Chapter 4.2, Problem 52AYU

To determine

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

Answer to Problem 52AYU

$f\left(x\right)=2x^4+11x^3-5x^2-43x+35=\left(x-1\right)\left(2x^3+13x^2+8x-35\right)$

Explanation of Solution

Given:

$f\left(x\right)=2x^4+11x^3-5x^2-43x+35$

$F$ 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

$f\left(x\right)=2x^4+11x^3-5x^2-43x+35$

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

There will be 2 or 0 positive real zeros.

$f\left(-x\right)=2x^4-11x^3-5x^2+43x+35$

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

$f\left(x\right)=2x^4+11x^3-5x^2-43x+35$

Here $a_0=35$ and $a_{\mathrm{n}}=2$

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

Zeros of 35, $p=\pm 1,\pm 5,\pm 7,\pm 35$

The potential rational zeros of $f= \pm 1,\pm 2,\pm 5,\pm 7,\pm 35,\pm \frac{1}{2},\pm \frac{5}{2},\pm \frac{7}{2},\pm \frac{35}{2}$.

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

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

Use long division or Synthetic division of $2x^4+11x^3-5x^2-43x+35$.

$f\left(x\right)=2x^4+11x^3-5x^2-43x+35=\left(x-1\right)\left(2x^3+13x^2+8x-35\right)$

The depressed equation $=\left(2x^3+13x^2+8x-35\right)$.

$f\left(x\right)=2x^4+11x^3-5x^2-43x+35=\left(x-1\right)\left(2x^3+13x^2+8x-35\right)$