# For Common Assessment Chapter 10 Review

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For Common Assessment Chapter 10 Review
For Common Assessment Adding and Subtracting Polynomials Multiplying Polynomials Factoring 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
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 + 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)
4a – 14b

Subtract the following polynomials (4x2 – 2xy + 3y2) – (-3x2 – xy + 2y2)
7x2 – xy + y2

Subtract (5x2 + 3a2 – 5x) – (2x2 – 5a2 + 7x) 5x2 + 3a2 – 5x
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)

Find the sum or difference. (5a – 3b) – (2a + 6b)

(5x2 - 3x + 7) + (2x2 + 5x - 7) = 7x2 + 2x (3x3 + 6x - 8) + (4x2 + 2x - 5) = 3x3 + 4x2 + 8x - 13 Example… this is called the horizontal method…where you keep the problem written horizontally and mentally group the terms together by degree. You can also write the rearrangement down (by commuting the terms). Just be sure to combine the terms with the correct terms. For example, the second example, it is common for student to accidentally combine the cubic term and the quadratic term.

(2x3 + 4x2 - 6) – (3x3 + 2x - 2) (2x3 + 4x2 - 6) + (-3x3 + -2x - -2)
More examples… this time with subtraction. You have to remember that every term of that second polynomial is being subtracted. So it is useful to change the problem to an addition one before moving on. You basically have to change every sign in the polynomial that you are subtracting. In essence, you are distributing a negative one. (7x3 - 3x + 1) + (-x3 - -4x2 - -2) = 6x3 + 4x2 - 3x + 3

7y2 – 3y y2 + 3y – 4 2x3 – 5x2 + 3x – 1 – (8x3 – 8x2 + 4x + 3) = 15y2 Now, some people like to arrange their addition and subtracting in the “old school” way… which is how you were taught to add and subtract in elementary school. You just write the problem so that the like terms line up with each other vertically… then add or subtract as necessary. I put the subtraction in parenthesis to remind myself that every term needs to be subtracted. Sometimes if you don’t , you just think that the coefficient of the first term is negative. –6x3 + 3x2 – x – 4

(7y3 +2y2 + 5y – 1) + (5y3 + 7y) 12y3 + 2y2 + 12y – 1
More examples… the zeroes are placed only as place holders. They are unnecessary

(b4 – 6 + 5b + 1) + (8b4 + 2b – 3b2) = 9b4 – 3b2 + 7b – 5
Even more examples

MULTIPLYING POLYNOMIALS
Remember that when you multiply two powers with the same bases, you add the exponents. To multiply two monomials, multiply the coefficients and add the exponents of the variables that are the same. (5m2n3)(6m3n6) 5 · 6 · m2+3n3+6 30m5n9 Pre-Algebra

Multiplying Monomials
A. (2x3y2)(6x5y3) (2x3y2)(6x5y3) Multiply coefficients and add exponents. 12x8y5 B. (9a5b7)(–2a4b3) (9a5b7)(–2a4b3) Multiply coefficients and add exponents. –18a9b10 Pre-Algebra

–21x6y7 Try This Multiply. A. (5r4s3)(3r3s2) (5r4s3)(3r3s2)
Multiply coefficients and add exponents. 15r7s5 B. (7x3y5)(–3x3y2) (7x3y5)(–3x3y2) Multiply coefficients and add exponents. –21x6y7

Multiplying a Polynomial by a Monomial
A. 3m(5m2 + 2m) 3m(5m2 + 2m) Multiply each term in parentheses by 3m. 15m3 + 6m2 B. –6x2y3(5xy4 + 3x4) –6x2y3(5xy4 + 3x4) Multiply each term in parentheses by –6x2y3. –30x3y7 – 18x6y3

Multiplying a Polynomial by a Monomial
C. –5y3(y2 + 6y – 8) –5y3(y2 + 6y – 8) Multiply each term in parentheses by –5y3. –5y5 – 30y4 + 40y3 Pre-Algebra

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Try This: Example 2A & 2B Multiply. A. 4r(8r3 + 16r) 4r(8r3 + 16r) Multiply each term in parentheses by 4r. 32r4 + 64r2 B. –3a3b2(4ab3 + 4a2) –3a3b2(4ab3 + 4a2) Multiply each term in parentheses by –3a3b2. –12a4b5 – 12a5b2

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Example 2 Multiply. C. –2x4(x3 + 4x + 3) –2x4(x3 + 4x + 3) Multiply each term in parentheses by –2x4. –2x7 – 8x5 – 6x4 Pre-Algebra

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 ( x + 4 ) ( x + 8 ) = ( x + 5 ) ( x – 6) = x2 + 8x + 4x + 32
+ 32 x2 + 12x + 32 x2 − 6x + 5x − 30 x2 − x − 30

