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3 years ago

What is the product? 2x+8 x2 + 3x-4 x(x-4)(x-1) 20x+4) 2 (x+4)(x-4) 2x(x-1) 2x(x+4)

Mathematics

2 answers:

guajiro [1.7K]3 years ago

7 0

=−640x10+1280x9+19904x8−40728x7−144488x6+323904x5−162304x4+1024x3+2048x2

step by step

(2x+8x2+3x−4x(x−4)(x−1)(20)x+4)(2)(x+4)(x−4)(2)x(x−1)(2)x(x+4)

=((2x+8x2+3x−4x(x−4)(x−1)(20)x+4)(2)(x+4)(x−4)(2)x(x−1)(2)x)(x+4)

=((2x+8x2+3x−4x(x−4)(x−1)(20)x+4)(2)(x+4)(x−4)(2)x(x−1)(2)x)(x)+((2x+8x2+3x−4x(x−4)(x−1)(20)x+4)(2)(x+4)(x−4)(2)x(x−1)(2)x)(4)

=−640x10+3840x9+4544x8−58904x7+91128x6−40608x5+128x4+512x3−2560x9+15360x8+18176x7−235616x6+364512x5−162432x4+512x3+2048x2

=−640x10+1280x9+19904x8−40728x7−144488x6+323904x5−162304x4+1024x3+2048x2

Oduvanchick [21]3 years ago

4 0

The product is the same as the answer

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It should be noted that integers are the whole numbers that can be either positive, negative, or zero.

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Step-by-step explanation:

see the attached figure to better understand the problem

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substitute the given values

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Answers:

15) x = 8, $\frac{1}{3}, - \frac{2}{5}$

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15x³ - 119x² - 10x + 16 = 0

$\frac{p}{q}: \frac{16}{15}:+/- \frac{1*2*4*8*16}{1*3*5*15}$

So, the possible rational roots are: +/- $1, 2, 4, 8, 16,\frac{1}{3},\frac{2}{3},\frac{4}{3},\frac{8}{3},\frac{16}{3},\frac{1}{5},\frac{2}{5},\frac{4}{5},\frac{8}{5},\frac{16}{5},\frac{1}{15},\frac{2}{15},\frac{4}{15},\frac{8}{15},\frac{16}{15}$

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Now, solve for x.

x = 8, x = $\frac{1}{3}$ , x = $-\frac{2}{5}$