Chapter 4.3, Problem 37AYU

To determine

To find: The Complex zeros of the polynomial function $f\left(x\right)$ whose coefficients are real numbers and to write in factored form.

Answer to Problem 37AYU

$\text{The complex zeros }=\pm 5i$.

$\text{Factored form of }f\left(x\right)=x^4+2x^3+22x^2+50x-75$.

$=\left[x-5i)\right]\left[x+5i\right]\left[x-1\right]\left[x+3\right]$

Explanation of Solution

Given:

$f\left(x\right)=x^4+2x^3+22x^2+50x-75$

Degree of the polynomial function $f\left(x\right)=x^4+2x^3+22x^2+50x-75=4$ .

Therefore $f\left(x\right)$ have 4 zeros.

From Descartes’ rule of signs.

$f\left(x\right)=x^4+2x^3+22x^2+50x-75$

$+\text{ To }-$

Therefore $f\left(x\right)=x^4+2x^3+22x^2+50x-75$ has one positive real zero.

$f\left(-x\right)=x^4-2x^3+22x^2-50x-75$

$+\text{To}-\text{ - to }+\text{ + to }-$

Therefore $f\left(x\right)=x^4+2x^3+22x^2+50x-75$ have 3 or 1 negative real zeros.

The real negative zero $=-3$.

The depressed equation $=x^3-x^2+25x-25$.

The real positive zero $=1$.

Depressed equation $=x^2+25$.

$x^2=-25$

The complex zeros $=\pm 5i$.

Factored form of $f\left(x\right)=x^4+2x^3+22x^2+50x-75$.

$=\left[x-\left(+5i\right)\right]\left[x-\left(-5i\right)\right]\left[x-1\right]\left[x-\left(-3\right)\right]=\left[x-5i)\right]\left[x+5i\right]\left[x-1\right]\left[x+3\right]$