91Ó°ÊÓ

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.

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" $p:Â\pm 1,Â\pm 3,Â\pm 5,Â\pm 9,Â\pm 15,Â\pm 45$

The factors of the leading coefficient $2$are

$q:Â\pm 1,Â\pm 2$

So the possible rational zeros are

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

The graph of the function is given as

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

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

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

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

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