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Factorising - Numeracy Workshop...workshopExpressions and Expansion). Topics include extracting common factors, factorising quadratic expressions and polynomials. Workshop resources:These
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Please Note
These pdf slides are configured for viewing on a computer screen.
Viewing them on hand-held devices may be difficult as they require a“slideshow” mode.
Do not try to print them out as there are many more pages than thenumber of slides listed at the bottom right of each screen.
IntroductionThese slides extend on a basic knowledge of algebra (such as the previous Algebraworkshop Expressions and Expansion). Topics include extracting common factors,factorising quadratic expressions and polynomials.
Workshop resources: These slides are available online:
www.studysmarter.uwa.edu.au → Numeracy and Maths → Online Resources
Next Workshop: See your Workshop Calendar →
www.studysmarter.uwa.edu.au
Drop-in Study Sessions: Monday, Wednesday, Friday, 10am-12pm, Room 2202,Second Floor, Social Sciences South Building, every week.
IntroductionThese slides extend on a basic knowledge of algebra (such as the previous Algebraworkshop Expressions and Expansion). Topics include extracting common factors,factorising quadratic expressions and polynomials.
Workshop resources: These slides are available online:
www.studysmarter.uwa.edu.au → Numeracy and Maths → Online Resources
Next Workshop: See your Workshop Calendar →
www.studysmarter.uwa.edu.au
Drop-in Study Sessions: Monday, Wednesday, Friday, 10am-12pm, Room 2202,Second Floor, Social Sciences South Building, every week.
IntroductionThese slides extend on a basic knowledge of algebra (such as the previous Algebraworkshop Expressions and Expansion). Topics include extracting common factors,factorising quadratic expressions and polynomials.
Workshop resources: These slides are available online:
www.studysmarter.uwa.edu.au → Numeracy and Maths → Online Resources
Next Workshop: See your Workshop Calendar →
www.studysmarter.uwa.edu.au
Drop-in Study Sessions: Monday, Wednesday, Friday, 10am-12pm, Room 2202,Second Floor, Social Sciences South Building, every week.
IntroductionThese slides extend on a basic knowledge of algebra (such as the previous Algebraworkshop Expressions and Expansion). Topics include extracting common factors,factorising quadratic expressions and polynomials.
Workshop resources: These slides are available online:
www.studysmarter.uwa.edu.au → Numeracy and Maths → Online Resources
Next Workshop: See your Workshop Calendar →
www.studysmarter.uwa.edu.au
Drop-in Study Sessions: Monday, Wednesday, Friday, 10am-12pm, Room 2202,Second Floor, Social Sciences South Building, every week.
IntroductionThese slides extend on a basic knowledge of algebra (such as the previous Algebraworkshop Expressions and Expansion). Topics include extracting common factors,factorising quadratic expressions and polynomials.
Workshop resources: These slides are available online:
www.studysmarter.uwa.edu.au → Numeracy and Maths → Online Resources
Next Workshop: See your Workshop Calendar →
www.studysmarter.uwa.edu.au
Drop-in Study Sessions: Monday, Wednesday, Friday, 10am-12pm, Room 2202,Second Floor, Social Sciences South Building, every week.
A factor of a term is a number, variable or combination that divides into it evenly.
Example: The term 12x has a factor of 4 since 3x × 4 = 12x .
Important point: the variable x is also a factor of 12x since 12× x = 12x . (Thisis true even though x could be any number, including non-whole numbers.)
Factors of 12x can be made up of combinations of other factors, such as 2, 6x, 1,3x, 12, etc.
A factor of a term is a number, variable or combination that divides into it evenly.
Example: The term 12x has a factor of 4 since 3x × 4 = 12x .
Important point: the variable x is also a factor of 12x since 12× x = 12x . (Thisis true even though x could be any number, including non-whole numbers.)
Factors of 12x can be made up of combinations of other factors, such as 2, 6x, 1,3x, 12, etc.
A factor of a term is a number, variable or combination that divides into it evenly.
Example: The term 12x has a factor of 4 since 3x × 4 = 12x .
