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Warm Up
Warm Up
Lesson Presentation
Lesson Presentation
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1. Simplify the following expressions a. -2x(4x – 3) + 8x – (10 – 12x2)
b. 6x (x + 2) – (x+3) – 5(x2 + 6x)
c. 5(2x2 + 4) – (x2 – 6x + 4)
2. Set up the following systems of equations. Do not Solve
12 Boys and Girls shirts are sold. Boys shirts cost $11 and Girls shirts cost $7. How many of each type of shirt was sold if the total money collected was $116.
3. The base of a triangle is (x + 2) inches long. The height of the triangle is
(2x -1) inches high. Solve for the area.
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Factor quadratic trinomials of the form x2 + bx + c and ax2 + bx + c.
Factoring special products
Objective
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A polynomial is in its factored form when it is written as a product of monomials and polynomials that cannot be
factored further. The polynomial 2(3x – 4x) is not fully factored because the terms in the parentheses have a common factor of x.
Distributive Property a(b + c) = ab + ac.
What operation is used during distribution?
Factoring is opposite of the distributive property ab + ac = a(b + c)
What operation do you think is used here and why?
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Factorizations of 12
The order of factors does not change the product, but there is only one example below that cannot be factored further. The circled factorization is the prime factorization because all the factors are prime numbers. The prime factors can be
written in any order, and except for changes in the order, there is only one way to write the prime factorization of a number.
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Teacher Example: Finding the GCF of Monomials
Find the GCF of each pair of monomials. 15x3 and 9x2
15x3 = 3 5 x x x
9x2 = 3 3 x x
3 x x = 3x2
Write the prime factorization of each coefficient and write powers as products.
Align the common factors.
Find the product of the common factors.
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Student Example 1
Find the GCF of each pair of monomials.
18g2 and 27g3
18g2 = 2 3 3 g g
27g3 = 3 3 3 g g g
3 3 g g
The GCF of 18g2 and 27g3 is 9g2.
Write the prime factorization of each coefficient and write powers as products. Align the common factors.
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Student Example 2
Find the GCF of each pair of monomials. 8x2 and 7y3
8x2 = 2 2 2 x x
7y3 = 7 y y y
Write the prime
factorization of each coefficient and write powers as products. Align the common
factors.
There are no
common factors other than 1.
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Teacher Example: Factoring Polynomials GCF
Factor each polynomial. Check your answer. 2x2 – 4
2x2 = 2 x x
4 = 2 2
2
Find the GCF.
The GCF of 2x2 and 4 is 2.
Write terms as products using the GCF as a factor.
2x2 – (2 2)
2(x2 – 2)
Check 2(x2 – 2)
2x2 – 4
Multiply to check your answer. The product is the original
polynomial.
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Student Example
Factor each polynomial. Check your answer. 5b + 9b3
5b = 5 b
9b = 3 3 b b b
b 5(b) + 9b2(b)
b(5 + 9b2)
b(5 + 9b2)
Check 5b + 9b3
Find the GCF.
The GCF of 5b and 9b3 is b.
Multiply to check your answer. The product is the original
polynomial.
Write terms as products using the GCF as a factor.
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Student Example
Factor each polynomial. Check your answer.
–18y3 – 7y2
– 1(18y3 + 7y2) Both coefficients are negative.
Factor out –1. Find the GCF.
The GCF of 18y3 and 7y2 is y2.
18y3 = 2 3 3 y y y
7y2 = 7 y y
y y = y2
Write each term as a product using the GCF.
Use the Distributive Property to factor out the GCF..
–1[18y(y2) + 7(y2)]
–1[y2(18y + 7)]
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Student Example
Factor each polynomial. Check your answer. 8x3 – 4x2 – 16x
2x2(4x) – x(4x) – 4(4x)
4x(2x2 – x – 4)
4x(2x2 – x – 4)
8x3 – 4x2 – 16x
8x3 = 2 2 2 x x x
4x2 = 2 2 x x
16x = 2 2 2 2 x
2 2 x = 4x
Find the GCF.
The GCF of 8x3, 4x2, and 16x is
4x.
Write terms as products using the GCF as a factor.
Use the Distributive Property to factor out the GCF.
Multiply to check your answer. The product is the original
polynomials.
Check
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Student Example
Factor each polynomial. Check your answer.
3x3 + 2x2 – 10
10 = 2 5
Find the GCF.
There are no common factors other than 1.
The polynomial cannot be factored further. 3x3 + 2x2 – 10
3x3 = 3 x x x
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When factoring trinomials you first divide out the GCF and then factor the
trinomial into two binomials.
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Teacher Example: Factoring Trinomials by Guess and Check
Factor x2 + 15x + 36 by guess and check.
( + )( + ) (x + )(x + )
Write two sets of parentheses.
