College Algebra, Chapter 4, 4.4, Section 4.4, Problem 42

Determine all rational zeros of the polynomial $P(x) = x^5 - 4x^4 - 3x^3 + 22x^2 - 4x - 24$, and write the polynomial in factored form.

The leading coefficient of $P$ is $1$, so all the rational zeros are integers. They are divisors of the constant term $-24$. Thus, the possible candidates are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24$.

Using Synthetic Division







We find that $1$ is not a zeros but that $2$ is a zero and $P$ factors as

$x^5 - 4x^4 - 3x^3 + 22x^2 - 4x - 24 = (x - 2)(x^4 - 2x^3 - 7x^2 + 8x + 12)$

We now factor the quotient $x^4 - 2x^3 - 7x^2 + 8x + 12$. Its possible zeros are the divisors of $12$, namely

$\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$

Using Synthetic Division







We find that $-1$ is a zero and $P$ factors as

$\displaystyle x^5 - 4x^4 - 3x^3 + 22x^2 - 4x - 24 = (x - 2)(x + 1)\left( x^3 - 3x^2 - 4x + 12 \right)$

We now factor the quotient $x^3 - 3x^2 - 4x + 12$. Its possible zeros are the divisors of $12$, namely

$\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$

Using Synthetic Division







We find that $2$ is a zero and $P$ factors as

$x^5 - 4x^4 - 3x^3 +22x^2 - 4x - 24 = (x - 2)(x + 1)(x - 2)(x ^2 - x - 6)$

We know factor the quotient $x^2 - x - 6$ using trial and error. We get

$x^5 - 4x^4 - 3x^3 + 22x^2 - 4x - 24 = (x - 2)(x + 1)(x - 2) (x - 3)(x + 2)$

The zeros of $P$ are $2, 3, -1$ and $2$.