Chapter 4.2, Problem 49AYU

To determine

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

Answer to Problem 49AYU

$21x^4+22x^3-99x^2-72x+28=7\left(x+1\right)\left(x-2\right)\left(7x+2\right)\left(3x+7\right)$

$\text{The real zeros }=-1,+2,-\frac{2}{7},-\frac{7}{3}$.

Explanation of Solution

Given:

$f\left(x\right)=21x^4+22x^3-99x^2-72x+28$

$F$ 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

$f\left(x\right)=21x^4+22x^3-99x^2-72x+28$

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

There will be 2 or 0 positive real zeros.

$f\left(-x\right)=21x^4-22x^3-99x^2+72x+28$

$+$ 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 $a_{\mathrm{n}}$.

$f\left(x\right)=21x^4+22x^3-99x^2-72x+28$

Here $a_0=28$ and $a_{\mathrm{n}}=21$

Zeros of 21, $q=\pm 1,\pm 3,\pm 7,\pm 21$

Zeros of 28, $p=\pm 1,\pm 2,\pm 4,\pm 7,\pm 14,\pm 28$

The potential rational zeros of $f= \pm 1,\pm 2,\pm 3,\pm 4,\pm 7,\pm 14,\pm 21,\pm \frac{1}{3},\pm \frac{1}{7},\pm \frac{1}{21},\pm \frac{2}{3},\pm \frac{2}{7},\pm \frac{2}{21}\dots \dots$.

From the graph of $f$, we have 3 real zeros near $-2$, $-1$, 0 and 2.

$x=-1$ is zero of $f$.

Use long division or Synthetic division of $21x^4+22x^3-99x^2-72x+28$.

$f\left(x\right)=\left(x+1\right)\left(21x^3+x^2-100x+28\right)$

The depressed equation $=\left(21x^3+x^2-100x+28\right)$.

From factorizing further

$\left(21x^3+x^2-100x+28\right)=\left(x-2\right)\left(21x^2+43x-14\right)=\left(x-2\right)\left(7x+2\right)\left(21x+49\right)$

$21x^4+22x^3-99x^2-72x+28=7\left(x+1\right)\left(x-2\right)\left(7x+2\right)\left(3x+7\right)$

The real zeros $=-1,+2,-\frac{2}{7},-\frac{7}{3}$.