Important point: the variable x is also a factor of 12x since 12× x = 12x . (Thisis true even though x could be any number, including non-whole numbers.)
Factors of 12x can be made up of combinations of other factors, such as 2, 6x, 1,3x, 12, etc.
A factor of a term is a number, variable or combination that divides into it evenly.
Example: The term 12x has a factor of 4 since 3x × 4 = 12x .
Important point: the variable x is also a factor of 12x since 12× x = 12x . (Thisis true even though x could be any number, including non-whole numbers.)
Factors of 12x can be made up of combinations of other factors, such as
A factor of a term is a number, variable or combination that divides into it evenly.
Example: The term 12x has a factor of 4 since 3x × 4 = 12x .
Important point: the variable x is also a factor of 12x since 12× x = 12x . (Thisis true even though x could be any number, including non-whole numbers.)
Factors of 12x can be made up of combinations of other factors, such as 2, 6x, 1,3x, 12, etc.
Here is a common type of double bracket expression. Let’s expand the brackets tosee whether there are any useful patterns we can use to reverse the process:
Here is a common type of double bracket expression. Let’s expand the brackets tosee whether there are any useful patterns we can use to reverse the process:
Here is a common type of double bracket expression. Let’s expand the brackets tosee whether there are any useful patterns we can use to reverse the process:
Here is a common type of double bracket expression. Let’s expand the brackets tosee whether there are any useful patterns we can use to reverse the process:
Here is a common type of double bracket expression. Let’s expand the brackets tosee whether there are any useful patterns we can use to reverse the process:
Here is a common type of double bracket expression. Let’s expand the brackets tosee whether there are any useful patterns we can use to reverse the process:
Here is a common type of double bracket expression. Let’s expand the brackets tosee whether there are any useful patterns we can use to reverse the process:
Here is a common type of double bracket expression. Let’s expand the brackets tosee whether there are any useful patterns we can use to reverse the process:
Here is a common type of double bracket expression. Let’s expand the brackets tosee whether there are any useful patterns we can use to reverse the process:
Here is a common type of double bracket expression. Let’s expand the brackets tosee whether there are any useful patterns we can use to reverse the process:
Here is a common type of double bracket expression. Let’s expand the brackets tosee whether there are any useful patterns we can use to reverse the process:
Here is a common type of double bracket expression. Let’s expand the brackets tosee whether there are any useful patterns we can use to reverse the process:
Here we see that the multiplier of x2 is 5. We also notice that all multipliers inthe above expression are divisible by 5. Hence, we can factor out this common
factor as follows:
5x2 + 40x + 60 = 5(x2 + 8x + 12)
Now we look for two numbers which add to 8 and multiply to 12. This gives us:
Here we see that the multiplier of x2 is 5. We also notice that all multipliers inthe above expression are divisible by 5. Hence, we can factor out this common
factor as follows:
5x2 + 40x + 60 = 5(x2 + 8x + 12)
Now we look for two numbers which add to 8 and multiply to 12. This gives us:
Here we see that the multiplier of x2 is 5. We also notice that all multipliers inthe above expression are divisible by 5. Hence, we can factor out this common
factor as follows:
5x2 + 40x + 60 = 5(x2 + 8x + 12)
Now we look for two numbers which add to 8 and multiply to 12. This gives us:
Here we see that the multiplier of x2 is 5. We also notice that all multipliers inthe above expression are divisible by 5. Hence, we can factor out this common
factor as follows:
5x2 + 40x + 60 = 5(x2 + 8x + 12)
Now we look for two numbers which add to 8 and multiply to 12. This gives us:
then we are being asked to write the above in the form
(x + a)(x + b)
It looks different to the expressions we have been factorising because it appears tohave no x term. However, we can make it the same if we include an x term with a
multiplier of 0:
x2+0x − 9
So we know that we need two numbers which add up to 0 and multiply to give−9. Can you find two numbers which do this?
