Factoring $(x+a)(x+b)(x+c)+(b+c)(c+a)(a+b)$ and use the result to solve an equation

I managed to prove that $$(x+a+b+c)$$ is a factor of $$(x+a)(x+b)(x+c)+(b+c)(c+a)(a+b)$$

Then I was asked to use the result to solve $$(x+2)(x-3)(x-1)+4=0$$

I know by comparison, $$a=2, b=-3, c=-1$$, and thus $$(x-2)$$ is a factor, but I can't really figure out how to solve the equation without expanding the brackets.

• Considering the other roots are $\pm \sqrt 5$, I can't imagine you'd be intended to do anything else but to expand the expression, divide by $(x-2)$, and find the roots of the quadratic that results. – Eevee Trainer May 24 at 19:37
• And the other factor is given by $$x^2-5$$ – Dr. Sonnhard Graubner May 24 at 19:38
• Also, the solutions are $x=2, x=\sqrt{5}, x=-\sqrt{5}$ – Loo Soo Yong May 24 at 19:38
• Yes, this is true. – Dr. Sonnhard Graubner May 24 at 19:38
• How hard is expanding the brackets - you have identified a factor of a cubic and what remains will be a quadratic. Practically that is a quick way through. – Mark Bennet May 24 at 20:28

Unfortunately, you do have to expand the brackets.

Fortunately, it's not so messy after all:

$$(x+2)(x-3)(x-1)+4=0$$

so

$$x^3-2x^2-5x+10=0$$ and factoring our $$x-2$$, $$(x-2)(x^2-5)=0$$

from which we can readily read the answer as $$x=2,\pm\sqrt{5}$$

• You actually don't have to expand the brackets if you spot a few tricks. I explain it in my answer. – J.G. May 24 at 19:51
• @J.G. Strictly speaking, your answer does expand most of the brackets, it just expands them one at a time after dividing $x-2$. – auscrypt May 24 at 19:54

Since the $$x^2$$ coefficient in your cubic is $$a+b+c$$, the quadratic factor is of the form $$x^2+k$$, with the roots being $$\pm\sqrt{-k}$$. The $$x=0$$ case gives $$k=\frac{abc+(b+c)(c+a)(a+b)}{a+b+c}=ab+bc+ca.$$In your case $$k=-5$$.

Simplifying $$k$$ as above looks like it requires tedious algebra, but things aren't as bad as they seem. It's the ratio of two fully symmetric polynomials in $$a,\,b,\,c$$, one of degree $$3$$, the other $$1$$. This doesn't prove on its own that $$k$$ is a polynomial; but if it is, it must be fully symmetric and of degree $$2$$, and hence proportional to $$ab+bc+ca$$. The case $$a=b=c$$ gives $$k=\frac{9a^3}{3a}=3a^2$$, so it'll have to be $$ab+bc+ca$$ itself. So it makes sense to double-check whether $$(ab+bc+ca)(a+b+c)=abc+(b+c)(c+a)(a+b)$$. But of course it does, because both sides are fully symmetric cubic functions, so they have a fixed ratio. Again, the case $$a=b=c$$ strengthens this to equality.

It is good that you managed to factor!

Here is one way to do it: $$(x+a)(x+b)(x+c)+(b+c)(c+a)(a+b)=\\ x^3+(a+b+c)x^2+(ab+bc+ca)x+abc+\\ 2abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2=\\ x^2(x+a+b+c)+(ab+bc+ca)x+\\ ab(a+b+c)+ac(a+b+c)+bc(a+b+c)=\\ x^2(x+a+b+c)+(ab+bc+ca)x+\\ (a+b+c)(ab+bc+ca)=\\ x^2(x+a+b+c)+(ab+bc+ca)(x+a+b+c)=\\ (x+a+b+c)(x^2+ab+bc+ca).$$ Now, write the given equation in this form: $$(x+2)(x-3)(x-1)+4=0 \iff \\ (x+2)(x-3)(x-1)+(2-3)(-3-1)(-1+2)=0 \iff \\ (x+2-3-1)(x^2+2(-3)+(-3)(-1)+(-1)2)=0 \iff \\ (x-2)(x^2-5)=0 \Rightarrow x_1=2, x_{2,3}=\pm \sqrt{5}.$$

• This is a good and complete answer. – NoChance May 24 at 20:09

Probably not more efficient than straightforwardly factoring out, but it works as well:

$$\begin{eqnarray}\frac{(x+2)(x−3)(x−1)+4}{x-2} & = & \frac{(x-2+4)(x−3)(x−1)}{x-2}+\frac{4}{x-2} \\ & = & (1+\frac{4}{x-2})(x−3)(x−1)+\frac{4}{x-2} \\ & = & (x−3)(x−1) + \frac{4(x−3)(x−2+1)}{x-2}+\frac{4}{x-2}\\ & = & (x−3)(x−1) + 4(x−3)(1+\frac{1}{x-2})+\frac{4}{x-2}\\ & = & (x−3)(x−1) + 4(x−3) + \frac{4(x−2-1)}{x-2}+\frac{4}{x-2}\\ & = & (x−3)(x−1) + 4(x−3) + 4\\ & = & (x−3)(x+3) + 4\\ & = & x^2 - 5 \; . \end{eqnarray}$$