For Common Assessment
Adding and Subtracting Polynomials
Multiplying Polynomials
Factoring Polynomials
Adding & Subtracting Polynomials
To add or subtract polynomials,
1) Align The Like Terms
2) Add/Subtract The Like Terms
*Subtracting is the same as adding the opposite!!
** When adding or subtracting,
EXPONENTS STAY THE SAME!!
There are two ways to add and subtract polynomials. You
can do it horizontally or vertically.
Horizontal example:
•Simplify (2z + 5y) + (3z – 2y)
(2z + 5y) + (3z – 2y)
= 2z + 5y + 3z – 2y
= 2z + 3z + 5y – 2y
= 5z + 3y
Add the following polynomials
(9y – 7x + 15a) + (-3y + 8x – 8a)
Line up your like terms.
9y – 7x + 15a
+ -3y + 8x – 8a
_________________________
6y + x + 7a
Add the following polynomials
(3a2 + 3ab – b2) + (4ab + 6b2)
+
3a2 + 3ab – b2
4ab + 6b2
_________________________
3a2 + 7ab + 5b2
Add the following polynomials
(4x2 – 2xy + 3y2) + (-3x2 – xy + 2y2)
Line up your like terms.
4x2 – 2xy + 3y2
+ -3x2 – xy + 2y2
_________________________
x2 - 3xy + 5y2
Subtract the following polynomials
(9y – 7x + 15a) – (-3y +8x – 8a)
Line up your like terms and add the
opposite.
9y – 7x + 15a
+
(+
3y
–
8x
+
8a)
--------------------------------------
12y – 15x + 23a
Subtract the following polynomials
(7a – 10b) – (3a + 4b)
7a – 10b
+ (– 3a – 4b)
--------------------------------------
4a – 14b
Subtract the following polynomials
(4x2 – 2xy + 3y2) – (-3x2 – xy + 2y2)
4x2 – 2xy + 3y2
+ (+ 3x2 + xy – 2y2)
--------------------------------------
7x2 – xy + y2
Subtract
2
2
2
2
(5x + 3a – 5x) – (2x – 5a + 7x)
5x2 + 3a2 – 5x
+ (- 2x2 + 5a2 – 7x)
-------------------------------------3x2 + 8a2 – 12x
Subtract
(3x2 + 8x + 4) – (5x2 – 4)
3x2 + 8x + 4
+ (- 5x2
+ 4)
--------------------------------------2x2 + 8x + 8
Find the sum or difference.
(5a – 3b) + (2a + 6b)
1.
2.
3.
4.
3a – 9b
3a + 3b
7a + 3b
7a – 3b
Find the sum or difference.
(5a – 3b) – (2a + 6b)
1.
2.
3.
4.
3a – 9b
3a + 3b
7a + 3b
7a – 9b
(5x2 - 3x + 7) + (2x2 + 5x - 7)
= 7x2 + 2x
(3x3 + 6x - 8) + (4x2 + 2x - 5)
= 3x3 + 4x2 + 8x - 13
(2x3 + 4x2 - 6) – (3x3 + 2x - 2)
(2x3 + 4x2 - 6) + (-3x3 + -2x - -2)
= -x3 + 4x2 - 2x - 4
(7x3 - 3x + 1) – (x3 - 4x2 - 2)
(7x3 - 3x + 1) + (-x3 - -4x2 - -2)
= 6x3 + 4x2 - 3x + 3
7y2 – 3y + 4 + 8y2 + 3y – 4
= 15y2
2x3 – 5x2 + 3x – 1 – (8x3 – 8x2 + 4x + 3)
–6x3 + 3x2 – x – 4
(7y3 +2y2 + 5y – 1) + (5y3 + 7y)
12y3 + 2y2 + 12y – 1
(b4 – 6 + 5b + 1) + (8b4 + 2b – 3b2)
= 9b4 – 3b2 + 7b – 5
MULTIPLYING POLYNOMIALS
Remember that when you multiply two powers
with the same bases, you add the exponents.
(5m2n3)(6m3n6)
5 · 6 · m2+3n3+6
30m5n9
Pre-Algebra
To multiply two monomials,
multiply the coefficients and
add the exponents of the
variables that are the same.
Multiplying Monomials
Multiply.
A. (2x3y2)(6x5y3)
(2x3y2)(6x5y3)
12x8y5
Multiply coefficients and add
exponents.
B. (9a5b7)(–2a4b3)
(9a5b7)(–2a4b3)
–18a9b10
Pre-Algebra
Multiply coefficients and add
exponents.
Try This
Multiply.
A. (5r4s3)(3r3s2)
(5r4s3)(3r3s2)
15r7s5
Multiply coefficients and add
exponents.
B. (7x3y5)(–3x3y2)
(7x3y5)(–3x3y2)
–21x6y7
Multiply coefficients and add
exponents.
Multiplying a Polynomial by a Monomial
Multiply.
A. 3m(5m2 + 2m)
3m(5m2 + 2m)
15m3 + 6m2
Multiply each term in
parentheses by 3m.
B. –6x2y3(5xy4 + 3x4)
–6x2y3(5xy4 + 3x4)
–30x3y7 – 18x6y3
Multiply each term in
parentheses by –6x2y3.
Multiplying a Polynomial by a Monomial
Multiply.
C. –5y3(y2 + 6y – 8)
–5y3(y2 + 6y – 8)
–5y5 – 30y4 + 40y3
Pre-Algebra
Multiply each term in
parentheses by –5y3.
Insert Lesson Title Here
Try This: Example 2A & 2B
Multiply.
A. 4r(8r3 + 16r)
4r(8r3 + 16r)
32r4 + 64r2
Multiply each term in
parentheses by 4r.
B. –3a3b2(4ab3 + 4a2)
–3a3b2(4ab3 + 4a2)
–12a4b5 – 12a5b2
Multiply each term in
parentheses by –3a3b2.
Insert Lesson Title Here
Example 2
Multiply.
C. –2x4(x3 + 4x + 3)
–2x4(x3 + 4x + 3)
–2x7 – 8x5 – 6x4
Pre-Algebra
Multiply each term in
parentheses by –2x4.
Multiply. (2x + 3)(5x + 8)
Using the Distributive property, multiply
2x(5x + 8) + 3(5x + 8).
10x2 + 16x + 15x + 24
Combine like terms.
10x2 + 31x + 24
Another option is called the FOIL method.
F.O.I.L
F irst
O uter
I nner
L ast
(x+2)(x+5)
EXAMPLES
x2 + 8x + 4x + 32
( x + 4 ) ( x + 8 ) =x2 + 12x + 32
( x + 5 ) ( x – 6) =
x2 − 6x + 5x − 30
x2 − x − 30
PRACTICE
( x + 10 ) ( x + 3 ) =
x2 + 3x + 10x + 30
x2 + 13x + 30
(x−7)(x−4)=
x2 − 4x − 7x + 28
