Now that we know how to factor using various methods, let's try to apply them to different examples and figure out which would work for each one! Here's a list of all we've learnt so far (click on the links to read more about them):
- GCF factoring: $AB+AC+\dots=A\left(B+C+\dots\right)$AB+AC+…=A(B+C+…), can be with any number of terms, not just two. Just a case of finding the GCF.
- Difference of two squares: $A^2-B^2=\left(A+B\right)\left(A-B\right)$A2−B2=(A+B)(A−B), so look for the difference of two terms which are both perfect squares.
- Grouping in pairs: Look for four terms where you can split them up into two pairs and factor separately, then finally factor using basic factoring afterwards.
- Perfect squares: $A^2+2AB+B^2=\left(A+B\right)^2$A2+2AB+B2=(A+B)2, $A^2-2AB+B^2=\left(A-B\right)^2$A2−2AB+B2=(A−B)2, so look for three terms where the first and third terms are perfect squares, and the middle term is twice the product of their square roots.
- Monic Quadratics: Look for three terms of the form $x^2+Px+Q$x2+Px+Q, where $P$P and $Q$Q are any numbers (and $x$x could be another variable!). Try and see if you can solve using the perfect square method first, otherwise find two numbers $A$A and $B$B that have a sum of $P$P and a product of $Q$Q, and the answer would be $\left(x+A\right)\left(x+B\right)$(x+A)(x+B). OR you could use the cross method as well.
The key when facing questions involving these techniques is to figure out which to use and when. Remember to always check if your answer can be further factored to finish answering the question! Let's have a look at some examples:
Examples
Question 1
Factor $xy-5y-2x+10$xy−5y−2x+10 completely
Think about whether to take out positive or negative factors
Do: These four terms have no common factors so let's try grouping in pairs.
$y$y goes into the first two pairs and $2$2 goes into the last two.
So $xy-5y=y\left(x-5\right)$xy−5y=y(x−5) but $-2x+10=2\left(-x+5\right)$−2x+10=2(−x+5)
Therefore taking $2$2 out does not help us factor further so let's try $-2$−2 instead
| $xy-5y-2x+10$xy−5y−2x+10 | $=$= | $y\left(x-5\right)-2\left(x-5\right)$y(x−5)−2(x−5) | |
| $=$= | $\left(x-5\right)\left(y-2\right)$(x−5)(y−2) | using the basic factoring where $x-5$x−5 is the GCF |
Question 2
Factor $3x^2-3x-90$3x2−3x−90 completely
Think about which 3 term method to use here, and whether you'll need to use another method first
Do: This is a quadratic but non-monic.
We can not use the perfect squares technique here, but the three terms do have an GCF of $3$3, so let's factor that out first.
$3x^2-3x-90=3\left(x^2-x-30\right)$3x2−3x−90=3(x2−x−30)
The expression in the brackets is a monic quadratic that is also not a perfect square.
Factor pairs of $-30$−30 are $1$1 & $-30$−30, $-1$−1 & $30$30, $2$2 & $-15$−15, $-2$−2 & $15$15, $3$3 & $-10$−10, $-3$−3 & $10$10, $5$5 & $-6$−6, and $-5$−5 & $6$6
The only pair with a sum of $-1$−1 is $5$5 & $-6$−6.
Therefore:
$3\left(x^2-x-30\right)=3\left(x+5\right)\left(x-6\right)$3(x2−x−30)=3(x+5)(x−6)
Question 3
Factor $10x+50$10x+50.
Question 4
Factor $k^2-81$k2−81.
Question 5
Factor $x^2+12x+36$x2+12x+36.