then we are being asked to write the above in the form
(x + a)(x + b)
It looks different to the expressions we have been factorising because it appears tohave no x term. However, we can make it the same if we include an x term with a
multiplier of
0:
x2+0x − 9
So we know that we need two numbers which add up to 0 and multiply to give−9. Can you find two numbers which do this?
then we are being asked to write the above in the form
(x + a)(x + b)
It looks different to the expressions we have been factorising because it appears tohave no x term. However, we can make it the same if we include an x term with a
multiplier of 0:
x2+0x − 9
So we know that we need two numbers which add up to 0 and multiply to give−9. Can you find two numbers which do this?
then we are being asked to write the above in the form
(x + a)(x + b)
It looks different to the expressions we have been factorising because it appears tohave no x term. However, we can make it the same if we include an x term with a
multiplier of 0:
x2+0x − 9
So we know that we need two numbers which add up to 0 and multiply to give−9. Can you find two numbers which do this?
then we are being asked to write the above in the form
(x + a)(x + b)
It looks different to the expressions we have been factorising because it appears tohave no x term. However, we can make it the same if we include an x term with a
multiplier of 0:
x2+0x − 9
So we know that we need two numbers which add up to 0 and multiply to give−9. Can you find two numbers which do this?
then we are being asked to write the above in the form
(x + a)(x + b)
It looks different to the expressions we have been factorising because it appears tohave no x term. However, we can make it the same if we include an x term with a
multiplier of 0:
x2+0x − 9
So we know that we need two numbers which add up to 0 and multiply to give−9. Can you find two numbers which do this?
We need to write this in the form (ax + b)(cx + d).
The numbers a and c must multiply up to 2, and so one of them must be 1 andthe other must be 2. (It doesn’t matter which is which because multiplication is
commutative.)
(2x + b)(x + d)
The numbers b and d must multiply up to 1, and so one of them must be 1 andthe other must be 1.
We need to write this in the form (ax + b)(cx + d).
The numbers a and c must multiply up to 2, and so one of them must be 1 andthe other must be 2. (It doesn’t matter which is which because multiplication is
commutative.)
(2x + b)(x + d)
The numbers b and d must multiply up to 1, and so one of them must be 1 andthe other must be 1.
We need to write this in the form (ax + b)(cx + d).
The numbers a and c must multiply up to 2, and so one of them must be 1 andthe other must be 2. (It doesn’t matter which is which because multiplication is
commutative.)
(2x + b)(x + d)
The numbers b and d must multiply up to 1, and so one of them must be 1 andthe other must be 1.
We need to write this in the form (ax + b)(cx + d).
The numbers a and c must multiply up to 2, and so one of them must be 1 andthe other must be 2. (It doesn’t matter which is which because multiplication is
commutative.)
(2x + b)(x + d)
The numbers b and d must multiply up to 1, and so one of them must be 1 andthe other must be 1.
We need to write this in the form (ax + b)(cx + d).
The numbers a and c must multiply to 7, and so one of them must be 1 and theother must be 7. It doesn’t matter which is which.
(7x + b)(x + d)
The numbers b and d must multiply to 2, so one of them must be 1 and the othermust be 2. The question is, which one is which? There are two possibilities:
We need to write this in the form (ax + b)(cx + d).
The numbers a and c must multiply to 7, and so one of them must be 1 and theother must be 7. It doesn’t matter which is which.
(7x + b)(x + d)
The numbers b and d must multiply to 2, so one of them must be 1 and the othermust be 2. The question is, which one is which? There are two possibilities:
We need to write this in the form (ax + b)(cx + d).
The numbers a and c must multiply to 7, and so one of them must be 1 and theother must be 7. It doesn’t matter which is which.
(7x + b)(x + d)
The numbers b and d must multiply to 2, so one of them must be 1 and the othermust be 2. The question is, which one is which? There are two possibilities:
We need to write this in the form (ax + b)(cx + d).
The numbers a and c must multiply to 7, and so one of them must be 1 and theother must be 7. It doesn’t matter which is which.
