Question

Find the real zeros off. Use the real zeros to factorf. $f\left(x\right)=2x^4+17x^3+35x^2-9x-45$

Short answer

The real zeros of the function are $-5,-3,-\frac{3}{2},1$.

The factored form of the function is role="math" localid="1646117935993" $f\left(x\right)=(x+5)(x+3)(2x+3)(x-1)$.

Step-by-step solution

Step 1. Find the possible number of zeros      

The given function$f\left(x\right)=2x^4+17x^3+35x^2-9x-45$is of degree four, so it has at most four real zeros.

Step 2. Use the rational zero theorem  

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

The factors of the constant term $-45$are

role="math" localid="1646118029584" $p:\pm 1,\pm 3,\pm 5,\pm 9,\pm 15,\pm 45$

The factors of the leading coefficient $2$are

$q:\pm 1,\pm 2$

So the possible rational zeros are

$\frac{p}{q}:\pm 1,\pm 3,\pm 5,\pm 9,\pm 15,\pm 45,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{5}{2},\pm \frac{9}{2},\pm \frac{15}{2},\pm \frac{45}{2}$

Step 3. Graph the function

The graph of the function is given as

So the graph has three turning points and thus has four real roots.

Step 4. Find the first factor 

From the graph, it appears that $-5$is a zero of the function. On performing synthetic division we get

As the remainder is zero so $-5$is a zero of the function.

And the quotient $2x^3+7x^2-9$is the depressed polynomial. So the given function is factored as

role="math" localid="1646118365223" $f\left(x\right)=(x+5)(2x^3+7x^2-9)$

Step 5. Find the second factor 

From the graph, it can be seen that $-3$may also be a zero. On performing synthetic division with the depressed polynomial we get

As the remainder is zero so $-3$is a zero of the function.

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

role="math" localid="1646118498157" $$\begin{gathered} f\left(x\right)=(x+5)(2x^3+7x^2-9) \\ f\left(x\right)=(x+5)(x+3)(2x^2+x-3) \end{gathered}$$

Step 6. Factor the depressed polynomial 

Equate the depressed polynomial equal to zero and factor by grouping.

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

So the other zeros are

$\begin{array}{rcl}2x+3& =& 0\\ 2x& =& -3\\ x& =& -\frac{3}{2}\end{array}$

and

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

So the final factored form is

$$\begin{gathered} f\left(x\right)=(x+5)(x+3)(2x^2+x-3) \\ f\left(x\right)=(x+5)(x+3)(2x+3)(x-1) \end{gathered}$$