1. Factor the trinomial

$a^2+a-12$

First, I see $a^2$, so I know the first factors must be $a$, and I see the last is a negative, so I know the signs are different

$a^2+a-12 = (a+?) (a-?)$

but what comes next? Do I have to try every combination of factors, and use FOIL until I find the correct combination? This is possible when I am working with these small numbers, because 12 only has a few factors (1*12,2*6,3*4), but what if it is made of larger numbers which could have dozens of different factors? It could take a whole day to finish a single problem.

This one is even more difficult, I also have to find the correct factors of 4

$4x^2-8x-21$

Aren't there any tricks to make this more simple?

2. Originally Posted by bryang
$a^2+a-12$

First, I see $a^2$, so I know the first factors must be $a$, and I see the last is a negative, so I know the signs are different

$a^2+a-12 = (a+?) (a-?)$

but what comes next? Do I have to try every combination of factors, and use FOIL until I find the correct combination? This is possible when I am working with these small numbers, because 12 only has a few factors (1*12,2*6,3*4), but what if it is made of larger numbers which could have dozens of different factors? It could take a whole day to finish a single problem.

This one is even more difficult, I also have to find the correct factors of 4

$4x^2-8x-21$

Aren't there any tricks to make this more simple?
In general, when factoring you want to find sums or differences that give you the coefficient of the first-order term. So your first case, you want to get 1, which is the coefficient of $a$. And in the second case, you want to get -8, the coefficient of $-8x$

For the first problem, you look at the factors of 12 and see the following possibilities:
(1,12), (2,6), (3,4)

The only pair whose sum or difference generates a 1 is the pair (3,4). So your factorization is $(x-4)(x+3)$

For the second problem, look at the factors for 4:
(1,4) and (2,2)

The factors for 21 are:
(1,21) and (3,7)

So you have four possible pairings. We see that -7(2) + 2(3) = -8, so the pairs that we seek are: (2,2) and (3,7). Your factorization gives you: $(2x - 7)(2x + 3)$

It comes with experience.

Hello bryang
Originally Posted by bryang
Aren't there any tricks to make this more simple?
Try this if you want a trick. I'll give the method first, then show you an example or two.

To factorise an expression like $ax^2+bx+c$:

• Step 1: Multiply the 'outside' numbers together (that's $a \times c$), and note the sign of the answer.
• Step 2: If the answer is positive, you now need two factors of the answer whose sum is the middle number (that's $b$); if it's negative, you need two factors whose difference is the 'middle' number.
• Step 3: Write the $x$-term (the 'middle' number) as a sum or difference, using the numbers you found in Step 2. (Be careful with minus signs!)
• Step 4: Take out common factors to complete the factorisation.

Example 1

$6x^2 +25x+14$

(Step 1) $6 \times 14 = 84$

(Step 2) Positive; $4 \times 21 = 84$; the sum of $4$ and $21$ is $25$

(Step 3) $6x^2 \textcolor{red}{+25x}+14$

$= 6x^2 \textcolor{red}{ + 4x + 21x} +14$

(Step 4) $= \textcolor{blue}{2x}(3x+2) \textcolor{blue}{+7}(3x+2)$

$= \textcolor{blue}{(2x+7)}(3x+2)$

Example 2

$10x^2-31x-36$

(Step 1) $10 \times -36 = -360$

(Step 2) Negative; $9 \times 40 = 360$; the difference between $9$ and $40$ is $31$

(Step 3) $10x^2\textcolor{red}{-31x}-36$

$= 10x^2$ $\textcolor{red}{ + 9x - 40x} - 36$
(Be careful with minus signs!)

(Step 4) $=$ $\textcolor{blue}x(10x+9)\textcolor{blue}{-4}(10x+9)$
(Be extra careful with minus signs!!)

$=$ $\textcolor{blue}{(x-4)}(10x +9)$

At first, you may think that this method makes things more complicated - which it may do for simple cases, where there aren't many factors. But when you have lots of factors to consider, you'll find it involves much less trial and error!

Give it a try.