(7x + b)(x + d)
The numbers b and d must multiply to 2, so one of them must be 1 and the othermust be 2. The question is, which one is which? There are two possibilities:
We need to write this in the form (ax + b)(cx + d).
The numbers a and c must multiply to 7, and so one of them must be 1 and theother must be 7. It doesn’t matter which is which.
(7x + b)(x + d)
The numbers b and d must multiply to 2, so one of them must be 1 and the othermust be 2. The question is, which one is which? There are two possibilities:
We need to write this in the form (ax + b)(cx + d).
The numbers a and c must multiply to 7, and so one of them must be 1 and theother must be 7. It doesn’t matter which is which.
(7x + b)(x + d)
The numbers b and d must multiply to 2, so one of them must be 1 and the othermust be 2. The question is, which one is which? There are two possibilities:
We need to write this in the form (ax + b)(cx + d).
The numbers a and c must multiply to 7, and so one of them must be 1 and theother must be 7. It doesn’t matter which is which.
(7x + b)(x + d)
The numbers b and d must multiply to 2, so one of them must be 1 and the othermust be 2. The question is, which one is which? There are two possibilities:
We need to write this in the form (ax + b)(cx + d).
The numbers a and c must multiply to 7, and so one of them must be 1 and theother must be 7. It doesn’t matter which is which.
(7x + b)(x + d)
The numbers b and d must multiply to 2, so one of them must be 1 and the othermust be 2. The question is, which one is which? There are two possibilities:
We need to write this in the form (ax + b)(cx + d).
The problem now is that both 6 and 10 have multiple possible factorisations!There are in fact 16 potential answers to test.
The best way to navigate through these options is with a combination of educatedguessing and trial-and-error. Start by writing out the potential factorisations of 6
and 10 as follows:
3
2
5
2
10
1
-
-
���
���*HHHHHHj-
-
In this case, the number term (−10) is negative so we need to get a pair whosedifference is 11.
We need to write this in the form (ax + b)(cx + d).
The problem now is that both 6 and 10 have multiple possible factorisations!There are in fact 16 potential answers to test.
The best way to navigate through these options is with a combination of educatedguessing and trial-and-error. Start by writing out the potential factorisations of 6
and 10 as follows:
3
2
5
2
10
1
-
-
���
���*HHHHHHj-
-
In this case, the number term (−10) is negative so we need to get a pair whosedifference is 11.
We need to write this in the form (ax + b)(cx + d).
The problem now is that both 6 and 10 have multiple possible factorisations!There are in fact 16 potential answers to test.
The best way to navigate through these options is with a combination of educatedguessing and trial-and-error. Start by writing out the potential factorisations of 6
and 10 as follows:
3
2
5
2
10
1
-
-
���
���*HHHHHHj-
-
In this case, the number term (−10) is negative so we need to get a pair whosedifference is 11.
We need to write this in the form (ax + b)(cx + d).
The problem now is that both 6 and 10 have multiple possible factorisations!There are in fact 16 potential answers to test.
The best way to navigate through these options is with a combination of educatedguessing and trial-and-error. Start by writing out the potential factorisations of 6
and 10 as follows:
3
2
5
2
10
1
-
-
���
���*HHHHHHj-
-
In this case, the number term (−10) is negative so we need to get a pair whosedifference is 11.
We need to write this in the form (ax + b)(cx + d).
The problem now is that both 6 and 10 have multiple possible factorisations!There are in fact 16 potential answers to test.
The best way to navigate through these options is with a combination of educatedguessing and trial-and-error. Start by writing out the potential factorisations of 6
and 10 as follows:
3
2
6
1
5
2
10
1
-
-
���
���*HHHHHHj-
-
In this case, the number term (−10) is negative so we need to get a pair whosedifference is 11.
We need to write this in the form (ax + b)(cx + d).
The problem now is that both 6 and 10 have multiple possible factorisations!There are in fact 16 potential answers to test.
The best way to navigate through these options is with a combination of educatedguessing and trial-and-error. Start by writing out the potential factorisations of 6
and 10 as follows:
3
2
6
1
5
2
10
1
-
-
���
���*HHHHHHj-
-
In this case, the number term (−10) is negative so we need to get a pair whosedifference is 11.
