Chapter 4.2, Problem 47AYU

In Problems 51-68, find the real zeros of f. Use the real zeros to factor f

59. $f\left(x\right)\text{ = }{x}^4\text{ + }{x}^3\text{ - 3}{x}^2\text{ - }x\text{ + 2}$

To determine

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

Answer to Problem 47AYU

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

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

Explanation of Solution

Given:

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

$f$ is a polynomial function.

The degree of the polynomial is 4. Therefore the number of real zeros by real zero theorem can be at most $=4$.

From Descartes’ rule of signs,

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

$+$ to $-$, $-$ to $+$

There will be 2 or 0 positive real zeros.

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

$+$ 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 $a_n$.

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

Here $a_0=\text{ 2}$ and $a_n=1$.

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

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

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

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

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

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

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

The depressed equation $=\left(x^3-x^2-x+1\right)$.

From factorizing further,

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

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

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

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