The first term is x2, so the variable
terms have a coefficient of 1.
The constant term in the trinomial is 36. (x + 1)(x + 36) = x2 + 37x + 36
(x + 2)(x + 18) = x2 + 20x + 36
(x + 3)(x + 12) = x2 + 15x + 36
Try factors of 36 for the constant
terms in the binomials.
The factors of x2 + 15x + 36 are (x + 3)(x + 12).
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Student Example
Factor each trinomial
x2 + 10x + 24 ( + )( + )
(x + )(x + )
The factors of x2 + 10x + 24 are (x + 4)(x + 6).
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Student Example
Factor each trinomial
x2 + 7x + 12
( + )( + ) (x + )(x + )
Write two sets of Parentheses.
The first term is x2, so the variable
terms have a coefficient of 1.
x2 + 10x + 24 = (x + 4)(x + 6)
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Teacher Example 2: Negative B Value
Factor each trinomial. Check your answer.
x2 – 8x + 15
b = –8 and c = 15; look for factors of 15 whose sum is –8.
The factors needed are –3 and –5 .
Factors of 15 Sum –1 and –15 –16
–3 and –5 –8
(x – 3)(x – 5)
Check (x – 3)(x – 5 ) = x2 – 3x – 5x + 15 Use the FOIL method.
The product is the original polynomial.
= x2 – 8x + 15
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Factor each trinomial. Check your answer.
x2 – 5x + 6
(x + )(x+ ) b = –5 and c = 6; look for factors of 6 whose sum is –5.
The factors needed are –2 and –3.
Factors of 6 Sum –1 and –6 –7
–2 and –3 –5
(x – 2)(x – 3)
Check (x – 2)(x – 3) = x2 –2x – 3x + 6 Use the FOIL method.
The product is the original polynomial.
= x2 – 5x + 6
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Teacher Example: Negative C Value
Factor each trinomial.
x2 + x – 20
(x + )(x + ) b = 1 and c = –20; look for
factors of –20 whose sum is 1. The factor with the greater absolute value is positive.
The factors needed are +5 and –4.
Factors of –20 Sum
–1 and 20 19 –2 and 10 8
–4 and 5 1
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Factor each trinomial. Check your answer.
Student Example: Negative C Value
x2 + 2x – 15 (x + )(x + )
Factors of –15 Sum
–1 and 15 14
–3 and 5 2
(x – 3)(x + 5)
b = 2 and c = –15; look for
factors of –15 whose sum is 2. The factor with the greater
absolute value is positive.
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Factor each trinomial.
x2 – 3x – 18
b = –3 and c = –18; look for factors of –18 whose sum is –3. The factor with the
greater absolute value is negative.
Factors of –18 Sum
1 and –18 –17
2 and – 9 – 7
3 and – 6 – 3
The factors needed are 3 and –6.
(x – 6)(x + 3)
Teacher Example: B and C Value are Negative
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X2 – 8x – 20
Factor each trinomial. Check your answer.
Student Example: B and C Value are negative
(x – 10)(x + 2)
Factors of –20 Sum
1 and –20 –19
2 and –10 –8
b = –8 and c = –20; look for factors of –20 whose sum is –8. The factor with the
greater absolute value is negative.
The factors needed are –10 and 2.
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(
X
+ )(
x
+ ) =
ax
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To factor a2 + bx + c, check the factors of a and
the factors of c in the binomials. The sum of the
products of the outer and inner terms should be
b.
Sum of outer and inner products = b
Product = c
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Since you need to check all the factors of a
and the factors of c, it may be helpful to make
a table. Then check the products of the outer
and inner terms to see if the sum is b. You can
multiply the binomials to check your answer.
(
X
+ )(
x
+ ) =
ax
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c
Sum of outer and inner products = b
Product = c
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Factoring Polynomials when the Coefficient of a is > 1
Factor each trinomial. Check your answer. 2x2 + 17x + 21
( x + )( x + ) a = 2 and c = 21, Outer + Inner = 17.
(x + 7)(2x + 3)
Factors of 2 Factors of 21 Outer + Inner
1 and 2 1 and 21 1(21) + 2(1) = 23
1 and 2 21 and 1 1(1) + 2(21) = 43
1 and 2 3 and 7 1(7) + 2(3) = 13
1 and 2 7 and 3 1(3) + 2(7) = 17
Check (x + 7)(2x + 3) = 2x2 + 3x + 14x + 21
= 2x2 + 17x + 21
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Student Examples
2x2 + 9x – 18
–6x2 – 17x – 12
6x2 + 7x – 3
–2x2 – 5x – 3.
Teacher Examples
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1.2m2 - 24m + 40
2. 2x2 - 3x – 14
WITH A PARTNER
3. 4y2 - 18y + 14
4. 3x2 - 13x + 12
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Exit Ticket 1
Factor each trinomial.
