2.8 Warm Up Factor. 1.8x² + 26x + 15 2.2x² + 15x + 7 3.5x² - 18x - 8.
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Transcript of 2.8 Warm Up Factor. 1.8x² + 26x + 15 2.2x² + 15x + 7 3.5x² - 18x - 8.
![Page 1: 2.8 Warm Up Factor. 1.8x² + 26x + 15 2.2x² + 15x + 7 3.5x² - 18x - 8.](https://reader035.fdocuments.in/reader035/viewer/2022062518/56649f055503460f94c1a081/html5/thumbnails/1.jpg)
2.8 Warm Up2.8 Warm UpFactor.1. 8x² + 26x + 15
2. 2x² + 15x + 7
3. 5x² - 18x - 8
![Page 2: 2.8 Warm Up Factor. 1.8x² + 26x + 15 2.2x² + 15x + 7 3.5x² - 18x - 8.](https://reader035.fdocuments.in/reader035/viewer/2022062518/56649f055503460f94c1a081/html5/thumbnails/2.jpg)
2.8 Factor 2.8 Factor Special Special
ProductsProducts
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EXAMPLE 1 Factor the difference of two squares
Factor the polynomial.
a. y2 – 16 =
b. 25m2 – 36
c. x2 – 49y2
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EXAMPLE 2 Factor the difference of two squares
Factor the polynomial 8 – 18n2.
8 – 18n2 = 2(4 – 9n2)
= 2(2 + 3n)(2 – 3n)
Factor out common factor.
Difference of two squares pattern
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GUIDED PRACTICE for Examples 1 and 2
Factor the polynomial.
1. 4y2 – 64 = (2y + 8)(2y – 8)
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EXAMPLE 3 Factor perfect square trinomials
Factor the polynomial.
a.
n2 – 12n + 36
b. 9x2 – 12x + 4
c. 4s2 + 4st + t2
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EXAMPLE 4 Factor a perfect square trinomial
Factor the polynomial –3y2 + 36y – 108.
–3y2 + 36y – 108 Factor out –3.
= –3(y – 6)2 Perfect square trinomial pattern
= –3(y2 – 12y + 36)
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GUIDED PRACTICE for Examples 3 and 4
Factor the polynomial.
= (h + 2)22. h2 + 4h + 4
3. 2y2 – 20y + 50 = 2(y – 5)2
4. 3x2 + 6xy + 3y2= 3(x + y)2
![Page 9: 2.8 Warm Up Factor. 1.8x² + 26x + 15 2.2x² + 15x + 7 3.5x² - 18x - 8.](https://reader035.fdocuments.in/reader035/viewer/2022062518/56649f055503460f94c1a081/html5/thumbnails/9.jpg)
EXAMPLE 5 Solve a polynomial equation
Solve the equation x2 + + = 0.23
x 19
Write original equation.
9x2 + 6x + 1 = 0 Multiply each side by 9.
(3x + 1)2 = 0
x = – 13
Solve for x.
x2 + + = 019
23
x
ANSWER
The solution of the equation is – .13
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GUIDED PRACTICE for Examples 5 and 6
Solve the equation
5. a2 + 6a + 9 = 0 a = –3
6. w2 – 14w + 49 = 0 w = 7
7. n2 – 81= 0 n = – 9 or n = 9
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FALLING OBJECT
SOLUTION
EXAMPLE 6 Solve a vertical motion problem
A window washer drops a wet sponge from a height of 64 feet. After how many seconds does the sponge land on the ground?
Use the vertical motion model to write an equation for the height h (in feet) of the sponge as a function of the time t (in seconds) after it is dropped.
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EXAMPLE 6 Solve a vertical motion problem
The sponge was dropped, so it has no initial vertical velocity. Find the value of t for which the height is 0.
h = –16t2 + vt + s Vertical motion model
0 = –16t2 + (0)t + 64 Substitute 0 for h, 0 for v, and 64 for s.
0 = –16(t2 – 4) Factor out –16.
0 = –16(t – 2)(t +2) Difference of two squares pattern
t – 2 = 0 or t + 2 = 0 Zero-product property
t = 2 or t = –2 Solve for t.
Disregard the negative solution of the equation.
ANSWERThe sponge lands on the ground 2 seconds after it is dropped.