Chapter 4.2, Problem 50AYU

To determine

To find: The real zeros of $f$ and to use the real zeros to factor $f$.

Answer to Problem 50AYU

$54x^4-57x^3-323x^2+20x+12=\left(x-3\right)\left(x+2\right)\left(9x-2\right)\left(6x+1\right)$

$\text{The real zeros }=3,-2,\frac{2}{9},-\frac{1}{6}$.

Explanation of Solution

Given:

$f\left(x\right)=54x^4-57x^3-323x^2+20x+12$

$F$ is a polynomial function.

The degree of the polynomial function is 4. Therefore the number of real zeros by real zero theorem can be at most $=4$.

From Descartes’ rule of signs

$f\left(x\right)=54x^4-57x^3-323x^2+20x+12$

$+$ To $-$ $-$ to $+$

There will be 2 or 0 positive real zeros.

$f\left(-x\right)=54x^4+57x^3-323x^2-20x+12$

$+$ To $-$ $-$ to $+$

There will be 2 or 0 negative real zeros.

Rational zeros theorem provides information about the potential rational zeros of a polynomial function with integer coefficients.

If $\frac{p}{q}$ in its lowest terms is a rational zero of $f$, then $p$ is a factor of $a_0$ and $q$ is the factor of $a_{\mathrm{n}}$.

$f\left(x\right)=54x^4-57x^3-323x^2+20x+12$

Here $a_0=12$ and $a_{\mathrm{n}}=54$

Zeros of 54, $q=\pm 1,\pm 2,\pm 3,\pm 6,\pm 9,\pm 18,\pm 27,\pm 54$

Zeros of 12, $p=\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12$

The potential rational zeros of $f= \pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 9,\pm 12,\pm 18,\pm 27,\pm 54,\pm \frac{1}{2},\pm \frac{1}{3},\pm \frac{1}{6},\pm \frac{1}{9},\pm \frac{2}{3},\pm \frac{2}{9},\pm \frac{2}{27}\dots \dots$.

From the graph of $f$, we have 3 real zeros near $-2$, 0 and 3.

$x=3$ is zero of $f$.

Use long division or Synthetic division of $54x^4-57x^3-323x^2+20x+12$.

$f\left(x\right)=\left(x-3\right)\left(54x^3+105x^2-8x-4\right)$

The depressed equation $=\left(54x^3+105x^2-8x-4\right)$.

From factorizing further

$\left(54x^3+105x^2-8x-4\right)=\left(x+2\right)\left(54x^2-3x-2\right)=\left(x+2\right)\left(9x-2\right)\left(6x+1\right)$

$54x^4-57x^3-323x^2+20x+12=\left(x-3\right)\left(x+2\right)\left(9x-2\right)\left(6x+1\right)$

The real zeros $=3,-2,\frac{2}{9},-\frac{1}{6}$.