Chapter 4.2, Problem 69AYU

To determine

To find: The real solutions of $f$.

Answer to Problem 69AYU

$\text{The real solution }=-2,-3,1\text{ multiplicity }2$.

Explanation of Solution

Given:

$f\left(x\right)=x^4+4x^3+2x^2-x+6=0$

$F$ 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

$f\left(x\right)=x^4+4x^3+2x^2-x+6$

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

There will be 2 or 0 positive real zeros.

$f\left(-x\right)=x^4-4x^3+2x^2+x+6$

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

There will be 2 or 0 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 $a_{\mathrm{n}}$.

$f\left(x\right)=x^4+4x^3+2x^2-x+6$

Here $a_0=6$ and $a_{\mathrm{n}}=1$

Zeros of 2, $q=\pm 1$

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

The potential rational zeros of $f= \pm 1,\pm 2,\pm 3,\pm 6$.

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

Use long division or Synthetic division of $x^4+4x^3+2x^2-x+6$.

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

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

Factorized further

$\left(x^3+2x^2-2x+3\right)=\left(x+3\right)\left(x^2-2x+1\right)=\left(x+3\right)\left(x-1\right)\left(x-1\right)$

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

The real zeros $=-2,-3,1$ multiplicity 2.