Question

In Problems 23–30, use the given zero to find the remaining zeros of each function.

$g\left(x\right)=2x^5-3x^4-5x^3-15x^2-207x+108;\mathrm{zero}:3i$

Short answer

The remaining answer is$\pm 3i,-\frac{1}{2},3,4.$

Step-by-step solution

Step 1. Given Information.

The given polynomial is

$g\left(x\right)=2x^5-3x^4-5x^3-15x^2-207x+108$

Step 2. Performing Long Division.

Here we apply the conjugate the Pairs Theorem. The degree of the polynomial is 5.

3i is zero of polynomial then it's conjugate -3i also zero of the polynomial. Now we have two zeros.

Since all the coefficients are $g\left(x\right)$are real and $3i$is a root of $g\left(x\right)$then $-3i$is also a root of $g\left(x\right)$. Then $\left(x+3i\right)\left(x-3i\right)$are factors of $g\left(x\right)$. So by divides $\left(x-3i\right)\left(x+3i\right)=x^2+9$to $g\left(x\right)$.

$\frac{(2x^5-3x^4-5x^3-15x^2-207x+108)}{(x^2+9)}=2x^3-3x^2-23x+12$

Step 3. Factors of the finding polynomial.

Now, find the factors and apply the zero methods.

$2x^3-3x^2-23x+12=2x^2(x-4)-5x(x-4)-3(x-4)$

And,

$\begin{array}{rcl}(x-4)(2x^2-5x-3)& =& (x-4)(2x^2-6x+x-3)\\ & =& (x-4)\left\{2x\right(x-3)+1(x-3\left)\right\}\\ & =& (x-4)(2x+1)(x-3)\end{array}$

The zeros are$\pm 3i,-\frac{1}{2},3,4.$