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Section 5.4 : Finding Zeroes of Polynomials

2. Find all the zeroes of the following polynomial.

\[P\left( x \right) = {x^4} - 3{x^3} - 5{x^2} + 3x + 4\]

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Start Solution

We’ll need all the factors of 4 and 1.

\[\begin{align*}4 & : \pm 1, \pm 2, \pm 4\\ 1 & : \pm 1\end{align*}\] Show Step 2

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.

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