Chapter 4.2, Problem 68AYU

To determine

To find: The real solutions of $f$.

Answer to Problem 68AYU

$\text{The real solution}=2,4,\frac{-1}{2}$.

Explanation of Solution

Given:

$f\left(x\right)=2x^3-11x^2+10x+8=0$

$F$ is a polynomial function.

The degree of the polynomial function is 3. Therefore the number of real zeros by real zero theorem can be at most $=3$.

From Descartes’ rule of signs

$f\left(x\right)=2x^3-11x^2+10x+8$

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

There will be 2 or 0 positive real zeros.

$f\left(-x\right)=-2x^3-11x^2-10x+8$

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

$f\left(x\right)=2x^3-11x^2+10x+8$

Here $a_0=8$ and $a_{\mathrm{n}}=2$

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

Zeros of 8, $p=\pm 1,\pm 2,\pm 4,\pm 8$

The potential rational zeros of $f= \pm 1,\pm 2,\pm 4,\pm 8,\pm \frac{1}{2}$.

$x=2$ is zero of $f$.

Use long division or Synthetic division of $2x^3-11x^2+10x+8$.

$f\left(x\right)=\left(x-2\right)\left(2x^2-7x-4\right)$

The depressed equation $=\left(2x^2-7x-4\right)$

Factorized further

$\left(2x^2-7x-4\right)=2x^2-8x+x-4=\left(2x+1\right)\left(x-4\right)$

$2x^3-11x^2+10x+8=\left(x-2\right)\left(2x+1\right)\left(x-4\right)$

The real zeros $=2,4,\frac{-1}{2}$.