Chapter 4.2, Problem 55AYU

To determine

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

Answer to Problem 55AYU

$x=2, \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\text{are real zeros}$.

$\text{The factored form }=\left(x-2\right)\left(\sqrt{2}x-1\right)\left(\sqrt{2}x+1\right)\left(2x^2+1\right)$.

Explanation of Solution

Given:

$f\left(x\right)=4x^5-8x^4-x+2$

$F$ is a polynomial function.

The degree of the polynomial function is 5. Therefore the number of real zeros by real zero theorem can be at most $=5$.

From Descartes’ rule of signs

$f\left(x\right)=4x^5-8x^4-x+2$

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

There will be 2 or 0 positive real zeros.

$f\left(-x\right)=-4x^5-8x^4+x+2$

$-$ To $+$

There will be 1 negative real zero.

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 $a_{\mathrm{n}}$.

$f\left(x\right)=4x^5-8x^4-x+2$

Here $a_0=2$ and $a_{\mathrm{n}}=4$

Zeros of 4, $q=\pm 1,\pm 2,\pm 4$

Zeros of 2, $p=\pm 1,\pm 2$

The potential rational zeros of $f=\pm 1,\pm 2,\pm 4,\pm \frac{1}{4},\pm \frac{1}{2}$.

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

$f\left(x\right)=4x^5-8x^4-x+2$

The real zero $=2$.

$f\left(x\right)=4x^5-8x^4-x+2=\left(x-2\right)\left(4x^4-1\right)$

The depressed equation $=\left(4x^4-1\right)$.

Further factorized

$\left(4x^4-1\right)=\left(2x^2-1\right)\left(2x^2+1\right)=\left(\sqrt{2}x-1\right)\left(\sqrt{2}x+1\right)\left(2x^2+1\right)$

$x=2,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}$ are real zeros.

The factored form $=\left(x-2\right)\left(\sqrt{2}x-1\right)\left(\sqrt{2}x+1\right)\left(2x^2+1\right)$.