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.

  • $\begingroup$ 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. $\endgroup$ – Eevee Trainer May 24 at 19:37
  • $\begingroup$ And the other factor is given by $$x^2-5$$ $\endgroup$ – Dr. Sonnhard Graubner May 24 at 19:38
  • $\begingroup$ Also, the solutions are $x=2, x=\sqrt{5}, x=-\sqrt{5}$ $\endgroup$ – Loo Soo Yong May 24 at 19:38
  • $\begingroup$ Yes, this is true. $\endgroup$ – Dr. Sonnhard Graubner May 24 at 19:38
  • 1
    $\begingroup$ 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. $\endgroup$ – Mark Bennet May 24 at 20:28

Unfortunately, you do have to expand the brackets.

Fortunately, it's not so messy after all:



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

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

  • $\begingroup$ You actually don't have to expand the brackets if you spot a few tricks. I explain it in my answer. $\endgroup$ – J.G. May 24 at 19:51
  • $\begingroup$ @J.G. Strictly speaking, your answer does expand most of the brackets, it just expands them one at a time after dividing $x-2$. $\endgroup$ – 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: $$(x+2)(x-3)(x-1)+4=0 \iff \\ (x+2)(x-3)(x-1)+(2-3)(-3-1)(-1+2)=0 \iff \\ (x+2-3-1)(x^2+2(-3)+(-3)(-1)+(-1)2)=0 \iff \\ (x-2)(x^2-5)=0 \Rightarrow x_1=2, x_{2,3}=\pm \sqrt{5}.$$

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

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

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


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