Chapter 4.2, Problem 63AYU

In Problems 75-84, find the real solutions of each equation.

$x^4-x^3+2x^2-4x-8=0$

To determine

To find: The real solutions of $f$.

Answer to Problem 63AYU

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

Explanation of Solution

Given:

$f\left(x\right) = x^4 - x^3 + 2x^2 - 4x - 8 = 0$

$\text{F}$ 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

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

$+\text{ To }– -\text{ to }+ +\text{ to }-$

There will be 3 or 1 positive real zeros.

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

$+\text{ To }-$

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 $\text{f}$,then $\text{p}$ is a factor of ${\text{a}}_{\text{0}}$ and $\text{q}$ is the factor of ${\text{a}}_{\text{n}}$.

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

Here ${\text{a}}_{\text{0}}\text{ = }-8$ and ${\text{a}}_{\text{n}}\text{ = 1}$

Zeros of 1, $\text{q =} \pm 1$

Zeros of $-8$, $\text{p = }\pm 1, \pm 2, \pm 4, \pm 8$

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

$x = 2$ is zero of $\text{f}$.

Use long division or Synthetic division of $x^4 - x^3 + 2x^2 - 4x - 8$

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

The depressed equation $= \left(x^3 + x^2 + 4x + 4\right)$

From factorizing further

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

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