# Different ways to factor

I'm interested to know some different methods to factor equations of the form $ax^2+bx+c$, where $a \ne 0$, other than pure guess and check.

• British Method is one way Commented Sep 22, 2016 at 23:13
• What? Please explain Commented Sep 22, 2016 at 23:13
• @fleablood I only know how to use the binomial theorem to expand binomials Commented Sep 22, 2016 at 23:18
• Compute the roots $x_0$ and $x_1$ (poss.ibly equal) with the quadratic formula and factor as $a(x-x_0)(x-x_1)$. Commented Sep 22, 2016 at 23:21
• @fleablood Didn't you say the binomial theorem earlier? What do you mean? Commented Sep 22, 2016 at 23:23

One method is The British Method. Here are the steps.

$1)$ Multiply $a*c$.

$2)$ Find a factor pair of $ac$ that adds up to $b$

$3)$ Replace $bx$ in the equation with the two factors, both multiplied by x.

$4)$ Factor by grouping.

Here is an annotated example. $$2x^2-x-6$$ $$a=2, b=-1, c=-6$$ Here we do step $1$, getting $$ac=-12$$ Then we do step $2$. The factor pair is $$-4,3$$ Now we do step $3$. The equation becomes $$2x^2-4x+3x-6$$ Then we factor by grouping $$2x^2-4x+3x-6$$ $$2x(x-2)+3(x-2)$$ $$(2x+3)(x-2)$$

Another method is shown below.

Steps:

$1)$ Multiply $a*c$.

$2)$ Find a factor pair of $ac$ that adds up to $b$, call them $x_1$ and $x_2$

$3)$ Write the following equation: $(ax+x_1)(ax+x_2)$.

$4)$ Factor out the GCF of each individual equation. The part remaining after factoring out the GCF is your answer.

Annotated example: $$2x^2-x-6$$ $$a=2, b=-1, c=-6$$ Here we do step $1$, getting $$ac=-12$$ Then we do step $2$. The factor pair is $$-4,3$$ Now we do step $3$, getting the equation: $$(2x-4)(2x+3)$$ Factor out a $2$ from the first part, where you get the final answer of $$(x-2)(2x+3)$$

• Yes, there is an element of guess and check involved, but OP wrote "pure guess and check" which I interpreted as methods that only used guess and check. Commented Sep 22, 2016 at 23:30
• If we aren't going to use some arithmetic guess and check step then quadratic formula or completing the square become your only options and for many questions those techniques are over the top. Commented Sep 22, 2016 at 23:32
• Thanks for the two methods! Commented Sep 22, 2016 at 23:54
• @fleablood Chill. I never said these methods were the best methods or the only methods. I'm just giving some interesting methods to answer OP's question. Commented Sep 23, 2016 at 0:30
• Thanks again, amazing answer =) Commented Sep 27, 2016 at 1:48

Assuming your question is in earnest and you aren't trolling the answer is

$ax^2 + bx + c = a(x + \frac{b + \sqrt{b^2 - 4ac}}{2a})(x + \frac{b - \sqrt{b^2 - 4ac}}{2a} )$

No guessing. If the equation factors at all (which it won't if $4ac > b^2$) then it will always factor to that.

• The OP isn't asking for different factorizations, but for different methods of arriving at a factorization. Certainly when you see $x^2 + 6x - 8$ you don't factor it using the quadratic formula, do you? Commented Sep 22, 2016 at 23:23
• The op said "other than guessing and checking" I never guess. I solve and factor from that. If the OP knew how to solve there'd be no reason to ask the question. No, I personally find completely the square to be easier than the quadratic equation but .. why on earth would you be surprised if I did... Factoring is the same thing as solving. Commented Sep 22, 2016 at 23:27
• I apologize for presuming, but I think most people look at $x^2 + 6x - 8$ and think to themselves "What two numbers multiply to $-8$ and sum to $6$?" Commented Sep 22, 2016 at 23:29
• So ... given $x^2 + 6x - 8$ I'd divide the 6 to get 3 and square it to get 9 subtract 9 from -8 to to get -17. Switch signs and take the two square roots and and three to get $(x - 3+ \sqrt(17))(x - 3 - \sqrt(17))$ ... which is the quadratic formula. Commented Sep 22, 2016 at 23:35
• No two rational numbers sum to -8 and sum to 6. If it wer $x^2 + 6x + 8$ I'd take a look at that but if before I go grubbing and guessing I want to know if there is a potato in that mud. $x^2 + 6x + 10$ won't have any and though $x^2 + 6 + 8$ will it's just as easy to square 3 and subtract 8 to get 1. That way I know it will factor. $x^2 + 6x + 10$ won't. Commented Sep 22, 2016 at 23:39

Variation of the method mentioned by @suomnonA

1) Multiply $a*c$.

2) Find a factor pair of $ac$ that adds up to $b$. Let them be $x_1$ and $x_2$.

3) Write factors as:

$$\frac{(ax-x_1)(ax-x_2)}{a}$$

The factors on top will factorize down to cancel out with the $a$ on the bottom.

Example:

$$6x^2+17x-14$$ $$a=6, b=17, c=-14$$ Here we do step $1$, getting $$ac=-84$$ Then we do step $2$. The factor pair is $$-4,21$$ Now we do step $3$.

The equation becomes $$\frac{(6x-4)(6x+21)}{2}$$

which factorizes to

$$\frac{2(3x-2)\cdot3(2x+7)}{6}$$

$$(3x-2)(2x+7)$$

• Apparently this is referred to as the Airplane Method. Commented Sep 22, 2016 at 23:38

Another option:

$$a\,{{x}^{2}}+bx+c$$ $$\downarrow$$ $$n\,{{x}^{2}}-2knx+{{k}^{2}}n-m$$ $$\downarrow$$ $${{x}^{2}}-2kx-\frac{m}{n}+{{k}^{2}}$$ $$\downarrow$$ $$\left( x-\sqrt{\frac{m}{n}}-k\right) \,\left( x+\sqrt{\frac{m}{n}}-k\right)$$