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

$$\frac{{ \pm 1}}{{ \pm 1}} = \pm 1\hspace{0.25in}\frac{{ \pm 2}}{{ \pm 1}} = \pm 2\hspace{0.25in}\frac{{ \pm 4}}{{ \pm 1}} = \pm 4$$

So, we have a total of 6 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.

$$\,\,\begin{array}{r|rrrrl} {} & 1 & -3 & -5 & 3 & 4 \\ \hline -1 & 1 & -4 & -1 & 4 & 0 =P\left( -1 \right)=0!! \\ \end{array}$$

Okay we now know that $x = - 1$ is a zero and we can write the polynomial as,

$$P\left( x \right) = {x^4} - 3{x^3} - 5{x^2} + 3x + 4 = \left( {x + 1} \right)\left( {{x^3} - 4{x^2} - x + 4} \right)$$ Show Step 4

So, now we need to continue the process using $Q\left( x \right) = {x^3} - 4{x^2} - x + 4$. The possible zeroes of this polynomial are the same as the original polynomial and so we won’t write them back down.

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

$$\begin{array}{r|rrrl} {} & 1 & -4 & -1 & 4 \\ \hline -1 & 1 & -5 & 4 & 0 =Q\left( -1 \right)=0!! \\ \end{array}$$

Therefore, $x = - 1$ is also a zero of $Q\left( x \right)$ and the factored form of $Q\left( x \right)$ is,

$$Q\left( x \right) = {x^3} - 4{x^2} - x + 4 = \left( {x + 1} \right)\left( {{x^2} - 5x + 4} \right)$$

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

$$P\left( x \right) = {x^4} - 3{x^3} - 5{x^2} + 3x + 4 = \left( {x + 1} \right)\left( {x + 1} \right)\left( {{x^2} - 5x + 4} \right) = {\left( {x + 1} \right)^2}\left( {{x^2} - 5x + 4} \right)$$ 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,

$$P\left( x \right) = {x^4} - 3{x^3} - 5{x^2} + 3x + 4 = {\left( {x + 1} \right)^2}\left( {x - 4} \right)\left( {x - 1} \right)$$ Show Step 6

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

$$\require{bbox} \bbox[2pt,border:1px solid black]{{x = - 1\,\,\,({\mbox{multiplicity 2}})\hspace{0.25in}\hspace{0.25in}x = 1\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}x = 4}}$$