A Quadratic Equation in Standard Form
(a, b, and c can have any value, except that a can't be 0.)
To "Factor" (or "Factorise" in the UK) a Quadratic is to:
find what to multiply to get the Quadratic
It is called "Factoring" because you find the factors (a factor is something you multiply by)
The factors of x2 + 3x - 4 are:
(x+4) and (x-1)
Why? Well, let us multiply them to see:
|(x+4)(x-1)||= x(x-1) + 4(x-1)|
|= x2 - x + 4x - 4|
|= x2 + 3x - 4|
Multiplying (x+4)(x-1) together is called Expanding.
In fact, Expanding and Factoring are opposites:
Expanding is easy, but Factoring can often be tricky
It can be like trying to find out what ingredients went into a cake to make it so delicious.
It is sometimes not obvious at all!
So let us try an example where we don't know the factors yet:
Guess and Check
Example: what are the factors of 2x2 + 7x + 3 ?
First, let us try to guess an answer, and then check if we are right ... we might get lucky!
We could guess (2x+3)(x+1):
(2x+3)(x+1) = 2x2 + 2x + 3x + 3 = 2x2 + 5x + 3 (WRONG)
How about (2x+7)(x-1):
(2x+7)(x-1) = 2x2 - 2x + 7x - 7 = 2x2 + 5x - 7 (WRONG AGAIN)
OK, how about (2x+9)(x-1):
(2x+9)(x-1) = 2x2 - 2x + 9x - 9 = 2x2 + 7x - 9 (WRONG AGAIN)
Oh No! We could be guessing for a long time before we get lucky!
So let us try something else:
A Method For Simple Cases
Luckily there is a method that works in simple cases.
Step 1: Find two numbers that multiply to give a×c, and add to give b.
Example: 2x2 + 7x + 3
ac is 2×3 = 6 and b is 7
So we want two numbers that multiply together to make 6, and add up to 7
In fact 6 and 1 do that (6×1=6, and 6+1=7)
How do you find 6 and 1?
It helps to list the factors of a×c=6, and then try adding some to get b=7.
Factors of 6 include 1, 2, 3 and 6. Aha! 1 and 6 add to 7, and 6×1=6.
Step 2: Rewrite the middle with those numbers:
Rewrite 7x with 6x and 1x:
2x2 + 6x + x + 3
Step 3: Factor the first two and last two terms separately:
The first two terms 2x2 + 6x factor into 2x(x+3)
The last two terms x+3 don't actually change in this case
So we get:
2x(x+3) + (x+3)
Step 4: If you've done this correctly, your two new terms should have a clearly visible common factor.
In this case you can see that (x+3) is common to both terms
So we can now rewrite it like this:
2x(x+3) + (x+3) = (2x+1)(x+3)
Check: (2x+1)(x+3) = 2x2 + 6x + x + 3 = 2x2 + 7x + 3 (Yes)
Much better than guessing!
Let us try another example:
Example: 6x2 + 5x - 6
Step 1: a×c is 6×(-6) = -36, and b is 5
List the positive factors of a×c = -36: 1, 2, 3, 4, 6, 9, 12, 18, 36
One of the numbers has to be negative to make -36, so by playing with a few different numbers I find that -4 and 9 work nicely:
-4×9 = -36 and -4+9 = 5
Step 2: Rewrite 5x with -4x and 9x:
6x2 - 4x + 9x - 6
Step 3: Factor first two and last two:
2x(3x - 2) + 3(3x -2)
Step 4: Common Factor is (3x - 2):
(2x+3)(3x - 2)
Check: (2x+3)(3x - 2) = 6x2 - 4x + 9x - 6 = 6x2 + 5x - 6 (Yes)
Finding Those Numbers
The hardest part is finding two numbers that multiply to give a×c, and add to give b.
It is partly guesswork, and it helps to list out all the factors.
Here is another example to assist you:
Example: a×c=-120 and b=7
What two numbers multiply to -120 and add to 7 ?
The factors of 120 are (plus and minus):
1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, and 120
We can try pairs of factors (start near the middle!) and see if they add to 7:
- -10 x 12 = -120, and -10+12 = 2 (no)
- -8 x 15 = -120 and -8+15 = 7 (YES!)
Well, one of the big benefits of factoring is that we can find the roots of the quadratic equation (where the equation is zero).
All we need to do (after factoring) is find where each of the two factors becomes zero
Example: what are the roots (zeros) of 6x2 + 5x - 6 ?
We already know (from above) the factors are
(2x + 3)(3x - 2)
And we can figure out that
(2x + 3) would be zero when x = -3/2
(3x - 2) would be zero when x = 2/3
So the roots of 6x2 + 5x - 6 are:
-3/2 and 2/3
Here is a plot of 6x2 + 5x - 6, can you see where it equals zero?
And you can also check it using a bit of arithmetic:
At x = -3/2: 6(-3/2)2 + 5(-3/2) - 6 = 6×(9/4) - 15/2 - 6 = 54/4 - 15/2 - 6 = 6-6 = 0
At x = 2/3: 6(2/3)2 + 5(2/3) - 6 = 6×(4/9) + 10/3 - 6 = 24/9 + 10/3 - 6 = 6-6 = 0
The General Solution
There is also a general solution (useful when the above method fails), which uses the quadratic formula:
Use that formula to get the two answers x+ and x− (one is for the "+" case, and the other is for the "−" case in the "±"), and you can get this factoring:
a(x - x+)(x - x−)
Let us use the previous example to see how that works:
Example: what are the roots of 6x2 + 5x - 6 ?
Substitute a=6, b=5 and c=-6 into the formula:
x = (-b ± √[b2 - 4ac]) / 2a
x = (-5 ± √[52 - 4·6·(-6)]) / 2·6
= (-5 ± √[25 + 144]) / 12
= (-5 ± √169) / 12
= (-5 ± 13) / 12
So the two roots are:
x+ = (-5 + 13) / 12 = 8/12 = 2/3,
x− = (-5 − 13) / 12 = -18/12 = -3/2
(Notice that we get the same result we did with the factoring we used before)
Now put those values into a(x - x+)(x - x−):
6(x - 2/3)(x + 3/2)
We can rearrange that a little to simplify it:
3(x - 2/3)·2(x + 3/2) = (3x - 2)(2x + 3)
And we get the same factors as we did before!
(Thanks to "mathsyperson" for parts of this article)