Question

Find all zeroes of h(x)=2x⁴-x³-18x²+9x

Answer 1

When you have a polynomial with no independent term you can start by factorizing the polynomial, as follows:

$h(x)=2x^4-x^3-18x^2+9x$

The common factor is x, then:

$h(x)=x(2x^3-x^2-18x+9)$

Now you have a third-degree polynomial inside the ( ), let's called it j(x):

$j(x)=2x^3-x^2-18x+9$

You need to factorize this polynomial, then find all the numbers that divide the independent term 9.

These numbers are:

$1,-1,3,-3$

Let's probe if any of these numbers makes j(x)=0:

$\begin{gathered} j(1)=2(1)^3-(1)^2-18(1)+9 \\ j(1)=2-1-18+9_{} \\ j(1)=-8 \end{gathered}$

Then 1 is not a root, let's try -1:

$\begin{gathered} j(-1)=2(-1)^3-(-1)^2-18(-1)+9 \\ j(-1)=-2-1+18+9 \\ j(-1)=24 \end{gathered}$

Then -1 is not a root, let's try 3:

$\begin{gathered} j(3)=2(3)^3-(3)^2-18(3)+9 \\ j(3)=2\cdot27-9-18\cdot3+9 \\ j(3)=54-9-54+9 \\ j(3)=0 \end{gathered}$

Then 3 is a root and you can apply synthetic division to find the others:

The new polynomial is:

$2x^2+5x-3$

Now you can factorize it as follows:

$2x^2+5x-3=(2x-1)(x+3)$

And finally, your 4th-grade polynomial can be written as:

$h(x)=x(x-3)(2x-1)(x+3)$

Thus, the zeros are at:

$\begin{gathered} x=0 \\ (x-3)=0\text{ then x=3} \\ (2x-1)=0\text{ then x=1/2} \\ (x+3)=0\text{ then x=-3} \end{gathered}$

The answer is 0, 3, 1/2 and -3