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

\[\begin{align*}\frac{{ \pm 1}}{{ \pm 1}} & = \pm 1 & \hspace{0.25in}\frac{{ \pm 2}}{{ \pm 1}} & = \pm 2 & \hspace{0.25in}\frac{{ \pm 4}}{{ \pm 1}} & = \pm 4\\ \\ \frac{{ \pm 1}}{{ \pm 2}} & = \frac{{ \pm 1}}{{ \pm 2}} & \hspace{0.25in}\frac{{ \pm 2}}{{ \pm 2}} & = \pm 1 & \hspace{0.25in}\frac{{ \pm 4}}{{ \pm 2}} & = \pm 2\\ \\ \frac{{ \pm 1}}{{ \pm 4}} & = \frac{{ \pm 1}}{{ \pm 4}} & \hspace{0.25in}\frac{{ \pm 2}}{{ \pm 4}} & = \frac{{ \pm 1}}{{ \pm 2}} & \hspace{0.25in}\frac{{ \pm 4}}{{ \pm 4}} & = \pm 1\\ \\ \frac{{ \pm 1}}{{ \pm 8}} & = \frac{{ \pm 1}}{{ \pm 8}} & \hspace{0.25in}\frac{{ \pm 2}}{{ \pm 8}} & = \frac{{ \pm 1}}{{ \pm 4}} & \hspace{0.25in}\frac{{ \pm 4}}{{ \pm 8}} & = \frac{{ \pm 1}}{{ \pm 2}}\end{align*}\]

So, we have a total of 12 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.

\[\begin{array}{r|rrrrrl} {} & 8 & 36 & 46 & 7 & -12 & -4 \\ \hline -1 & 8 & 28 & 18 & -11 & -1 & -3 =g\left( -1 \right)\ne 0 \\ 1 & 8 & 44 & 90 & 97 & 85 & 81 =g\left( 1 \right)\ne 0 \\ \end{array}\]

Okay, notice that we have opposite signs for the two function evaluations listed above. Recall that means that we know we have a zero somewhere between them.

So, let’s take a look at some of the fractions from our list and give them a try in the synthetic division table. We’ll start with the fractions with the smallest denominators.

\[\begin{array}{r|rrrrrl} {} & 8 & 36 & 46 & 7 & -12 & -4 \\ \hline -\frac{1}{2} & 8 & 32 & 30 & -8 & -8 & 0 =g\left( -1 \right)=0!! \\ \end{array}\]

We now know that \(x = - \frac{1}{2}\) is a zero and we can write the polynomial as,

\[g\left( x \right) = 8{x^5} + 36{x^4} + 46{x^3} + 7{x^2} - 12x - 4 = \left( {x + \frac{1}{2}} \right)\left( {8{x^4} + 32{x^3} + 30{x^2} - 8x - 8} \right)\] Show Step 4

So, now we need to continue the process using \(Q\left( x \right) = 8{x^4} + 32{x^3} + 30{x^2} - 8x - 8\). Here is a list of all possible zeroes of \(Q\left( x \right)\).

\[\begin{align*}\frac{{ \pm 1}}{{ \pm 1}} & = \pm 1 & \hspace{0.25in}\frac{{ \pm 2}}{{ \pm 1}} & = \pm 2 & \hspace{0.25in}\frac{{ \pm 4}}{{ \pm 1}} & = \pm 4 & \hspace{0.25in}\frac{{ \pm 8}}{{ \pm 1}} & = \pm 8\\ \\ \frac{{ \pm 1}}{{ \pm 2}} & = \frac{{ \pm 1}}{{ \pm 2}} & \hspace{0.25in}\frac{{ \pm 2}}{{ \pm 2}} & = \pm 1 & \hspace{0.25in}\frac{{ \pm 4}}{{ \pm 2}} & = \pm 2 & \hspace{0.25in}\frac{{ \pm 8}}{{ \pm 2}} & = \pm 4\\ \\ \frac{{ \pm 1}}{{ \pm 4}} & = \frac{{ \pm 1}}{{ \pm 4}} & \hspace{0.25in}\frac{{ \pm 2}}{{ \pm 4}} & = \frac{{ \pm 1}}{{ \pm 2}} & \hspace{0.25in}\frac{{ \pm 4}}{{ \pm 4}} & = \pm 1 & \hspace{0.25in}\frac{{ \pm 8}}{{ \pm 4}} & = \pm 2\\ \\ \frac{{ \pm 1}}{{ \pm 8}} & = \frac{{ \pm 1}}{{ \pm 8}} & \hspace{0.25in}\frac{{ \pm 2}}{{ \pm 8}} & = \frac{{ \pm 1}}{{ \pm 4}} & \hspace{0.25in}\frac{{ \pm 4}}{{ \pm 8}} & = \frac{{ \pm 1}}{{ \pm 2}} & \hspace{0.25in}\frac{{ \pm 8}}{{ \pm 8}} & = \pm 1\end{align*}\]

