Vikolers6

2021-12-26

Form a polynomial whose zeros and degree are given.
​Zeros:
−2​,
2​,
8​;
​ degree: 3
Type a polynomial with integer coefficients and a leading coefficient of 1 in the box below.

### Answer & Explanation

Jenny Sheppard

Step 1
We have to find the polynomial whose zeros and degree are as follows -
Zeros = -2, 2, 8
Degree = 3
And leading coefficient is 1.
The general form of a polynomial function is as follows-
$f\left(x\right)=a\left(x-{c}_{1}\right)\left(x-{c}_{2}\right)\left(x-{c}_{3}\right)\dots \left(x-{c}_{n}\right)$
Step 2
Given that the zeros are -2, 2, 8 therefore the factors of the required polynomial are - (x+2), (x-2) and (x-8).
Since the degree is 3 and the leading coefficient is 1, therefore, the required polynomial is written as -
f(x)=1(x+2)(x-2)(x-8)
$f\left(x\right)=\left({x}^{2}-4\right)\left(x-8\right)$
$f\left(x\right)={x}^{3}-8{x}^{2}-4x+32$
This is the required polynomial.

vicki331g8

Let f(x)=x(x+2)(x-2)(x-8)
$\therefore f\left(x\right)=\left(x+2\right)\left(x-2\right)\left(x-8\right)$
$=\left({x}^{2}-4\right)\left(x-8\right)$
$={x}^{3}-8{x}^{2}-4x+32$

Vasquez

Given, a polynomial f(x) with zeros -2,2,8.
$\begin{array}{}⇒f\left(x\right)=\left(x-\left(-2\right)\right)\left(x-2\right)\left(x-8\right)\\ =\left(x+2\right)\left(x-2\right)\left(x-8\right)\\ =\left({x}^{2}-4\right)\left(x-8\right)\\ ={x}^{2}\left(x-8\right)-4\left(x-8\right)\\ f\left(x\right)={x}^{3}-8{x}^{2}-4x+32\end{array}$
This is the required polynomial.

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