Chapter 4.2, Problem 40AYU

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

52. $f\left(x\right)\text{ = }{x}^3\text{ + 8}{x}^2\text{ + 11}x\text{ - 20}$

To determine

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

Answer to Problem 40AYU

The real zeros $=-4,-5\text{ and }1$.

$f\left(x\right)=x^3+8x^2+11x-20=\left(x-1\right)\left(x+4\right)(x+5)$

Explanation of Solution

Given:

$f\left(x\right)=x^3+8x^2+11x-20$

$f$ is a polynomial function.

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

From Descartes’ rule of signs,

$f\left(x\right)=x^3+8x^2+11x-20$

$+$ to $-$

There will be 1 positive real zero.

$f\left(-x\right)=-x^3+8x^2-11x-20$

$-$ 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 $a_n$.

$f\left(x\right)=x^3+8x^2+11x-20$

Here $a_0=\text{ −20}$ and $a_n=1$.

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

Zeros of $-20$, $p=\pm 1,\pm 2,\pm 4,\pm 5,\pm 10,\pm 20$.

The potential rational zeros of $f=\pm 1,\pm 2,\pm 4,\pm 5,\pm 10,\pm 20$.

From the graph of $f$, we have 3 real zeros near $-5$, $-4$ and 1.

$x\text{ = 1}$ is zero of $f$.

Use long division or Synthetic division of $x^3+8x^2+11x-20$.

$f\left(x\right)=\left(x-1\right)\left(x^2+9x+20\right)$

The depressed equation $=\left(x^2+9x+20\right)$ which is a quadratic equation.

From factorizing further,

$\left(x^2+9x+20\right)=\left(x+4\right)(x+5)$

The real zeros $=-4,-5$ and 1.

$f\left(x\right)=x^3+8x^2+11x-20=\left(x-1\right)\left(x+4\right)(x+5)$