1. x2 – 11x + 30
2. x2 + 10x + 9
3. 5x2 + 17x + 6
4. 4x2 + 6x – 12
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Do Now
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1. Determine if there is a Greatest Common Factor (GCF) of a polynomial. If there is a GCF divide it out.
2. If all terms are negative factor out a -1.
3. Factor a trinomial of the form x2 + bx + c (factor*product*sum)
4. Factor a trinomial of the form ax2 + bx + c (swing-divide*factor*product*sum)
5. If it cannot be factored use quadratic formula.
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6x2 + 10x + 4
Student Examples
2x2 + 9x – 18
–6x2 – 17x – 12
4x2 + 10x + 6
6x2 + 7x – 3
–2x2 – 5x – 3.
Teacher Examples
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Do now
Factor each trinomial.
1. x2 – 11x + 30
2. x2 + 10x + 9
3. 5x2 + 17x + 6
4. Find the dimensions of a rectangle with an area = 3x 2 + 13x + 10
(x-5)(x-6)
(x+1)(x+9) (5x+2)(x+3)
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A trinomial is a perfect square if:
• The first and last terms are perfect squares.
• The middle term is two times one factor from the first term and one factor from the last term.
9x2 + 12x + 4
3x • 3x 2(3x • 2) 2 • 2
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Teacher Example: Recognizing and Factoring Perfect-Square Trinomials
Determine whether each trinomial is a perfect square. If so, factor. If not explain.
81x2 + 90x + 25
81x2 + 90x + 25
The trinomial is a perfect square. Factor.
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Teacher Example Continued
Determine whether each trinomial is a perfect square. If so, factor. If not explain.
Method 2 Use the rule.
81x2 + 90x + 25 a = 9x, b = 5
(9x)2 + 2(9x)(5) + 52
(9x + 5)2
Write the trinomial as a2 + 2ab + b2.
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Student Example 1
Determine whether each trinomial is a perfect square. If so, factor. If not explain.
x2 + 4x + 4
The trinomial is a perfect square. Factor.
x x 2(x 2) 2 2
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Student Example 2
Determine whether each trinomial is a perfect square. If so, factor. If not explain.
x2 – 14x + 49
The trinomial is a perfect square. Factor.
x2 – 14x + 49
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Determine whether each trinomial is a perfect square. If so, factor. If not explain.
Student Example 2 Continued
Method 2 Use the rule.
a = 1, b = 7
(x)2 – 2(x)(7) + 72
(x – 7)2
Write the trinomial as a2 – 2ab + b2.
Write the trinomial as (a – b)2.
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Recognize a difference of two squares: the
coefficients of variable terms are perfect squares, powers on variable terms are even, and constants are perfect squares.
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Teacher Example: Recognizing and Factoring the Difference of Two Squares
Determine whether each binomial is a difference of two squares. If so, factor. If not, explain.
100x2 – 4y2
Write the polynomial as (a + b)(a – b).
a = 10x, b = 2y
The polynomial is a difference of two squares.
100x2 – 4y2
2y 2y
10x 10 x
(10x)2 – (2y)2
(10x + 2y)(10x – 2y)
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Teacher Example 2: Recognizing and Factoring the Difference of Two Squares
Determine whether each binomial is a difference of two squares. If so, factor. If not, explain.
x4 – 25y6
Write the polynomial as (a + b)(a – b).
a = x2, b = 5y3
The polynomial is a difference of two squares.
(x2)2 – (5y3)2
(x2 + 5y3)(x2 – 5y3)
x4 – 25y6 = (x2 + 5y3)(x2 – 5y3)
5y3 5y 3
x2 x2
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Student Example 1
Determine whether each binomial is a difference of two squares. If so, factor. If not, explain.
1 – 4x2
Write the polynomial as (a + b)(a – b).
a = 1, b = 2x
The polynomial is a difference of two squares.
(1) – (2x)2
(1 + 2x)(1 – 2x)
1 – 4x2 = (1 + 2x)(1 – 2x)
2x 2x
1 1
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Student Example 2
Determine whether each binomial is a difference of two squares. If so, factor. If not, explain.
p8 – 49q6
Write the polynomial as (a + b)(a – b).
a = p4, b = 7q3
The polynomial is a difference of two squares.
(p4)2 – (7q3)2
(p4 + 7q3)(p4 – 7q3)
p8 – 49q6 = (p4 + 7q3)(p4 – 7q3)
7q3 7q● 3
p4 ● p4
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1. 121x2 – 44x + 4
2. 49x2 + 140x + 100
(11x – 2)2
(7x + 10)2
3. 9x2 – 144y4 (3x + 12y2)(3x – 12y2)
4. 121x2 – 4y8
Exit Ticket 2