Nann

2021-02-25

Find a polynomial of the specified degree that has the given zeros. Degree 4, zeros -2, 0, 2, 4

### Answer & Explanation

comentezq

Recall, according to the factor theorem the expression $\left(x–c\right)$ is a factor of a polynomial P(x) if and only if P(c) = 0.
1) Let a polynomial function P(x):
If c = -2 is a zero of P(x) the expression $\left(x+2\right)$ is a factor of P(x).
If c = 0 is a zero of P(x) the expression $\left(x–0\right)=$ is a factor of P(x).
If c = 2 is a zero of P(x) the expression $\left(x–2\right)=$ is a factor of P(x).
If c = 4 is a zero of P(x) the expression $\left(x–4\right)=$ is a factor of P(x).
2) The polynomial P(x) can be written in factored form as:
P(x) $=x\left(x–4\right)\left(x+2\right)\left(x–2\right)$
P(x) $=\left({x}^{2}–4x\right)\left(x+2\right)\left(x–2\right)$
3) Apply the difference of squares identity: ${a}^{2}–{b}^{2}=\left(a+b\right)\left(a–b\right)$
P(x) $=\left({x}^{2}–4x\right)\left({x}^{2}–4\right)$
P(x) $={x}^{4}–4{x}^{3}–4{x}^{2}+16x$

Eliza Beth13

Result:
$f\left(x\right)={x}^{4}-4{x}^{3}-4{x}^{2}+16x$ or $f\left(x\right)=\left(x+2\right)\left(x\right)\left(x-2\right)\left(x-4\right)$
Solution:
Given the zeros -2, 0, 2, and 4, we can write the factors as follows:
$\left(x-\left(-2\right)\right)\left(x-0\right)\left(x-2\right)\left(x-4\right)$
Simplifying each factor, we have:
$\left(x+2\right)\left(x\right)\left(x-2\right)\left(x-4\right)$
Now, let's expand this expression:
$\left(x+2\right)\left(x\right)\left(x-2\right)\left(x-4\right)=\left({x}^{2}+2x\right)\left(x-2\right)\left(x-4\right)$
Expanding further:
$\left({x}^{2}+2x\right)\left(x-2\right)\left(x-4\right)=\left({x}^{2}+2x\right)\left({x}^{2}-2x-4x+8\right)$
Combining like terms:
$\left({x}^{2}+2x\right)\left({x}^{2}-2x-4x+8\right)=\left({x}^{2}+2x\right)\left({x}^{2}-6x+8\right)$
Now, we can expand once again:
$\left({x}^{2}+2x\right)\left({x}^{2}-6x+8\right)={x}^{2}\left({x}^{2}-6x+8\right)+2x\left({x}^{2}-6x+8\right)$
Expanding further:
${x}^{2}\left({x}^{2}-6x+8\right)+2x\left({x}^{2}-6x+8\right)={x}^{4}-6{x}^{3}+8{x}^{2}+2{x}^{3}-12{x}^{2}+16x$
Combining like terms:
${x}^{4}-6{x}^{3}+8{x}^{2}+2{x}^{3}-12{x}^{2}+16x={x}^{4}-4{x}^{3}-4{x}^{2}+16x$
Thus, the polynomial of degree 4 with zeros at -2, 0, 2, and 4 is:
$f\left(x\right)={x}^{4}-4{x}^{3}-4{x}^{2}+16x$
You can also express it in factored form as:
$f\left(x\right)=\left(x+2\right)\left(x\right)\left(x-2\right)\left(x-4\right)$

Nick Camelot

To find a polynomial of degree 4 with zeros -2, 0, 2, and 4, we can use the factored form of a polynomial. The factored form is obtained by multiplying the factors (x - zero) for each zero of the polynomial.
Therefore, the polynomial is given by:
$P\left(x\right)=\left(x+2\right)\left(x-0\right)\left(x-2\right)\left(x-4\right)$

Step 1: Let's denote the polynomial as $P\left(x\right)$, and we know that it has four zeros. Therefore, we can write the polynomial as:
$P\left(x\right)=a\left(x-{r}_{1}\right)\left(x-{r}_{2}\right)\left(x-{r}_{3}\right)\left(x-{r}_{4}\right)$ where $a$ is a constant and ${r}_{1},{r}_{2},{r}_{3},{r}_{4}$ are the zeros $-2,0,2,4$, respectively.
Substituting these zeros into the equation, we get:
$P\left(x\right)=a\left(x-\left(-2\right)\right)\left(x-0\right)\left(x-2\right)\left(x-4\right)$
Simplifying further:
$P\left(x\right)=a\left(x+2\right)\left(x\right)\left(x-2\right)\left(x-4\right)$
Step 2: Expanding the equation:
$P\left(x\right)=a\left({x}^{2}+2x\right)\left({x}^{2}-2x-8\right)$
Multiplying the terms:
$P\left(x\right)=a\left({x}^{4}-4{x}^{3}-8{x}^{2}+16x\right)$
Therefore, a polynomial of degree 4 with the given zeros $-2,0,2,4$ can be expressed as:
$P\left(x\right)=a\left({x}^{4}-4{x}^{3}-8{x}^{2}+16x\right)$

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