It’s the same thing, just with the numbers in a different place.
3 -9 -54 0 3 |3 -18 -27 162 9 -27 -162
3 | 3 -18 -27 162 9 -27 -162 3 -9 -54 | 0
Is thesame as
Which of the following is a linear factor of f(x) = x3 - 6x2 + 21x - 26?
a) x - 2b) x + 2 c) x d) (a) and (b)e) None of the above
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