We need to write this in the form (ax + b)(cx + d).
The problem now is that both 6 and 10 have multiple possible factorisations!There are in fact 16 potential answers to test.
The best way to navigate through these options is with a combination of educatedguessing and trial-and-error. Start by writing out the potential factorisations of 6
and 10 as follows:
3
2
6
1
5
2
10
1
-
-
���
���*HHHHHHj-
-
In this case, the number term (−10) is negative so we need to get a pair whosedifference is 11.
We need to write this in the form (ax + b)(cx + d).
The problem now is that both 6 and 10 have multiple possible factorisations!There are in fact 16 potential answers to test.
The best way to navigate through these options is with a combination of educatedguessing and trial-and-error. Start by writing out the potential factorisations of 6
and 10 as follows:
3
2
6
1
5
2
10
1
-
-
���
���*HHHHHHj
-
-
In this case, the number term (−10) is negative so we need to get a pair whosedifference is 11.
We need to write this in the form (ax + b)(cx + d).
The problem now is that both 6 and 10 have multiple possible factorisations!There are in fact 16 potential answers to test.
The best way to navigate through these options is with a combination of educatedguessing and trial-and-error. Start by writing out the potential factorisations of 6
and 10 as follows:
3
2
6
1
5
2
10
1
-
-
���
���*HHHHHHj
-
-
In this case, the number term (−10) is negative so we need to get a pair whosedifference is 11.
2× 2 = 4 and 3× 5 = 15. Difference = 11. We have a [email protected] Factorising 25 / 43
Harder Factorisation: Example
Factorise 6x2 − 11x − 10.
We need to write this in the form (ax + b)(cx + d).
We have found that 2× 2 = 4 and 3× 5 = 15.
(2x5)(3x2)
All we need to do now is place the “+” and “−” signs in the appropriate brackets.
Factorising can also simplify algebraic fractions:
x2 + 3x + 2
x + 1=
(x + 1)(x + 2)
x + 1(factorise)
=1��
��(x + 1)(x + 2)
���x + 11(cancel common factors)
= x + 2
Note: Watch out when you cancel terms involving variables. The original fractionmakes it clear that there is a problem when x = −1 because the fraction becomes
00 , which is an indeterminate quantity. This problem is no longer obvious in the
simplified version. Usually, we would write the answer as
Factorising can also simplify algebraic fractions:
x2 + 3x + 2
x + 1=
(x + 1)(x + 2)
x + 1(factorise)
=1��
��(x + 1)(x + 2)
���x + 11(cancel common factors)
= x + 2
Note: Watch out when you cancel terms involving variables. The original fractionmakes it clear that there is a problem when x = −1 because the fraction becomes
00 , which is an indeterminate quantity. This problem is no longer obvious in the
simplified version. Usually, we would write the answer as
Factorising can also simplify algebraic fractions:
x2 + 3x + 2
x + 1=
(x + 1)(x + 2)
x + 1(factorise)
=1��
��(x + 1)(x + 2)
���x + 11(cancel common factors)
= x + 2
Note: Watch out when you cancel terms involving variables. The original fractionmakes it clear that there is a problem when x = −1 because the fraction becomes
00 , which is an indeterminate quantity. This problem is no longer obvious in the
simplified version. Usually, we would write the answer as
Factorising can also simplify algebraic fractions:
x2 + 3x + 2
x + 1=
(x + 1)(x + 2)
x + 1(factorise)
=1��
��(x + 1)(x + 2)
���x + 11(cancel common factors)
= x + 2
Note: Watch out when you cancel terms involving variables. The original fractionmakes it clear that there is a problem when x = −1 because the fraction becomes
00 , which is an indeterminate quantity. This problem is no longer obvious in the
simplified version. Usually, we would write the answer as
Factorising can also simplify algebraic fractions:
x2 + 3x + 2
x + 1=
(x + 1)(x + 2)
x + 1(factorise)
=1��
��(x + 1)(x + 2)
���x + 11(cancel common factors)
= x + 2
Note: Watch out when you cancel terms involving variables. The original fractionmakes it clear that there is a problem when x = −1 because the fraction becomes
00 , which is an indeterminate quantity. This problem is no longer obvious in the
simplified version. Usually, we would write the answer as
Factorising can also simplify algebraic fractions:
x2 + 3x + 2
x + 1=
(x + 1)(x + 2)
x + 1(factorise)
=1��
��(x + 1)(x + 2)
���x + 11(cancel common factors)
= x + 2
Note: Watch out when you cancel terms involving variables. The original fractionmakes it clear that there is a problem when x = −1 because the fraction becomes
00 , which is an indeterminate quantity. This problem is no longer obvious in the
simplified version. Usually, we would write the answer as
When we add or subtract polynomials we get a new polynomial (just like we dowith numbers!). This can be done by adding and subtracting their like terms.
