How to solve the following quadratic equation by factoring?

And what to do when the 2nd degree term has a non-1 coefficient?

$$ 3x^2 + 11x - 4 = 0 $$

Essentially, what I'm asking is how can I factor $3x^2 + 11x - 4$ as the product of two linear expressions?

  • $\begingroup$ I usually factor my second degree polynomials by solving the corresponding equation, not the other way around. $\endgroup$ – Arthur Jul 11 '18 at 8:40


Can you agree ?

  • $\begingroup$ I think this is what I was looking for. Can you explain how you arrived at this? Thank you! $\endgroup$ – Gnoliz Jul 11 '18 at 9:19

Let’s follow Sridhara, http://www-history.mcs.st-andrews.ac.uk/Biographies/Sridhara.html.

First multiply by $12$: $$36x^2+132x=48,$$ then complete the square by adding $11^2$: $$36x^2+132x+11^2=48+121\iff (6x+11)^2=169.$$ Finally take square root.

Alternatively, you may factor from here: $$0=(6x+11)^2-13^2=(6x+11+13)(6x+11-13).$$


If your quadratic polynomial has rational roots, you will be able to solve it like this:

$$A=3x^2 + 11x - 4$$ $$3A=9x^2 + 11(3x) - 12$$ $$3A=(3x)^2 + 11(3x) - 12$$

Now substitute $u=3x$ and you can see that $3A=(u+12)(u-1)$

Plug $u=3x$ back in the equation:

$$3A=3(x+4)(3x-1)$$ $$A=(x+4)(3x-1)$$

If your polynomial $p(x)=ax^2+bx+c$ does not have rational roots, you should first calculate its discriminant $b^2-4ac$ and then use the following formula that always works:

$$x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

  • $\begingroup$ Now that you are still needed to factor $u^2+11u-12$, why not directly factoring the original equation itself?? $\endgroup$ – mengdie1982 Jul 11 '18 at 8:53
  • $\begingroup$ @mengdie1982 It's easy to mentally find two numbers that add up to $11$ and are multiplied to give $-12$. It's not easy to factor the original equation mentally. $\endgroup$ – stressed out Jul 11 '18 at 8:57
  • $\begingroup$ No. It's easy to get that just by observation. $\endgroup$ – mengdie1982 Jul 11 '18 at 9:32
  • $\begingroup$ @mengdie1982 maybe it's easy for you, but not for me. It turned out that it wasn't easy for the OP to "just observe it" either. $\endgroup$ – stressed out Jul 11 '18 at 9:47



A bit of trial and error.

There are $2$ factors: $(ax +b)(cx +d)$.

Assume: $a,b,c,d \in \mathbb{Z}$.

1) Coefficient of $x^2$ is $3$:

$ac =3$, try $a=3$, and $c=1$:


2) The constant term :


Choices are : $(b,d)= (-2,2)$, $(b,d) =(2,-2)$, $(b,d)=(4,-1)$, or $(b,d)=(-1,4)$.

3) The coefficient of $x$ :


Hence $(b,d) =(-1,4)$ qualifies.




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