# Factorising polynomials and Equating Coefficients

I am studying Factorising polynomials and equating coefficients at the moment, and up until now all has been well. I have no problem working these out up as high as the third degree as there are plenty of examples. Questions like this:

Factorise the expression $x^3 - 17x^2 + 54x - 8$ given $(x-4)$ is a factor.

I can work this out using $(x-4)(\alpha x^2 + \beta x + \gamma)$ and then equating the coefficients.

However, in the end of block exercises some of the questions are 4th degree polynomials and I don't know where to begin. With the polynomial $x^4 -7x^3 + 3x^2 + 31x + 20$ where $(x+1)$ is a factor for example.

Do I use the same procedure but start a degree higher, like $(x+1)(\alpha x^3 + \beta x^2 + \gamma x + \delta)$ and move on down to figure out the 2nd degree after that?

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But $x-1$ is not a factor of $x^3-17x^2+54x-8$, and $x+1$ is not a factor of $x^4-7x^3+31x+20$. –  Gerry Myerson Apr 25 '12 at 10:50
mal, it would be helpful if you would address @Gerry's comment. Did you make up the "questions like this" yourself? If so, maybe we could suggest some that are coherent. –  The Chaz 2.0 Apr 25 '12 at 12:53
my apologies, that was a typo, it should have been (x-4), I've corrected it now. –  bot_bot Apr 25 '12 at 13:01
Good. But what about the other one in my comment? –  Gerry Myerson Apr 25 '12 at 13:05

With the polynomial $x^4 -7x^3 + 3x^2 + 31x + 20$ where $(x+1)$ is a factor for example.

Do I use the same procedure but start a degree higher, like $(x+1)(\alpha x^3 + \beta x^2 + \gamma x + \delta)$ and move on down to figure out the 2nd degree after that?

There are other techniques, namely the polynomial long division as mentioned in a comment by pedja, the Ruffini's rule and for polynomials with integer coefficients the rational root theorem, but you can apply the method of equating coefficients to

$$\begin{eqnarray*} P(x) &:&=x^{4}-7x^{3}+3x^{2}+31x+20 =\left( x+1\right) \left( \alpha x^{3}+\beta x^{2}+\gamma x+\delta \right). \end{eqnarray*}\;\; \tag{1}$$

Expanding the RHS and comparing with the LHS, where the coefficient of $x^{4}$ is $1$, you would conclude that $\alpha =1$ and are left with a simpler equality

$$\begin{eqnarray*} P(x) &:&=x^{4}-7x^{3}+3x^{2}+31x+20=\left( x+1\right) \left( x^{3}+\beta x^{2}+\gamma x+\delta \right) \\ &=&x^{4}+\left( \beta +1\right) x^{3}+\left( \beta +\gamma \right) x^{2}+\left( \gamma +\delta \right) x+\delta . \end{eqnarray*}$$ Equating coefficients you get the simple system of four linear equations $$\begin{equation*} \left\{ \begin{array}{c} \delta =20 \\ \gamma +\delta =31 \\ \beta +\gamma =3 \\ \beta +1=-7, \end{array} \right. \end{equation*}$$ whose solution is $\delta =20,\beta =-8,\gamma =11$. So

$$\begin{equation*} x^{4}-7x^{3}+3x^{2}+31x+20=\left( x+1\right) \left( x^{3}-8x^{2}+11x+20\right). \end{equation*}\tag{2}$$ Now by direct inspection (or by the rational root theorem, since $1,2,4$ and $5$ are the positive divisors of $20$) you can find that $x=-1$ is a zero of the cubic polynomial $Q(x):=x^{3}-8x^{2}+11x+20$, ie $Q(-1)=0$. Applying a similar method to $Q(x)$ you would find that

$$\begin{equation*} Q(x)=x^{3}-8x^{2}+11x+20=\left( x+1\right) \left( x-4\right) \left( x-5\right) . \end{equation*}$$

Consequently, $$\begin{equation*} x^{4}-7x^{3}+3x^{2}+31x+20=\left( x+1\right) ^{2}\left( x-4\right) \left( x-5\right). \end{equation*}\tag{3}$$

A shorter method is the Ruffini's rule applied to the polynomial division of $P(x)$ by $(x-r)$. This case (with $P(x)=x^{4}-7x^{3}+3x^{2}+31x+20$ and $r=-1,P(r)=0$, ) is shown bellow, where the remainder is $P(r)=s$.

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Exemplary answer! (+1). –  Quixotic Apr 26 '12 at 7:47
Many, many thanks Américo - I will study your comprehensive answer. –  bot_bot Apr 26 '12 at 9:33
@mal, You are welcome! –  Américo Tavares Apr 26 '12 at 10:25