So we have a list of 14 possible zeroes (lots of repeats), 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 \(Q\left( x \right)\) .

\[\begin{array}{r|rrrrl} {} & 8 & 32 & 30 & -8 & -8 \\ \hline -2 & 8 & 16 & -2 & -4 & 0 =Q\left( -2 \right)=0!! \\ \end{array}\]

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

\[Q\left( x \right) = 8{x^4} + 32{x^3} + 30{x^2} - 8x - 8 = \left( {x + 2} \right)\left( {8{x^3} + 16{x^2} - 2x - 4} \right)\]

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

\[g\left( x \right) = 8{x^5} + 36{x^4} + 46{x^3} + 7{x^2} - 12x - 4 = \left( {x + \frac{1}{2}} \right)\left( {x + 2} \right)\left( {8{x^3} + 16{x^2} - 2x - 4} \right)\] Show Step 5

So, it looks like we need to continue with the synthetic division. This time we’ll do it on the polynomial \(P\left( x \right) = 8{x^3} + 16{x^2} - 2x - 4\).

The possible zeroes of this are the same as the original polynomial and so we won’t write them down here. Again, however, we’ve already proved that \( \pm 1\) can’t be zeroes of the original polynomial and so can’t be zeroes of \(P\left( x \right)\) either.

Here is the synthetic division for \(P\left( x \right)\).

\[\begin{array}{r|rrrl} {} & 8 & 16 & -2 & -4 \\ \hline -2 & 8 & 0 & -2 & 0 =P\left( -2 \right)=0!! \\ \end{array}\]

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

\[P\left( x \right) = 8{x^3} + 16{x^2} - 2x - 4 = \left( {x + 2} \right)\left( {8{x^2} - 2} \right) = 8\left( {x + 2} \right)\left( {{x^2} - \frac{1}{4}} \right)\]

We factored an 8 out of the quadratic to make it a little easier for the next step.

The factored form of the original polynomial is now,

\[g\left( x \right) = 8{x^5} + 36{x^4} + 46{x^3} + 7{x^2} - 12x - 4 = 8\left( {x + \frac{1}{2}} \right){\left( {x + 2} \right)^2}\left( {{x^2} - \frac{1}{4}} \right)\] Show Step 6

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,

\[\begin{align*}g\left( x \right) = 8{x^5} + 36{x^4} + 46{x^3} + 7{x^2} - 12x - 4 & = 8\left( {x + \frac{1}{2}} \right){\left( {x + 2} \right)^2}\left( {x + \frac{1}{2}} \right)\left( {x - \frac{1}{2}} \right)\\ & = 8{\left( {x + \frac{1}{2}} \right)^2}{\left( {x + 2} \right)^2}\left( {x - \frac{1}{2}} \right)\end{align*}\] Show Step 7

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

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

Note that this problem was VERY long and messy. The point of this problem was really just to illustrate just how long and messy these can get. The moral, if there is one, is that we generally sit back and really hope that we don’t have to work these kinds of problems on a regular basis. They are just too long and it’s too easy to make a mistake with them.