# How to factor quadratic $ax^2+bx+c$?

How do I shorten this? How do I have to think? $$x^2 + x - 2$$

The answer is $$(x+2)(x-1)$$

I don't know how to get to the answer systematically. Could someone explain? Does anyone have a link to a site that teaches basic stuff like this? My book does not explain anything and I have no teacher; this is self-studies.

• You may not appreciate this answer, but, just as when young children first learn to count, few can do so "mentally" -- they need objects, or fingers, to help them grasp addition, for example. Years later, most of us can do reasonably well doing addition "mentally" (some better than others!), but what I'm trying to say is this: what you're experiencing is "temporary" provided you persist, and work through many such problems. That's how factoring becomes "second nature"... – Namaste May 18 '11 at 18:43
• @Amy J.M. Oh, sorry for the misunderstandings. what I meant was systematically. I don't know how to get to the answer, unless I spent a lot of time trying get to the answer by just guessing a lot. And I may not get to it. I want to know how to get there without guessing. Is that possible? There should be a rule for it? Thanks! – Muazam May 18 '11 at 19:14
• See some of the links provided in the answers below: e.g., I answered with a link to the Khan Academy: the free video lectures/demos, problems, etc, for algebra...etc, will give you a better handle on factoring techniques, and sooooo much more! – Namaste May 18 '11 at 19:21
• @Muazam: I just wanted to add that my first comment above wasn't intended to "hand-wave" away your situation. I simply meant that you should try not to be discouraged: that persisting with these kinds of problems and asking questions will pay off. – Namaste May 18 '11 at 19:31
• @Muazam: In this case you see by inspection that $x=1$ is a root of $x^2 + x - 2=0$. Then you can divide $x^2 + x - 2$ by $x-1$ to find the second factor of $x^2 + x - 2$. – Américo Tavares May 18 '11 at 22:51

You might appreciate the series of Khan Academy lectures (+ practice problems, etc).

Each lesson is given via a short 12 minute video, which you can replay if needed: the link I'm including below should take you to "Algebra" lessons; but you might also want to look at "Developmental math I and II".

This is a great way to learn via video demonstrations, practice, repetition, etc., and if you stumble, say, with negative exponents, e.g., you can go directly to a lesson addressing the issue at hand. I believe there is diagnostic testing available (this is all free, no fees for watching videos, doing practice sets, tests, etc.), and such testing can really be helpful in learning where you need to direct your energies to move to the next level.

Edit: I've also come across this site, devoted to tutoring/demonstrating how to factor polynomials: it's developmental in approach, in that it starts slowly, with very common patterns used when factoring, and progresses in difficulty, building on what you've already learned, so best to approach the tutorial in the order given.

• J.M. Thanks for the link, the videos are great and very well explained. – Muazam May 18 '11 at 19:41

Given

$$A: x^2 + x - 2$$

you're trying to do the 'magic' in your head in order to get backwards to

$$B: (x+2)(x-1)$$

What is it that you are trying to do backwards. It's the original multiplication of $(x+2)(x-1)$. Note that

• the -2 in $A$ comes from multiplying the +2 and -1 in $B$
• the +1 (it's kind of invisible it's the coefficient of $x$ ) in $B$ comes from:
• $x$ in the first part times -1 in the second, plus
• +2 in the first part times $x$ in the second

or $(-1)+2 = +1$.

So that's how the multiplication works going forward. Now you have to think of that to go backwards. In $x^2 + x - 2$:

• where does the -2 come from? From two things that multiply to get -2. What could those possibly be? Usually we assume integers so the only possibilities are the two pairs 2, -1, and -2, 1.
• of those two pairs, they have to -add- to the coefficient for $x$ or just plain positive 1. So the answer has to be the pair 2 and -1.

Another example might help: given

$$x^2-5x+6$$

what does this factor to? (that is, find $(x-a)(x-b)$ which equals $x^2 -5x + 6$).

So the steps are:

• what are the factors of 6? (you should get 2 pairs, all negative.
• for those pairs, which pair adds up to -5?

The main difficulty is keeping track in your head of what is multiplying, what is adding, and what is positive and negative.

The pattern for any sort of problem solving skill like this that seems like magic (but really is not) is to:

• Do more examples to get a speedier feel for it.
• Check your work. Since you're going backwards, once you get a possible answer, you can do the non-magic (multiplying) to see if you can get the original item in the question.

As Ross pointed out, and as was previously discussed, we know that

$$(x+a)(x+b) = x(x+b) + a(x+b) = x^2 + bx + ax + ab = x^2 + (a+b)x + ab.$$

Therefore, to factor a quadratic expression $x^2 + cx + d$, all one has to do is find two numbers that multiply to $d$ and add to $c$. Let $m$ and $n$ be those two numbers that add to $c$ and multiply $d$. Then:

$$x^2 + cx + d = (x+m)(x+n).$$

Note that it doesn't matter which order we pick the two numbers since $a+b = b+a$ and $ab = ba$.

In your case, to factor $x^2 + x - 2 = x^2 + 1x - 2$, we need to find two numbers that add to $1$ and multiply to $-2$. Just by picking and trying different numbers, we find the two numbers are $2$ and $-1$.

