# Factorizing Faster

As one of the steps in a calculation, I had to factorize the following quadratic polynomial:

$$12x^2 - 11x - 15$$

Using guess and check, and after roughly 10 minutes, I came to the following solution:

$$(4x + 3)(3x - 5)$$

I stumble upon factoring these kinds of quadratics all the time, and they take up a majority of the time in answering a multi-step question. Is there a method more efficient than guess and check which would help me get to the answer in less time?

Any help is greatly appreciated, thanks in advance.

• Do you know the quadratic formula? – Gerry Myerson Jul 23 '15 at 7:25
• Guessing is supposed to be a shortcut if you see it quickly. Otherwise, the quadratic formula is the generic approach. – orion Jul 23 '15 at 7:32
• $x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ always works. In this case, it would give you $x_{1,2}=-\frac34,\frac53$. Therefore, $12x^2-11x-15=(x+\frac34)(x-\frac53)$. – barak manos Jul 23 '15 at 7:39
• @barakmanos You forgot the coefficient $a$ at the beginning of your final expression. – thanasissdr Jul 23 '15 at 7:49
• @barakmanos $12x^2 -11x - 15 = \mathbf {12} \cdot (x+3/4) (x-5/3)$ – thanasissdr Jul 23 '15 at 7:52

I solve it like this $12x^2- 11x -15...(1)$ we will search for two numbers their product is $12(-15)$ and their sum is $-11$. I look to $12 ( -15) = 4(3) ( -5)(3)$ and I start to multiply them so I construct two numbers and check their sum. In this example it is clear that $(-20), (9)$ satisfies the conditions.

To factor $(1)$ we need to choose $a, b$ in $(ax + n)(bx + m)$ such that $ab = 12$ you may choose $3, 4$ or $2, 6$ or $1, 12$. If you choose $a=2, b=6$ . $(2x + n)(2x + m)$ now $n , m$ are $-20 , 9$ divided by $2, 6$. Finlay $\displaystyle \left( 2x + \frac{-20}{6}\right) \left( 6x + \frac{9}{2}\right)$ Or $\displaystyle \left(2x + \frac{9}{6} \right) \left(6x + \frac{-20}{2}\right)$. If you chose $a$ and $b$ to be $3 ,4$ we may write it like $\displaystyle \left(3x + \frac{-20}{4} \right) \left(4x + \frac{9}{3} \right)$

Adding an important notice. You see in my work for example in this factoring

$\displaystyle \left(3x + \frac{-20}{4} \right)\left( 4x + \frac{9}{3}\right)$ I divided $9$ by $3$ since $3$ is the coefficient of $x$ in the other brackets and I divided $-20$ by $4$ since it is the coefficient of $x$ is the other brackets you can see that in all the ways I end up with.

$$12x^2 - 11x - 15=(12x^2 - 20x)+(9x-15)=4x(3x-5)+3(3x-5)=(3x-5)(4x+3)$$

• This is great, if you are lucky enough to come up with that particular way of splitting the $-11x$ term up into two pieces. But you haven't really helped OP if you can't say how you found that. – Gerry Myerson Jul 23 '15 at 10:33

One method you might like to try is to list the ways the first coefficient can be factorized, and the last coefficient can be factorized: $$\binom{12}{1},\binom{6}{2},\binom{4}{3}.........\binom{15}{1},\binom{5}{3}$$

Now, after some obvious elimination of unlikely candidates, pick a pair on the left which can multiply a pair on the right (either way up) so that the difference is 11 (difference because of the minus sign).

In this case we can quickly spot the pair $$\binom{4}{3}...\binom{5}{3}$$ which does the trick, so you can construct the brackets accordingly as $$(4x+3)(3x-5)$$

$Ax^2 + Bx + C$

If you're good at arithmetic, you can use this method. Find two numbers, say $m$ and $n$, such that $mn = AC$ and $m + n = B$ Write the expression as

$(Ax^2 + mx) + (nx + C)$

and it will be factorable.

For example $12x^2 - 11x - 15$

$$AC = (12)(-15) = -180$$

That means we are looking for a positive and a negative number whos product is $-180$ and whos "sum" is $-11$.

For the moment, forget about the minus sign and make a table of the factors whos product is 180.

   1   180
2    90
3    60
4    45
5    36
6    30
9    20
we can stop here because we see that the difference
between 9 and 20 is 11.


So $m$ and $n$ are $9$ and $-20$.

\begin{array}{c} 12x^2 - 11x - 15 \\ (12x^2 + 9x) + (-20x - 15) \\ 3x(4x + 3) - 5(4x + 3) \\ (3x - 5)(4x + 3) \end{array}

It looks complicated, but with practice, it goes pretty fast. And there is no guessing required.