x2 − 11x + 28
EXAMPLES
( 2x2 + 4 ) ( 3x − 5 ) = 6x3 − 10x2 + 12x − 20
( 3x2 − 6x) (4x + 2) =
12x3 + 6x2 − 24x2 − 12x
12x3 − 18x2 − 12x
Example:
(x +3)(x+1)=(x)(x)+(x)(1)+(3)(x)+(3)((1)
x  x  3x  3
2
x  4x  3
2
1) Simplify: 5(7n - 2)
Use the distributive property.
5 • 7n
35n - 10
5•2
2)
3
Simplify: a(8a  12)
4
3
3
a 8a  a 12
4
4
6a2 + 9a
3) Simplify: 6rs(r2s - 3)
6rs • r2s - 6rs • 3
6r3s2 - 18rs
4) Simplify: 4t2(3t2 + 2t - 5)
12t4
+ 8t3- 20t2
5) Simplify: - 4m3(-3m - 6n + 4p)
12m4 + 24m3n - 16m3p
Simplify 4y(3y2 – 1)
1.
2.
3.
4.
7y2 – 1
12y2 – 1
12y3 – 1
12y3 – 4y
Simplify -3x2y3(y2 – x2 + 2xy)
1.
2.
3.
4.
-3x2y5 + 3x4y3 – 6x3y4
-3x2y6 + 3x4y3 – 6x2y3
-3x2y5 + 3x4y3 – 6x2y3
3x2y5 – 3x4y3 + 6x3y4
Try These.
1.) (x+2) (x+8) = x2+10x+16
2.) (x+5) (x-7) =
x2-2x-35
3.) (2x+4) (2x-3) = 4x2+2x-12
Examples:. Multiply: 2x(3x2 + 2x – 1).
= 2x(3x2 ) + 2x(2x) + 2x(–1)
= 6x3 + 4x2 – 2x
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Multiply: – 3x2y(5x2 – 2xy + 7y2).
= – 3x2y(5x2 ) – 3x2y(–2xy) – 3x2y(7y2)
= – 15x4y + 6x3y2 – 21x2y3
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Example: Multiply: (x – 1)(2x2 + 7x + 3).
= (x – 1)(2x2) + (x – 1)(7x) + (x – 1)(3)
= 2x3 – 2x2 + 7x2 – 7x + 3x – 3
= 2x3 + 5x2 – 4x – 3
40
Examples:
Multiply: (2x + 1)(7x – 5).
First
Outer
Inner
Last
= 2x(7x) + 2x(–5) + (1)(7x) + (1)(–5)
= 14x2 – 10x + 7x – 5
= 14x2 – 3x – 5
41
Multiply: (5x – 3y)(7x + 6y).
First
Outer
Inner
Last
= 5x(7x) + 5x(6y) + (– 3y)(7x) + (– 3y)(6y)
= 35x2 + 30xy – 21yx – 18y2
= 35x2 + 9xy – 18y2
42
Special Cases
The multiply the sum and difference of two terms,
use this pattern:
(a + b)(a – b) = a2 – ab + ab – b2
= a2 – b2
square of the second term
square of the first term
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Examples: (3x + 2)(3x – 2)
= (3x)2 – (2)2
= 9x2 – 4
(x + 1)(x – 1)
= (x)2 – (1)2
= x2 – 1
44
Special Cases
To square a binomial, use this pattern:
(a + b)2 = (a + b)(a + b) = a2 + ab + ab + b2
= a2 + 2ab + b2
square of the first term
twice the product of the two terms
square of the last term
45
Special Cases
Examples: Multiply: (2x – 2)2 .
= (2x)2 + 2(2x)(– 2) + (– 2)2
= 4x2 – 8x + 4
Multiply: (x + 3y)2 .
= (x)2 + 2(x)(3y) + (3y)2
= x2 + 6xy + 9y2
46
FACTORING
GCF
Sum + Product
Factor by Grouping 4 Terms
Special Products
Techniques of Factoring
Polynomials
1. Greatest Common Factor (GCF).
The GCF for a polynomial is the largest
monomial that divides each term of the
polynomial.
4y  2y
3
Factor out the GCF:
2
Factoring Polynomials - GCF
4y  2y
3
2
Write the two terms in the
form of prime factors…
22y y y 2 y y
2 yy ( 2 y
They have in common 2yy
1)
 2 y (2 y  1)
2
This process is basically the reverse of the distributive property.
Factoring - GCF
The simplest method of factoring a polynomial is to factor out
the greatest common factor (GCF) of each term.
Example: Factor 18x3 + 60x.
GCF = 6x
18x3 + 60x = 6x(3x2) + 6x(10)
= 6x(3x2 + 10)
Find the GCF.
Apply the distributive law
to factor the polynomial.
Check the answer by multiplication.
6x(3x2 + 10) = 6x(3x2) + 6x(10) = 18x3 + 60x
50
Factoring - GCF
Example: Factor 4x2 – 12x + 20.
Therefore, GCF = 4.
4x2 – 12x + 20 = 4x2 – 4 · 3x + 4 · 5
= 4(x2 – 3x + 5)
Check the answer.
4(x2 – 3x + 5) = 4x2 – 12x + 20
51
Factoring - GCF
A common binomial factor can be factored
out of certain expressions.
Example: Factor the expression 5(x + 1) – y(x + 1).
5(x + 1) – y(x + 1) = (5 – y)(x + 1)
Check.
(5 – y)(x + 1) = 5(x + 1) – y(x + 1)
52
Factoring Polynomials - GCF
3 terms
Factor the GCF:
4ab  12a b c  8ab c 
3
2
3 2
2
4 a b ( b - 3a c
One term
4 2
+
2
2b c
2
)
Factoring Polynomials - GCF
EXAMPLE:
5 x(2 x  4)  3(2 x  4) 
( 2 x  4) ( 5x - 3 )
Examples
Factor the following polynomial.
12 x  20 x  3  4  x  x  4  5  x  x  x  x
2
4
 4  x  x (3  5  x  x )
 4 x (3  5 x )
2
2
Examples
Factor the following polynomial.
15 x y  3 x y  3  5  x  y  3  x  y
3
5
2
4
3
5
 3  x  y (5  x  y  1)
2
4
 3 x y (5 xy  1)
2
4
2
4
Factoring – Sum and Product
To factor a trinomial of the form x2 + bx + c,
express the trinomial as the product of two
binomials. For example,
x2 + 10x + 24 = (x + 4)(x + 6).
4 and 6 add up to 10
4 and 6 multiply to 24
57
Factoring Trinomials
x  7x  6
2
( x  6 ) x  1
factors of 6 that add up to 7:
x  5x  6
2
6
x  1x  6
1
( x  6 ) x  1
factors of – 6 that add up to – 5: – 6
2
and
and