A good way to go about finding the numbers is to determine all the factors of the constant term, which for us is $-2$. Since $-2 = 2 \cdot (-1) = (-2) \cdot 1$, we only have to check whether $2 + (-1)$ or $(-2) + 1$ equals one. Since it is the first pair of numbers with this property, these are the two numbers we were looking for. Therefore,

$$x^2 + x - 2 = (x+2)(x+(-1)) = (x+2)(x-1).$$

The reason this works: Suppose that we want to factor $x^2 + cx + d$, and we have found the numbers $m$ and $n$ such that $m + n = c$ and $mn = d$. Then:

\begin{align} x^2 + cx + d &= x^2 + (m+n)x + mn \\ &= x^2 + mx + nx + mn \\ &= x(x+m) + n(x+m) \\ &= (x+n)(x+m) \end{align}

You can do this: Write $x = 2x-x$. And then you have $$x^{2} + 2x - x-2 = x(x+2) -1 (x+2) = (x+2)(x-1)$$

Suppose you have a quadratic like $x^{2} + (a+b)x + ab$, then you actually want to see how $ab$ can be split up into some numbers such that their product is $ab$ and their sum is $a+b$. Now you have $-2$ which can be written as $2 \times -1$ and $2 - 1=-1$ which the co-efficient of $x$. That's why you write it as $x^{2} + 2x -x -2$. Another example, would be consider

• $x^{2} + 8x+12$. You want to factorize this. $12$ can be split up as $4 \times 3$ or $2 \times 6$. But $4+3=7$ which is not our $x$ coefficient. So we have to go with $2$ and $6$ and see that $x^{2} + 6x + 2x +12 = (x+6)\cdot (x+2)$.
• Write x=2x−x... Sorry, but why should I? – Did May 18 '11 at 18:40
• @Didier: To get the answer as $(x+2)(x-1)$ :) – user9413 May 18 '11 at 19:03
• @Chandru: I think that Didier was trying to make the point that writing x = 2x-x isn't that intuitive for a first learner, meaning that they see no reason for doing that. – Andy May 18 '11 at 19:18
• @Andy: I know andy was just joking :) – user9413 May 18 '11 at 19:19
• @Andy: Thats why I edited my answer to become more clearer. Hopefully that will help. – user9413 May 18 '11 at 19:19

I always keep Viete's formula in mind, it's particularly easy to apply to quadratics. It states that given a polynomial of form $ax^{2}+bx+c$ the roots $x_1$ and $x_2$ will satisfy the following equations:

$$x_1+x_2=-\frac{b}{a}$$ and

$$x_1 \times x_2=\frac{c}{a}$$

This allows for a quick and easy 'mental' calculation.

$(x+a)(x+b)=x^2+(a+b)x+ab$, so if you factor the constant term you can find possibilities and look to see if the middle term works out. In your case, $-2=-2*1=-1*2$, so you can look to see if $-2+1$ or $-1+2$ add to $+1$. This leads you to $(x+2)(x-1)$. If the $x^2$ term has a constant multiplying it, you have to factor it and match things up to look for possibilities.

Still one more way: if you don't want to remember the quadratic formula, then just remember how it is proved. Let's do an example: Factor $3x^2 + 4x - 5$.

1) What gives $3x^2$ when squared? Answer: $\sqrt{3}x$.

2) What number $a$ will I have to add to this to get $4x$ as one of the terms in $(\sqrt{3}x + a)^2$? Answer: The coefficient of $x$ in $(\sqrt{3}x + a)^2$ is $2\sqrt{3}a$, so the answer is $\frac{2}{\sqrt{3}}$.

3) What is $\left(\frac{2}{\sqrt{3}}\right)^2$? It is $\frac{4}{3}$.

From these considerations we get $$3x^2 + 4x - 5 = (\sqrt{3}x + \frac{2}{\sqrt{3}})^2 - \frac{4}{3} - 5 = (\sqrt{3}x + \frac{2}{\sqrt{3}})^2 - \frac{19}{3}.$$

Finally you can use the factorization $a^2 - b^2 = (a+b)(a-b)$ to conclude $$3x^2 + 4x - 5 = (\sqrt{3}x + \frac{2+\sqrt{19}}{\sqrt{3}})(\sqrt{3}x + \frac{2-\sqrt{19}}{\sqrt{3}}).$$

I'm not sure why this question was edited and thus returned to the top, but I have noticed a different answer. Sometimes it's nice to know that there is a uniform method that will always work for quadratics, if you get frustrated by trying one of the slicker methods.

Don't forget the quadratic formula. For $ax^2 + bx + c$, one can use that the roots are $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Here, that means the roots are $\dfrac{-1 \pm \sqrt{1 + 8}}{2} = 1, -2$. So then we know that the polynomial is given by $(x + 2)(x-1)$.

Alternatively, you might look at $x^2 + x - 2$ and immediately see that 1 is a root, so that you know that it factors as $(x-1)*$(some linear term). You could then divide out or solve for the linear term.

Just giving lots of options ;p