Chapter 4.3, Problem 38AYU

In Problems 31-40, find the complex zeros of each polynomial function. Write $f$ in factored form.

$f\left(x\right)=x^4+3x^3-19x^2+27x-252$

To determine

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

Answer to Problem 38AYU

$\text{The complex zeros }=\pm 3\text{i}$,

$\text{Factored}\text{form}\text{of f}\text{(<mi>x</mi>)} = x^4 + 3x^3 - 19x^2 + 27x - 252 = [x-3i)][x + 3i][x - 4][x + 7]$

Explanation of Solution

Given:

$f\left(x\right) = x^4 + 3x^3 - 19x^2 + 27x - 252$

Degree of the polynomial function $f\left(x\right) = x^4 + 3x^3 - 19x^2 + 27x - 252 = 4$.

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

From Descartes’ rule of signs

$f\left(x\right) = x^4 + 3x^3 - 19x^2 + 27x - 252$

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

Therefore $f\left(x\right) = x^4 + 3x^3 - 19x^2 + 27x - 252$ have 3 or 1 positive real zeros

$f\left(-x\right) = x^4 - 3x^3 - 19x^2 - 27x - 252$

$+\text{ TO }-$

Therefore $f\left(x\right) = x^4 + 3x^3 - 19x^2 + 27x - 252$ have 1 negative real zero

By synthetic division

The real positive zero $=4$

The depressed equation $= x^3 + 7x^2 + 9x + 63$

The real negative zero $=-7$

Depressed equation $= x^2 + 9$,

$x^2 = -9$

The complex zeros $= \pm 3\text{i}$

Factored form of $\text{f}\left(x\right) = x^4 + 3x^3 - 19x^2 + 27x - 252$

$= \left[x - \left( + 3i\right)\right]\left[x - \left(-3i\right)\right]\left[x - 4\right]\left[x - \left(-7\right)\right]$

$= \left[x - 3i)\right]\left[x + 3i\right]\left[x - 4\right]\left[x + 7\right]$