1
( x  3 ) x  2 
factors of – 6 that add up to 1:
3
and – 2
Factoring – Sum and Product
Example: Factor x2 – 8x + 15 = (x + a)(x + b)
= x2 + (a + b)x + ab
Therefore a + b = –8
and ab = 15.
It follows that both a and b are negative.
x2 – 8x + 15 = (x – 3)(x – 5).
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Factoring – Sum and Product
Example: Factor x2 + 13x + 36. = (x + a)(x + b)
= x2 + (a + b)x + ab
Therefore a and b are two positive factors of 36 whose sum
is 13.
x2 + 13x + 36 = (x + 4)(x + 9)
60
Factoring 4 Terms by
Grouping
x  3x  2 x  6 
There is no GCF for
all
four terms.
x ( x  3)  2 ( x  3) 
In this problem we
factor GCF
by grouping the first
two
terms and the last two
terms.
3
2
2
( x  3) ( x  2)
2
Factoring – By Grouping 4
Terms
Some polynomials can be factored by grouping terms to
produce a common binomial factor.
Examples: Factor 2xy + 3y – 4x – 6.
= (2xy + 3y) – (4x + 6) Group terms.
= (2x + 3)y – (2x + 3)2
eachthe
pair
of terms.
= (2x + 3)( y – 2)
Factor out
common
binomial.
62
Factoring – By Grouping 4 Terms
2a2 + 3bc – 2ab – 3ac
Factor 2a2 + 3bc – 2ab – 3ac.
= 2a2 – 2ab + 3bc – 3ac
Rearrange terms.
= (2a2 – 2ab) + (3bc – 3ac)
Group terms.
= 2a(a – b) + 3c(b – a)
Factor.
= 2a(a – b) – 3c(a – b)
= (2a – 3c)(a – b)
b – a = – (a – b).
Factor.
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Factoring a trinomial when a ≠ 1
Factor 8b2 + 2b – 3
8b 2  6b  4b  3
2
8b
  6b4b  3



