Question

Use the given zero to find the remaining zeros of each function

$h\left(x\right)=3x^5+2x^4+15x^3+10x^2-528x-352;\text{zero:}-4i$

Short answer

$4i,-\frac{2}{3},\sqrt{11}$and role="math" localid="1648655458162" $-\sqrt{11}$ are the remaining zeros of h

Step-by-step solution

Given Information

The given value is:

$h\left(x\right)=3x^5+2x^4+15x^3+10x^2-528x-352$; zero;$-4i$

Apply the Conjugate Pairs Theorem

• As you observed, 5 is the degree of $h,$so we have $5$zeros We have, $-4i,$now by Conjugate Pairs Theorem, $4i$is also a zero of $h$Hence, We now have $2$zeros, so we need to find the remaining zeros
• Notice that, We already have $(x+4i)(x-4i)$as the factor of the zeros of $h$So, By simplifying $(x+4i)(x-4i)$We now have $x^2+16$ as the factor of$h$
  

Perform the long division method

$$\begin{gathered} x^2+163x^3+2x^2-33x-223x^5+2x^4+15x^3+10x^2-528x-3523x^5+0x+48x^3---------2x^4-33x^3+10x^2-528x-352 \\ 2x^4+0x^3+32x^2 \\ ------- \\ -33x^3-22x^2-528x-352 \\ 33x^3+0x^2-528x \\ --------- \\ -22x^2+0x-352 \\ -22x^2+0x-352 \\ -------- \\ 0 \end{gathered}$$

We now have

$$\begin{gathered} h\left(x\right)=3x^3+2x^2-33x-22+\frac{0}{x^2+16} \\ =3x^3+2x^2-33x-22 \end{gathered}$$

Find remaining zeros

Now Let's equate $3x^3+2x^2-33x-22$to zero, in order to find the remaining zeros

factor out $x^3$and $11$

$3x^3+2x^2-33x-22=0$

factor out $(3x+2),$and combine $x^2$and $11$

$x^2(3x+2)-11(3x+2)=0$

$(3x+2)\left(x^2-11\right)=0$

$$\begin{gathered} 3x+2=0 \\ 3x=-2 \\ x=-\frac{2}{3} \end{gathered}$$

$$\begin{gathered} x^2-11=0 \\ \sqrt{x^2}=\sqrt{11} \\ x=\pm \sqrt{11} \end{gathered}$$

Therefore, $-\frac{2}{3}$and $\pm \sqrt{11}$are the zeros of $h$So, $4i,-\frac{2}{3},\sqrt{11}$and $-\sqrt{11}$ are the remaining zeros of$h.$