College Algebra, Chapter 4, 4.4, Section 4.4, Problem 28

Determine all rational zeros of the polynomial $P(x) = x^4 - x^3 - 23x^2 - 3x + 90$, 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 $90$. Thus, the possible candidates are

$\pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 9, \pm 10, \pm 15, \pm 18, \pm 30, \pm 45, \pm 90$

Using Synthetic Division







We find that $1$ and $3$ are not zeros but that $2$ is a zero and that $P$ factors as

$x^4 - x^3 - 23x^2 - 3x + 90 = (x - 2)(x^3 + x^2 - 21x - 45)$

We now factor the quotient $x^3 + x^2 - 21x - 45$. Its possible zeros are the divisors of $-45$, namely

$\pm 1, \pm 3, \pm 5, \pm 9, \pm 15, \pm 45$

Using Synthetic Division







We find that $-3$ is a zero and that $P$ factors as

$x^4 - x^3 - 23x^2 - 3x + 90 = (x - 2)(x + 3)(x^2 - 2x - 15)$

We now factor $x^2 - 2x - 15$ using trial and error, so

$x^4 - x^3 - 23x^2 - 3x + 90 = (x - 2)(x + 3)(x + 3)(x - 5)$

The zeros of $P$ are $2, 5$ and $-3$.