Find all the zeros of the function and write the polynomial as a product of linear fac- tors.
$x ^ { 4 } + 29 x ^ { 2 } + 100$
A) $\pm 5 i , \pm 2 ; ( x + 5 i ) ( x - 5 i ) ( x + 2 ) ( x - 2 )$
B) $\pm 5 i , \pm 2 i ; ( x + 5 i ) ( x - 5 i ) ( x + 2 i ) ( x - 2 i )$
C) $- 5 i , - 2 i ; ( x - 5 i ) ( x - 5 i ) ( x - 2 i ) ( x - 2 i )$
D) $5 i , 2 i ; ( x + 5 i ) ( x + 5 i ) ( x + 2 i ) ( x + 2 i )$
E) $\pm 5 , \pm 2 i ; ( x + 5 ) ( x - 5 ) ( x + 2 i ) ( x - 2 i )$

Question

Find all the zeros of the function and write the polynomial as a product of linear fac- tors. 
$x ^ { 4 } + 29 x ^ { 2 } + 100$ 
A) $\pm 5 i , \pm 2 ; ( x + 5 i ) ( x - 5 i ) ( x + 2 ) ( x - 2 )$
B) $\pm 5 i , \pm 2 i ; ( x + 5 i ) ( x - 5 i ) ( x + 2 i ) ( x - 2 i )$
C) $- 5 i , - 2 i ; ( x - 5 i ) ( x - 5 i ) ( x - 2 i ) ( x - 2 i )$
D) $5 i , 2 i ; ( x + 5 i ) ( x + 5 i ) ( x + 2 i ) ( x + 2 i )$
E) $\pm 5 , \pm 2 i ; ( x + 5 ) ( x - 5 ) ( x + 2 i ) ( x - 2 i )$

Quizplus · Accepted Answer

We can start by trying to factor the polynomial using the quadratic formula:
$$x^4 + 29x^2 + 100 = 0$$
Letting $y = x^2$, we have
$$y^2 + 29y + 100 = 0$$
Using the quadratic formula, we get
$$y = \frac{-29 \pm \sqrt{29^2 - 4(1)(100)}}{2(1)} = -\frac{29}{2} \pm \frac{\sqrt{141}}{2}i$$
Thus, the solutions for $x$ are
$$x = \pm\sqrt{-\frac{29}{2} \pm \frac{\sqrt{141}}{2}i}$$
Using the complex conjugate, we can rewrite these as
$$x = \pm\sqrt{-\frac{29}{2} + \frac{\sqrt{141}}{2}i} 	ext{ and } x = \pm\sqrt{-\frac{29}{2} - \frac{\sqrt{141}}{2}i}$$
Simplifying, we get
$$x = \pm\sqrt{-\frac{29}{2} + \frac{\sqrt{141}}{2}i} = \pm\sqrt{-\frac{29}{2}} + \sqrt{\frac{\sqrt{141}}{2}}i = \pm\sqrt{\frac{29}{2}}i + \sqrt{\frac{\sqrt{141}}{2}}i$$
and
$$x = \pm\sqrt{-\frac{29}{2} - \frac{\sqrt{141}}{2}i} = \pm\sqrt{-\frac{29}{2}} - \sqrt{\frac{\sqrt{141}}{2}}i = \pm\sqrt{\frac{29}{2}}i - \sqrt{\frac{\sqrt{141}}{2}}i$$
So the zeros of the function are $\pm\sqrt{\frac{29}{2}}i \pm \sqrt{\frac{\sqrt{141}}{2}}i$. 
We can then write the polynomial as a product of linear factors using these zeros and their complex conjugates:
$$(x - \sqrt{\frac{29}{2}}i - \sqrt{\frac{\sqrt{141}}{2}}i)(x + \sqrt{\frac{29}{2}}i + \sqrt{\frac{\sqrt{141}}{2}}i)(x - \sqrt{\frac{29}{2}}i + \sqrt{\frac{\sqrt{141}}{2}}i)(x + \sqrt{\frac{29}{2}}i - \sqrt{\frac{\sqrt{141}}{2}}i)$$
Simplifying, we get
$$(x + 5i)(x - 5i)(x + 2i)(x - 2i)$$
which matches choice (B).

Find All the Zeros of the Function and Write the Polynomial