2b(4b  3) 1(4b  3)
(4b  3)(2b 1)
Multiply 8  -3, and break up
the middle term
Now factor by grouping
Factoring a trinomial when a ≠ 1
Factor 2x2 + 19x - 10
8b 2  6b  4b  3
2
8b
  6b4b  3



2b(4b  3) 1(4b  3)
(4b  3)(2b 1)
Multiply 8  -3, and break up
the middle term
Now factor by grouping
Factoring a trinomial when a ≠ 1
Factor 6y2 – 11y - 10
8b 2  6b  4b  3
2
8b
  6b4b  3



2b(4b  3) 1(4b  3)
(4b  3)(2b 1)
Multiply 8  -3, and break up
the middle term
Now factor by grouping
Factoring a trinomial when a ≠ 1
Factor 2x2 – x – 3
8b 2  6b  4b  3
2
8b
  6b4b  3



2b(4b  3) 1(4b  3)
(4b  3)(2b 1)
Multiply 8  -3, and break up
the middle term
Now factor by grouping
Factoring a trinomial when a ≠ 1
Factor 3t2 + 16t + 5
8b 2  6b  4b  3
2
8b
  6b4b  3



2b(4b  3) 1(4b  3)
(4b  3)(2b 1)
Multiply 8  -3, and break up
the middle term
Now factor by grouping
Factoring a trinomial when a ≠ 1
Factor 5x2 + 2x – 3
8b 2  6b  4b  3
2
8b
  6b4b  3



2b(4b  3) 1(4b  3)
(4b  3)(2b 1)
Multiply 8  -3, and break up
the middle term
Now factor by grouping
Factoring a trinomial when a ≠ 1
Factor 6b2 – 11b – 2
8b 2  6b  4b  3
2
8b
  6b4b  3



2b(4b  3) 1(4b  3)
(4b  3)(2b 1)
Multiply 8  -3, and break up
the middle term
Now factor by grouping
Factoring the Difference of Two
Squares
(a + b)(a – b) = a2– ab + ab – b2 = a2 – b2
FORMULA:
a2 – b2 = (a + b)(a – b)
The difference of two bases being squared,
factors as the product of the sum and
difference of the bases that are being
squared.
Factoring the difference of two squares
2
a
Factor
Difference
of two squares
–
= (a + b)(a – b)
2
b
x2 – 4y2
(x) 2
2
(2y)
(x – 2y)(x + 2y)
Factor 16r2 – 25
Difference
Of two squares
2
(4r)
2
(5)
(4r – 5)(4r + 5)
Difference of two squares
y  16 
2
 ( y )  (4)
2
2
 ( y  4)( y  4)
Difference of two squares
25 x  81 
2
 (5 x )  (9)
2
2
 (5 x  9)(5 x  9)
Difference of two squares
y  16 
4
 ( y )  ( 4)
2
2
2
 ( y  4)( y  4)
2
2
 ( y  2)( y  2)( y  4)
2
Factoring – Special Products
A difference of squares can be factored
using the formula a2 – b2 = (a + b)(a – b).
Example: Factor x2 – 9y2.
= (x)2 – (3y)2
= (x + 3y)(x – 3y)
Write terms as perfect
squares.
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