(4x2 + 3x + 7) + (2x2 + 5x + 2)
We may drop the brackets in this case (why?).
4x2 + 3x + 7 + 2x2 + 5x + 2
If you need to, shift the signed terms around so that like terms are next to eachother.
When we add or subtract polynomials we get a new polynomial (just like we dowith numbers!). This can be done by adding and subtracting their like terms.
(4x2 + 3x + 7) + (2x2 + 5x + 2)
We may drop the brackets in this case (why?).
4x2 + 3x + 7 + 2x2 + 5x + 2
If you need to, shift the signed terms around so that like terms are next to eachother.
When we add or subtract polynomials we get a new polynomial (just like we dowith numbers!). This can be done by adding and subtracting their like terms.
(4x2 + 3x + 7) + (2x2 + 5x + 2)
We may drop the brackets in this case (why?).
4x2 + 3x + 7 + 2x2 + 5x + 2
If you need to, shift the signed terms around so that like terms are next to eachother.
When we add or subtract polynomials we get a new polynomial (just like we dowith numbers!). This can be done by adding and subtracting their like terms.
(4x2 + 3x + 7) + (2x2 + 5x + 2)
We may drop the brackets in this case (why?).
4x2 + 3x + 7 + 2x2 + 5x + 2
If you need to, shift the signed terms around so that like terms are next to eachother.
There is a negative sign to the left of the second set of brackets. This iseffectively a multiplier of −1. From the previous workshop, we know that this −1
gets distributed to each term in the bracket.
3x3 − 4x2 + 5− (x3 + 3x − 4)
= 3x3 − 4x2 + 5− x3− 3x + 4
(In short, every sign in the 2nd bracket changes.)
If you need to, shift the signed terms around so that like terms are next to eachother.
There is a negative sign to the left of the second set of brackets. This iseffectively a multiplier of −1. From the previous workshop, we know that this −1
gets distributed to each term in the bracket.
3x3 − 4x2 + 5− (x3 + 3x − 4)
= 3x3 − 4x2 + 5− x3− 3x + 4
(In short, every sign in the 2nd bracket changes.)
If you need to, shift the signed terms around so that like terms are next to eachother.
There is a negative sign to the left of the second set of brackets. This iseffectively a multiplier of −1. From the previous workshop, we know that this −1
gets distributed to each term in the bracket.
3x3 − 4x2 + 5− (x3 + 3x − 4)
= 3x3 − 4x2 + 5
− x3− 3x + 4
(In short, every sign in the 2nd bracket changes.)
If you need to, shift the signed terms around so that like terms are next to eachother.
There is a negative sign to the left of the second set of brackets. This iseffectively a multiplier of −1. From the previous workshop, we know that this −1
gets distributed to each term in the bracket.
3x3 − 4x2 + 5− (x3 + 3x − 4)
= 3x3 − 4x2 + 5
− x3− 3x + 4
(In short, every sign in the 2nd bracket changes.)
If you need to, shift the signed terms around so that like terms are next to eachother.
There is a negative sign to the left of the second set of brackets. This iseffectively a multiplier of −1. From the previous workshop, we know that this −1
gets distributed to each term in the bracket.
