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### Section 5.4 : Finding Zeroes of Polynomials

3. Find all the zeroes of the following polynomial.

$A\left( x \right) = 2{x^4} - 7{x^3} - 2{x^2} + 28x - 24$

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We’ll need all the factors of -24 and 2.

\begin{align*} - 24 & : \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24\\ 2 & : \pm 1, \pm 2\end{align*} Show Step 2

Here is a list of all possible rational zeroes for the polynomial.

\begin{align*}\frac{{ \pm 1}}{{ \pm 1}} & = \pm 1 & \hspace{0.25in}\frac{{ \pm 2}}{{ \pm 1}} & = \pm 2 & \hspace{0.25in}\frac{{ \pm 3}}{{ \pm 1}} & = \pm 3 & \hspace{0.25in}\frac{{ \pm 4}}{{ \pm 1}} & = \pm 4 & \hspace{0.25in}\frac{{ \pm 6}}{{ \pm 1}} & = \pm 6 & \hspace{0.25in}\frac{{ \pm 8}}{{ \pm 1}} & = \pm 8\\ \\ & & \hspace{0.25in} & \hspace{0.25in} & \hspace{0.25in} & \hspace{0.25in} & \hspace{0.25in} & \hspace{0.25in} & \hspace{0.25in}\,\,\,\,\frac{{ \pm 12}}{{ \pm 1}} & = \pm 12 & \hspace{0.25in}\,\,\,\,\,\frac{{ \pm 24}}{{ \pm 1}} & = \pm 24\\ \\ \frac{{ \pm 1}}{{ \pm 2}} & = \frac{{ \pm 1}}{{ \pm 2}} & \hspace{0.25in}\frac{{ \pm 2}}{{ \pm 2}} & = \pm 1 & \hspace{0.25in}\frac{{ \pm 3}}{{ \pm 2}} & = \frac{{ \pm 3}}{{ \pm 2}} & \hspace{0.25in}\frac{{ \pm 4}}{{ \pm 2}} & = \pm 2 & \hspace{0.25in}\frac{{ \pm 6}}{{ \pm 2}} & = \pm 3 & \hspace{0.25in}\frac{{ \pm 8}}{{ \pm 2}} & = \pm 4\\ \\ & & \hspace{0.25in} & \hspace{0.25in} & \hspace{0.25in} & \hspace{0.25in} & \hspace{0.25in} & \hspace{0.25in} & \hspace{0.25in}\,\,\,\,\frac{{ \pm 12}}{{ \pm 2}} & = \pm 6 & \hspace{0.25in}\,\,\,\,\,\frac{{ \pm 24}}{{ \pm 2}} & = \pm 12\end{align*}

So, we have a total of 20 possible zeroes for the polynomial.

Show Step 3

We now need to start the synthetic division work. We’ll start with the “small” integers first.

$\begin{array}{r|rrrrl} {} & 2 & -7 & -2 & 28 & -24 \\ \hline -1 & 2 & -9 & 7 & 21 & -45=A\left( -1 \right)\ne 0 \\ 1 & 2 & -5 & -7 & 21 & -3=A\left( 1 \right)\ne 0 \\ -2 & 2 & -11 & 20 & -12 & 0 =A\left( -2 \right)=0!! \\ \end{array}$

Okay we now know that $$x = - 2$$ is a zero and we can write the polynomial as,

$A\left( x \right) = 2{x^4} - 7{x^3} - 2{x^2} + 28x - 24 = \left( {x + 2} \right)\left( {2{x^3} - 11{x^2} + 20x - 12} \right)$ Show Step 4

So, now we need to continue the process using $$Q\left( x \right) = 2{x^3} - 11{x^2} + 20x - 12$$. Here is a list of all possible zeroes of $$Q\left( x \right)$$.

\begin{align*}\frac{{ \pm 1}}{{ \pm 1}} & = \pm 1 & \hspace{0.25in}\frac{{ \pm 2}}{{ \pm 1}} & = \pm 2 & \hspace{0.25in}\frac{{ \pm 3}}{{ \pm 1}} & = \pm 3 & \hspace{0.25in}\frac{{ \pm 4}}{{ \pm 1}} & = \pm 4 & \hspace{0.25in}\frac{{ \pm 6}}{{ \pm 1}} & = \pm 6 & \hspace{0.25in}\frac{{ \pm 12}}{{ \pm 1}} & = \pm 12\\ \\ \frac{{ \pm 1}}{{ \pm 2}} & = \frac{{ \pm 1}}{{ \pm 2}} & \hspace{0.25in}\frac{{ \pm 2}}{{ \pm 2}} & = \pm 1 & \hspace{0.25in}\frac{{ \pm 3}}{{ \pm 2}} & = \frac{{ \pm 3}}{{ \pm 2}} & \hspace{0.25in}\frac{{ \pm 4}}{{ \pm 2}} & = \pm 2 & \hspace{0.25in}\frac{{ \pm 6}}{{ \pm 2}} & = \pm 3 & \hspace{0.25in}\frac{{ \pm 12}}{{ \pm 2}} & = \pm 6\end{align*}

So we have a list of 16 possible zeroes, but note that we’ve already proved that $$\pm 1$$ can’t be zeroes of the original polynomial and so can’t be zeroes of $$Q\left( x \right)$$ either.

Here’s the synthetic division work for this $$Q\left( x \right)$$.

$\,\,\begin{array}{r|rrrl} {} & 2 & -11 & 20 & -12 \\ \hline -2 & 2 & -15 & 50 & -112 =Q\left( -2 \right)\ne 0 \\ 2 & 2 & -7 & 6 & 0 =Q\left( 2 \right)=0!! \\ \end{array}$

Therefore, $$x = 2$$ is a zero of $$Q\left( x \right)$$ and the factored form of $$Q\left( x \right)$$ is,

$Q\left( x \right) = 2{x^3} - 11{x^2} + 20x - 12 = \left( {x - 2} \right)\left( {2{x^2} - 7x + 6} \right)$

This also means that the factored form of the original polynomial is now,

$A\left( x \right) = 2{x^4} - 7{x^3} - 2{x^2} + 28x - 24 = \left( {x + 2} \right)\left( {x - 2} \right)\left( {2{x^2} - 7x + 6} \right)$ Show Step 5

We’re down to a quadratic polynomial and so we can and we can just factor this to get the fully factored form of the original polynomial. This is,

$A\left( x \right) = 2{x^4} - 7{x^3} - 2{x^2} + 28x - 24 = \left( {x + 2} \right)\left( {x - 2} \right)\left( {x - 2} \right)\left( {2x - 3} \right)=\left( {x + 2} \right){\left( {x - 2} \right)^2}\left( {2x - 3} \right)$ Show Step 6

From the fully factored form we get the following set of zeroes for the original polynomial.

$\require{bbox} \bbox[2pt,border:1px solid black]{{x = - 2\hspace{0.25in}\hspace{0.25in}x = \frac{3}{2}\hspace{0.25in}\hspace{0.25in}x = 2\,\,\,({\mbox{multiplicity 2}})}}$