So far, most of the quadratics we've dealt with are monic, meaning their $$x2 term only has a coefficient of $$1. If the coefficient is not $$1, then we've usually found we can factor out that coefficient from the whole quadratic.
eg. $$2x2−4x+6=2(x2−2x+3).
But how do we factor quadratics that can't be simplified in this way? First let's have a look at how a non-monic quadratic is composed:
Now we are more familiar with these tricky quadratics let's have a look at the three different methods below.
Cross Method
We've already encountered the cross method once before with monic quadratics, and it's easy to see how this extends into non-monic territory.
For example, let's have a look at $$5x2+11x−12. We must draw a cross with a possible pair of factors of $$5x2 on one side and another possible factor pair of $$−12 on the other side.
Let's start with the factor pairs of $$5x & $$x on the left, and $$−6 & $$2 on the other:
$$5x×2+x×(−6)=4x, which is incorrect, so let's try again with another two pairs:
$$5x×3+x×(−4)=11x which is the right answer. By reading across in the two circles, the quadratic must then factor to $$(5x−4)(x+3).
Factor By Grouping Method
The PSF (Product, Sum, Factor) method uses a similar idea we had with monic quadratics where we think about sums and products, but slightly different.
For a quadratic in the form $$ax2+bx+c:
1. Find two numbers, $$m & $$n, that have a SUM of $$b and a PRODUCT of $$ac.
2. Rewrite the quadratic as $$ax2+mx+nx+c.
3. Use grouping in pairs to factor the four-termed expression.
Example
question 1
Using the same example as above, factor $$5x2+11x−12 using the Factor By Grouping Method.
Think about what the sum and product of $$m & $$n should be
Do
We want the sum of of $$m & $$n to be $$11, and the product to be $$5×(−12)=−60.
The two numbers work out to be $$4 & $$−15, so:
| $$5x2+11x−12 | $$= | $$5x2−4x+15x−12 |
| $$= | $$x(5x−4)+3(5x−4) | |
| $$= | $$(5x−4)(x+3) |
This is the same answer that we got before!
PSF Variation
The above two methods are the most often used. However, a slightly different method can also be used to factor directly if you can remember the formula.
$$ax2+bx+c=(ax+m)(ax+n)a, where $$m+n=b & $$mn=ac
Example
question 2
Factor $$5x2−36x+7 completely
Think about whether it is easier to consider the product or the sum of $$m & $$n first
Do
| $$m+n | $$= | $$b |
| $$= | $$−36 | |
| $$mn | $$= | $$ac |
| $$= | $$5×7 | |
| $$= | $$35 |
It's much easier to look at the product first as there're less possible pairs that multiply to give $$35 than those that add to give $$−36. We can easily see that $$m & $$n $$= $$−1 & $$−35. Then:
| $$5x2−36x+7 | $$= | $$(5x−1)(5x−35)5 |
| $$= | $$(5x−1)(x−7)×55 | |
| $$= | $$(5x−1)(x−7) |
Worked Examples
Question 3
Factor the trinomial:
$$7x2−75x+50
Question 4
Factor the following trinomial:
$$6x2+13x+6
Question 5
Factor $$−12x2−7x+12.