How do you factor by grouping 5a^3 - 10a^2 +25a- 50?

May 3, 2015

The answer is $5 \left(a - 2\right) \left({a}^{2} + 5\right)$

Let's call P(a)=5a^3−10a^2+25a−50

First, you notice that $5$ divides all the coefficients of $P \left(a\right)$, so $P \left(a\right) = 5 Q \left(a\right)$ with $Q \left(a\right) = {a}^{3} - 2 {a}^{2} + 25 a - 50$

This is factoring by grouping, you collect a $5$ outside the polynomial because you notice it divides every coefficient.

Then, if you want a prime factorisation, you can notice that $2$ is a root of $Q \left(a\right)$

So $\left(a - 2\right)$ divides the polynomial, so you have, using Ruffini's rule

$P \left(a\right) = 5 Q \left(a\right) = 5 \left(a - 2\right) \left({a}^{2} + 5\right)$ and you know it's a prime factorisation because $\left(a - 2\right)$ has degree 1 and $\left({a}^{2} + 5\right)$ doesn't have any roots in $\mathbb{Q}$ (nor in $\mathbb{R}$).

If you want the factorisation in $\mathbb{C}$, you have that $\left({a}^{2} + 5\right) = \left(a + 5 i\right) \left(a - 5 i\right)$, but I think this is a little out of your league.

PS: you notice $2$ is a root just trying, e.g. in the beginning I thought $Q \left(a\right)$ was irreducible, but I wasn't able to prove that with any criterion, and I found out that $2$ is a root when I was looking where the polynomial was positive or negative, hoping Gauss' lemma could prove it was irreducible in $\mathbb{Z}$ so in $\mathbb{Q}$.

There's Cardano's formula for roots of polynomials of degree $3$, but it's horrible and I usually avoid it.