College Algebra, Chapter 4, 4.4, Section 4.4, Problem 62

The polynomial $P(x) = x^5 - x^4 - 6x^3 + 14x^2 - 11x + 3$.

a.) Find all the real zeros of $P$

The leading coefficient of $P$ is $1$, so all rational zeros are integers. They are divisors of constant term $3$. Thus, the possible zeros are

$\displaystyle \pm 1, \pm 3$

Using Synthetic Division







We find that $1$ is a zeros and that $P$ factors as

$\displaystyle x^5 - x^4 - 6x^3 + 14x^2 - 11x + 3 = (x - 1) \left( x^4 - 6x^2 + 8x - 3 \right)$

We now factor the quotient $x^4 - 6x^2 + 8x - 3$. Its possible zeros are

$\pm 1, \pm 3$

Using Synthetic Division







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

$\displaystyle x^5 - x^4 - 6x^3 + 14x^2 - 11x + 3 = (x - 1)(x - 1) \left( x^3 + x^2 - 5x + 3 \right)$

We now factor the quotient $x^3 + x^2 - 5x + 3$. Its possible zeros are

$\pm 1, \pm 3$

Using Synthetic Division







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

$\displaystyle x^5 - x^4 - 6x^3 + 14x^2 - 11x + 3 = (x - 1)(x - 1)(x -1) (x^2 + 2x - 3)$

We now factor the quotient $x^2 + 2x - 3$ using trial and error. We get,

$\displaystyle x^5 - x^4 - 6x^3 + 14x^2 - 11x + 3 = (x - 1)^3 (x + 3)(x - 1)$

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

b.) Sketch the graph of $P$