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).

xandir307dc

Step 1
here given zeros are x=8, x=2, x=-5
so the factored form is
$f\left(x\right)=a\left(x-8\right)\left(x-2\right)\left(x+5\right)$...(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)
$f\left(x\right)=\left({x}^{2}-8x-2x+16\right)\left(x+5\right)$
$f\left(x\right)=\left({x}^{2}-10x+16\right)\left(x+5\right)$
$f\left(x\right)={x}^{3}-10{x}^{2}+16x+5{x}^{2}-50x+80$
$f\left(x\right)={x}^{3}-5{x}^{2}-34x+80$

Jenny Sheppard

Given, Degree: 3
Zeroes, (8,0),(2,0), and (-5,0)
y-intercept at (0,80)
so,
(x-8)(x-2)(x+5)=0
$\left({x}^{2}-8x-2x+16\right)\left(x+5\right)=0$
$\left({x}^{2}-10x+16\right)\left(x+5\right)=0$
${x}^{3}+5{x}^{2}-10{x}^{2}-50x+16x+80=0$
${x}^{3}-5{x}^{2}-34x+80=0$
So, $P\left(x\right)={x}^{3}-5{x}^{2}-34x+80$

Vasquez

Equation of polynomial is factored form is
$f\left(x\right)=a\left(x-8\right)\left(x-2\right)\left(x-\left(-5\right)\right)$
$⇒f\left(x\right)=a\left(x-8\right)\left(x-2\right)\left(x+5\right)$
Given at x=0 f(0)=80
$\begin{array}{}⇒a\left(0-8\right)\left(0-2\right)\left(0+5\right)=80\\ ⇒a\left(-8\right)\left(-2\right)\left(5\right)=80\\ ⇒80a=80\\ ⇒a=1\end{array}$
So, f(x)=(x-8)(x-2)(x+5)

Do you have a similar question?