Question

Find the real zeros of f. Use the real zeros to factor f.

$f\left(x\right)=x^3+2x^2-5x-6$

Short answer

The real zeros of the function are $-3,-1,2$.

And it is factored as$f\left(x\right)=(x+3)(x+1)(x-2)$

Step-by-step solution

Step 1. Find the possible number of zeros

The given function $f\left(x\right)=x^3+2x^2-5x-6$is of degree three, so it has at most three real zeros.

Step 2. Use the rational zero theorem

Now all the coefficients are integers so we use the rational zeros theorem.

The factors of the constant term $-6$are $p:\pm 1,\pm 2,\pm 3,\pm 6$

The factors of the leading coefficient $1$are

$q:\pm 1$.

So the possible rational zeros are

$\frac{p}{q}:\pm 1,\pm 2,\pm 3,\pm 6$

Step 3. Graph the function

The graph of the function is given as

The graph has two turning points, so it will have three real zeros.

Step 4. Find the first factor

From the graph, it appears that $2$is a zero of the function. On performing synthetic division we get

As the remainder is zero so $2$is a zero of the function.

And the quotient $x^2+4x+3$is the depressed polynomial. So the given function is factored as

$$\begin{gathered} f\left(x\right)=x^3+2x^2-5x-6 \\ f\left(x\right)=(x-2)(x^2+4x+3) \end{gathered}$$

Step 5. Factor the depressed equation

Equating the depressing polynomial with zero and factoring by grouping we get

$\begin{array}{rcl}{x}^2+4x+3& =& 0\\ x^2+3x+x+3& =& 0\\ x(x+3)+1(x+3)& =& 0\\ (x+3)(x+1)& =& 0\end{array}$

So the zeros are

$\begin{array}{rcl}x+3& =& 0\\ x& =& -3\end{array}$

and

$\begin{array}{rcl}x+1& =& 0\\ x& =& -1\end{array}$

Step 6. Factor the function

All the three real zeros of the function $f\left(x\right)=x^3+2x^2-5x-6$are $-3,-1,2$.

And it can be written in factored form as

$f\left(x\right)=(x+3)(x+1)(x-2)$