hadejada7x

2021-12-26

Write a polynomial, P(x), in factored form given the following requirements.

Degree: 3

Zeros at (8,0), (2,0), and (−5,0)

y-intercept at (0,80).

Degree: 3

Zeros at (8,0), (2,0), and (−5,0)

y-intercept at (0,80).

xandir307dc

Beginner2021-12-27Added 35 answers

Step 1

here given zeros are x=8, x=2, x=-5

so the factored form is

$f\left(x\right)=a(x-8)(x-2)(x+5)$ ...(1)

here y-interest is (0,80)

80=a(0-8)(0-2)(0+5)

80=a(-8)(-2)(5)

80=a*80

a=1...put it back in equation 1

Step 2

f(x)=a(x-8)(x-2)(x+5)

f(x)=1(x-8)(x-2)(x+5)

f(x)=(x-8)(x-2)(x+5)

$f\left(x\right)=({x}^{2}-8x-2x+16)(x+5)$

$f\left(x\right)=({x}^{2}-10x+16)(x+5)$

$f\left(x\right)={x}^{3}-10{x}^{2}+16x+5{x}^{2}-50x+80$

$f\left(x\right)={x}^{3}-5{x}^{2}-34x+80$

here given zeros are x=8, x=2, x=-5

so the factored form is

here y-interest is (0,80)

80=a(0-8)(0-2)(0+5)

80=a(-8)(-2)(5)

80=a*80

a=1...put it back in equation 1

Step 2

f(x)=a(x-8)(x-2)(x+5)

f(x)=1(x-8)(x-2)(x+5)

f(x)=(x-8)(x-2)(x+5)

Jenny Sheppard

Beginner2021-12-28Added 35 answers

Given, Degree: 3

Zeroes, (8,0),(2,0), and (-5,0)

y-intercept at (0,80)

so,

(x-8)(x-2)(x+5)=0

$({x}^{2}-8x-2x+16)(x+5)=0$

$({x}^{2}-10x+16)(x+5)=0$

${x}^{3}+5{x}^{2}-10{x}^{2}-50x+16x+80=0$

${x}^{3}-5{x}^{2}-34x+80=0$

So,$P\left(x\right)={x}^{3}-5{x}^{2}-34x+80$

Zeroes, (8,0),(2,0), and (-5,0)

y-intercept at (0,80)

so,

(x-8)(x-2)(x+5)=0

So,

Vasquez

Expert2022-01-09Added 669 answers

Equation of polynomial is factored form is

Given at x=0 f(0)=80

So, f(x)=(x-8)(x-2)(x+5)

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