3x3 − 4x2 + 5− (x3 + 3x − 4)
= 3x3 − 4x2 + 5− x3
− 3x + 4
(In short, every sign in the 2nd bracket changes.)
If you need to, shift the signed terms around so that like terms are next to eachother.
There is a negative sign to the left of the second set of brackets. This iseffectively a multiplier of −1. From the previous workshop, we know that this −1
gets distributed to each term in the bracket.
3x3 − 4x2 + 5− (x3 + 3x − 4)
= 3x3 − 4x2 + 5− x3
− 3x + 4
(In short, every sign in the 2nd bracket changes.)
If you need to, shift the signed terms around so that like terms are next to eachother.
There is a negative sign to the left of the second set of brackets. This iseffectively a multiplier of −1. From the previous workshop, we know that this −1
gets distributed to each term in the bracket.
3x3 − 4x2 + 5− (x3 + 3x − 4)
= 3x3 − 4x2 + 5− x3− 3x
+ 4
(In short, every sign in the 2nd bracket changes.)
If you need to, shift the signed terms around so that like terms are next to eachother.
There is a negative sign to the left of the second set of brackets. This iseffectively a multiplier of −1. From the previous workshop, we know that this −1
gets distributed to each term in the bracket.
3x3 − 4x2 + 5− (x3 + 3x − 4)
= 3x3 − 4x2 + 5− x3− 3x
+ 4
(In short, every sign in the 2nd bracket changes.)
If you need to, shift the signed terms around so that like terms are next to eachother.
There is a negative sign to the left of the second set of brackets. This iseffectively a multiplier of −1. From the previous workshop, we know that this −1
gets distributed to each term in the bracket.
3x3 − 4x2 + 5− (x3 + 3x − 4)
= 3x3 − 4x2 + 5− x3− 3x + 4
(In short, every sign in the 2nd bracket changes.)
If you need to, shift the signed terms around so that like terms are next to eachother.
There is a negative sign to the left of the second set of brackets. This iseffectively a multiplier of −1. From the previous workshop, we know that this −1
gets distributed to each term in the bracket.
3x3 − 4x2 + 5− (x3 + 3x − 4)
= 3x3 − 4x2 + 5− x3− 3x + 4
(In short, every sign in the 2nd bracket changes.)
If you need to, shift the signed terms around so that like terms are next to eachother.
There is a negative sign to the left of the second set of brackets. This iseffectively a multiplier of −1. From the previous workshop, we know that this −1
gets distributed to each term in the bracket.
3x3 − 4x2 + 5− (x3 + 3x − 4)
= 3x3 − 4x2 + 5− x3− 3x + 4
(In short, every sign in the 2nd bracket changes.)
If you need to, shift the signed terms around so that like terms are next to eachother.
There is a negative sign to the left of the second set of brackets. This iseffectively a multiplier of −1. From the previous workshop, we know that this −1
gets distributed to each term in the bracket.
3x3 − 4x2 + 5− (x3 + 3x − 4)
= 3x3 − 4x2 + 5− x3− 3x + 4
(In short, every sign in the 2nd bracket changes.)
If you need to, shift the signed terms around so that like terms are next to eachother.
There is a negative sign to the left of the second set of brackets. This iseffectively a multiplier of −1. From the previous workshop, we know that this −1
gets distributed to each term in the bracket.
3x3 − 4x2 + 5− (x3 + 3x − 4)
= 3x3 − 4x2 + 5− x3− 3x + 4
(In short, every sign in the 2nd bracket changes.)
If you need to, shift the signed terms around so that like terms are next to eachother.
There is a negative sign to the left of the second set of brackets. This iseffectively a multiplier of −1. From the previous workshop, we know that this −1
gets distributed to each term in the bracket.
3x3 − 4x2 + 5− (x3 + 3x − 4)
= 3x3 − 4x2 + 5− x3− 3x + 4
(In short, every sign in the 2nd bracket changes.)
If you need to, shift the signed terms around so that like terms are next to eachother.