# Factorizing 4th degree expression.

Hello I'm having trouble understanding the factorizing of a polynomial as

$$x^4-4x$$

After that, I turned it into $$x(x^3-8)$$

But I don't quite understand how it's factored (the process) as

$$x(x−2)(x^2+2x+4)$$

Thanks!

• The relevant search term is "difference of cubes." – user61527 Jan 24 '14 at 2:07
• Pardon if I'm not seeing something but I don't know why $x^4-4x=x(x^3-8)$ – AndreGSalazar Jan 24 '14 at 2:44
• Is it (x^4 - 4 x) or (x^4 - 8 x)? – Claude Leibovici Jan 24 '14 at 4:42

## 2 Answers

Notice that if $f(x) = x^3 - 8$, we can easily see that $f(2)= 0$ This means that $x - 2$ is a divisor of $x^3 - 8$. Do polynomial long division. You an easily generalize to factor $x^3 - a^3$ for a constant $a$.

• Or generalize further to $\ x-a\mid f(x)-f(a)\$ i.e. the Factor Theorem – Bill Dubuque Jan 24 '14 at 2:37

You can learn "difference of cubes" $$(x^3-y^3)=(x-y)(x^2+xy+y^2)$$ or the general form $$(x^n-y^n)=(x-y)(x^{n-1}y^0+x^{n-2}y^1+...+x^1y^{n-2}+x^0y^{n-1})$$

• Yes! I had learnt difference of cubes , I just didn't even realized this was the case! Thanks a lot! – Joel Hernandez Jan 24 '14 at 2:52