Factorize $x^4 + x^3 - 3x^2 - 4x - 4$ showing all work

Question

Factorize showing all workings $x^4 + x^3 - 3x^2 - 4x - 4$.

I've attempted this question from the textbook "Core Maths for Advanced Level" by L. Bostock and S. Chandler and I'm having difficulty factorizing it. I got to the point where I know that $(x-2)$ is a factor and I tried solving for the co-efficients of $x^2$ and $x$ to have it factorization in the form $(x-b)(x^3 + cx^2 + dx + e)$, from which I was going to further factorize using the factor theorem.

Answer 1

$$x^4+x^3-3x^2-\color{blue}{4x}-4=(x^4-\color{red}{4x^2})+(x^3-\color{blue}{4x})+(\color{red}{x^2}-4)=\\=x^2(x^2-4)+x(x^2-4)+(x^2-4)=(x-2)(x+2)(x^2+x+1)$$

Answer 2

Why don't you use the Rational root theorem ?

Let's start trying: $x=\pm1, \pm2, \pm4.$

You can easily deduce that, $x_1=2,x_2=-2$ are roots.

Then applying the method Polynomial long division

$$\begin{align} \dfrac {x^4 + x^3 - 3x^2 - 4x - 4}{(x-2)(x+2)}= \dfrac {x^4 + x^3 - 3x^2 - 4x - 4}{x^2-4}=x^2+x+1 \end{align}$$

So, you get

$$\begin{align} P(x)=x^4 + x^3 - 3x^2 - 4x - 4=(x - 2) (x + 2) (x^2 + x + 1). \end{align}$$

Answer 3

$$\begin{align} &x^4 + x^3 - 3x^2 - 4x - 4\\ =&x^4 + x^3 +x^2 - 4x^2 - 4x - 4\\ =&x^2(x^2+x+1)-4(x^2+x+1)\\ =&(x^2-4)(x^2+x+1)\\ =&(x-2)(x+2)(x^2+x+1)\\ \end{align}$$

Answer 4

I like the following way: $$x^4+x^3-3x^2-4x-4=\left(x^2+\frac{1}{2}x-\frac{3}{2}\right)^2-\left(\frac{1}{2}x+\frac{5}{2}\right)^2=$$ $$=(x^2-4)(x^2+x+1)=(x-2)(x+2)(x^2+x+1).$$ We can get this factoring by the following way.

For any real $k$ we have: $$x^4+x^3-3x^2-4x-4=$$ $$=\left(x^2+\frac{1}{2}x-k\right)^2-\frac{1}{4}x^2-k^2-2kx^2-kx-3x^2-4x-4=$$ $$=\left(x^2+\frac{1}{2}x-k\right)^2-\left(\left(2k+\frac{13}{4}\right)+(k+4)x+k^2+4\right).$$ Now, we'll choose a value of $k$, for which $2k+\frac{13}{4}>0$ and $$(k+4)^2-4\left(2k+\frac{13}{4}\right)(k^2+4)=0$$ or $$2k^3+3k^2+6k+9=0$$ or $$k^2(2k+3)+3(2k+3)=0$$ or $$(2k+3)(k^2+3)=0,$$ which gives $k=-\frac{3}{2}$ and we got our factoring.

Answer 5

If you develop $(x-2)(x^3+cx^2+dx+e)=x^4+(c-2)x^3+(d-2c)x^2+(e-2d)x-2e$

Now identifying the coefficients you get

$\begin{cases} c-2=1 & c=3\\ d-2c=-3 & d=2c-3=3 \\ e-2d=-4 & e=2d-4=2\end{cases}$

You now have to factorize $x^3+3x^2+3x+2$.

As indicated try first to find obvious solutions before going for Cardan formula. If there are rational roots they should be in $$\pm\dfrac{\text{divisors of }a_0}{\text{divisors of }a_3}=\{-2,2,-1,1\}$$

And you find $-2$ as a new root.

Now another method to divide, add terms of higher degree until you get everything:

$\begin{align} x^3+3x^2+3x+2 &=(x+2)(x^2+...) &=x^3+2x^2+...\\ &=(x+2)(x^2+x+...) &=x^3+3x^2+2x+...\\ &=(x+2)(x^2+x+1) &=x^3+3x^2+3x+2 \end{align}$