How to factorize a 4th degree polynomial?

Question

I need help to factorise the following polynomial:

$x^4 - 2x^3 + 8x^2 - 14x + 7$

The solution I need to reach is $(x-1)(x^3 - x^2 + 7x - 7)$. I need to factorize to this exactly as it is for a limit question where I cancel out the $(x-1)$ in the numerator and denominator. How do I proceed?

Answer 1

Notice that the coefficients add to $0$, hence $p(1) = 0$. (Instead of noticing this possibly hard to see fact, we might have also chosen to try $1$ because of the rational root theorem)

Now we can use polynomial long division using $x-1$. I'm partial to this method that is essentially the same thing but presented as factoring by grouping.

$x^3(x-1) - x^2 (x-1) + 7x(x - 1) - 7(x - 1) = (x-1)(x^3 - x^2 + 7x - 7)$.

You say this is the solution you need to reach, but we can go further:

We see that the roots still add to zero, so we can do the same process:

$(x-1)(x^2(x-1) + 7(x-1)) = (x-1)^2 (x^2 + 7)$

Answer 2

A brute force approach is to notice that

$$\begin{align} x^4-2x^3+8x^2-14x+7 =&(x^4-x^3)-(x^3-x^2)+7(x^2-2x+1)\\ =&x^3(x-1)-x^2(x-1)+7(x-1)^2\\ =&(x-1)(x^3-x^2+7x-7)\\ =&(x-1)\left(x^2(x-1)+7(x-1)\right)\\ =&(x-1)(x-1)(x^2+7) \end{align}$$

QED

Answer 3

One may recognize the factorization of the polynomial(#): $$y^2+8y+7=(y+1)(y+7)$$

Also(^) $$-2x^3-14x=-2x(x^2+7)$$

From (#), when $y=x^2$, we obtain(*) $$x^4+8x^2+7=(x^2+1)(x^2+7)$$

Combine(*) and (^), we have \begin{align} x^4-2x^3+8x^2-14x+7&=(x^2+1)(x^2+7)-2x(x^2+7)\\&=(x^2-2x+1)(x^2+7)\\&=(x-1)^2(x^2+7) \end{align}

Answer 4

Let $P(x)=(x-1)Q(x)$, or $Q(x)=\dfrac{P(x)}{x-1}$, where $Q(x)$ is the requested polynomial.

The independent term is $q_0=Q(0)=\dfrac{P(0)}{0-1}=\color{green}{-7}$.

The coefficient of the next term can be found by deriving, with $q_1=Q'(0)=\dfrac{P'(0)}{0-1}-\dfrac{P(0)}{(0-1)^2}=\color{green}{7}$.

Then $2!\,q_2=Q''(0)=\dfrac{P''(0)}{0-1}-2\dfrac{P'(0)}{(0-1)^2}+2\dfrac{P(0)}{(0-1)^3}=-2=2!\,(\color{green}{-1})$ and $3!\,q_3=Q'''(0)=\dfrac{P'''(0)}{0-1}-3\dfrac{P''(0)}{(0-1)^2}+6\dfrac{P'(0)}{(0-1)^3}-6\dfrac{P(0)}{(0-1)^4}=6=3!\,\color{green}{1}$.