PRACTICE ( x + 10 ) ( x + 3 ) = ( x − 7 ) ( x − 4 ) =

EXAMPLES ( 2x2 + 4 ) ( 3x − 5 ) = ( 3x2 − 6x) (4x + 2) = 6x3 − 10x2
− 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)

1) Simplify: 5(7n - 2) Use the distributive property. 5 • 7n 35n - 10
- 5 • 2

2) Simplify: 3) Simplify: 6rs(r2s - 3) 6a2 + 9a 6rs • r2s 6r3s2 - 18rs

5) Simplify: - 4m3(-3m - 6n + 4p)
4) Simplify: 4t2(3t2 + 2t - 5) 12t4 5) Simplify: - 4m3(-3m - 6n + 4p) 12m4 + 8t3 - 20t2 + 24m3n - 16m3p

Simplify 4y(3y2 – 1) 7y2 – 1 12y2 – 1 12y3 – 1 12y3 – 4y

Simplify -3x2y3(y2 – x2 + 2xy) -3x2y5 + 3x4y3 – 6x3y4

Try These. 1.) (x+2) (x+8) = x2+10x+16 2.) (x+5) (x-7) = x2-2x-35

Examples:. Multiply: 2x(3x2 + 2x – 1).
= 2x(3x2 ) + 2x(2x) + 2x(–1) = 6x3 + 4x2 – 2x

Multiply: – 3x2y(5x2 – 2xy + 7y2).
= – 3x2y(5x2 ) – 3x2y(– 2xy) – 3x2y(7y2) = – 15x4y + 6x3y2 – 21x2y3

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

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

= 5x(7x) + 5x(6y) + (– 3y)(7x) + (– 3y)(6y)
Multiply: (5x – 3y)(7x + 6y). First Outer Inner Last = 5x(7x) + 5x(6y) + (– 3y)(7x) + (– 3y)(6y) = 35x2 + 30xy – 21yx – 18y2 = 35x2 + 9xy – 18y2

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

Examples: (3x + 2)(3x – 2) (x + 1)(x – 1) = (3x)2 – (2)2 = (x)2 – (1)2 = 9x2 – 4 = x2 – 1

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

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

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. Factor out the GCF:

Factoring Polynomials - GCF
Write the two terms in the form of prime factors… They have in common 2yy 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 Find the GCF. 18x3 + 60x = 6x (3x2) + 6x (10) Apply the distributive law to factor the polynomial. = 6x (3x2 + 10) Check the answer by multiplication. 6x (3x2 + 10) = 6x (3x2) + 6x (10) = 18x3 + 60x

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

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)

Factoring Polynomials - GCF
3 terms Factor the GCF: a b 2 - 3a c 2 + 2b c 2 4 ( ) b One term

Factoring Polynomials - GCF
EXAMPLE: 5x 3

Examples Factor the following polynomial.

Examples Factor the following polynomial.

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

Factoring Trinomials factors of 6 that add up to 7: 6 and 1

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).

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)

Factoring 4 Terms by Grouping
There is no GCF for all four terms. In this problem we factor GCF by grouping the first two terms and the last two terms.

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 = (2x + 3)( y – 2) Factor each pair of terms. Factor out the common binomial.

Factoring – By Grouping 4 Terms
Factor 2a2 + 3bc – 2ab – 3ac. 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) b – a = – (a – b). = (2a – 3c)(a – b) Factor.

Factoring a trinomial when a ≠ 1
Multiply 8  -3, and break up the middle term Factor 8b2 + 2b – 3 Now factor by grouping

Factoring a trinomial when a ≠ 1
Multiply 8  -3, and break up the middle term Factor 2x2 + 19x - 10 Now factor by grouping

Factoring a trinomial when a ≠ 1
Multiply 8  -3, and break up the middle term Factor 6y2 – 11y - 10 Now factor by grouping

Factoring a trinomial when a ≠ 1
Multiply 8  -3, and break up the middle term Factor 2x2 – x – 3 Now factor by grouping

Factoring a trinomial when a ≠ 1
Multiply 8  -3, and break up the middle term Factor 3t2 + 16t + 5 Now factor by grouping

Factoring a trinomial when a ≠ 1
Multiply 8  -3, and break up the middle term Factor 5x2 + 2x – 3 Now factor by grouping

Factoring a trinomial when a ≠ 1
Multiply 8  -3, and break up the middle term Factor 6b2 – 11b – 2 Now factor by grouping

Factoring the Difference of Two Squares
a2– ab + ab – b2 = a2 – b2 (a + b)(a – b) = a2 – b2 = (a + b)(a – b) FORMULA: 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
a2 – b2 = (a + b)(a – b) Factor x2 – 4y2 Factor 16r2 – 25 (4r) 2 (5) 2 (2y) 2 (x) 2 Difference Of two squares Difference of two squares (x – 2y)(x + 2y) (4r – 5)(4r + 5)

Difference of two squares

Difference of two squares

Difference of two squares

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 Write terms as perfect squares. = (x + 3y)(x – 3y)

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