Now that we know how to factor polynomials 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 learned so far (click on the links to read more about them):
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 factorized to finish answering the question! The complete factorization of polynomials has occurred when each factor is a prime polynomial.
Let's have a look at some examples:
Factor $\left(y+11y^2\right)^2-y^2$(y+11y2)2−y2 completely
Think: We can treat the expression in the parentheses as one term.
Do:
What we can see here is the difference of two squares, where our $A=y+11y^2$A=y+11y2 and $B=y$B=y
Therefore:
$\left(y+11y^2\right)^2-y^2$(y+11y2)2−y2 | $=$= | $\left(\left(y+11y^2\right)+y\right)\left(\left(y+11y^2\right)-y\right)$((y+11y2)+y)((y+11y2)−y) |
$=$= | $\left(2y+11y^2\right)\times11y^2$(2y+11y2)×11y2 | |
$=$= | $y\left(2+11y\right)\times11y^2$y(2+11y)×11y2 | |
$=$= | $11y^3\left(2+11y\right)$11y3(2+11y) |
Factor and simplify $\left(3m-n\right)\left(4n+7m\right)-2n+6m$(3m−n)(4n+7m)−2n+6m
Think: The first term has always been factorized, so we can factorize the rest of the expression first. Also be aware of negatives.
Do:
Let's concentrate on the $-2n+6m$−2n+6m part first. We can either take out $-2$−2 or $2$2 using GCF factorization.
If we take out $-2$−2 then we get $-2\left(n-3m\right)$−2(n−3m). If we take out $2$2 then we get $2\left(-n+3m\right)=2\left(3m-n\right)$2(−n+3m)=2(3m−n).
Since we have $\left(3m-n\right)$(3m−n) in the first term, we should take $2$2 out.
$\left(3m-n\right)\left(4n+7m\right)-2\left(n-3m\right)$(3m−n)(4n+7m)−2(n−3m) | $=$= | $\left(3m-n\right)\left(4n+7m\right)+2\left(3m-n\right)$(3m−n)(4n+7m)+2(3m−n) |
$=$= | $\left(3m-n\right)\left(4n+7m+2\right)$(3m−n)(4n+7m+2) |
Factor and simplify completely: $j^3-27j^2k+50jk^2$j3−27j2k+50jk2
Think: What kind of expression do we have after factoring the GCF $j$j?
Do:
$j^3-27j^2k+50jk^2$j3−27j2k+50jk2 | $=$= | $j\left(j^2-27jk+50k^2\right)$j(j2−27jk+50k2) |
Notice that $\left(j^2-26jk+50k^2\right)$(j2−26jk+50k2) is a quadratic trinomial.
We need two numbers that have a product of $50k^2$50k2. We can either have a $k^2$k2 term and a number, or two $k$k terms. Looking at the fact that we need the sum to be $-27k$−27k, that means we're looking for $2$2 negative $k$k terms.
Possible factor pairs of $50k^2$50k2 are$-50k$−50k & $-k$−k, $-25k$−25k & $-2k$−2k, and $-10k$−10k & $-5k$−5k.
$-25k$−25k & $-2k$−2k give us our sum of $-27k$−27k, therefore:
$j\left(j^2-27jk+50k^2\right)$j(j2−27jk+50k2) | $=$= | $j\left(j-25k\right)\left(j-2k\right)$j(j−25k)(j−2k) |
Factor the following expression:
$x^2-6x+9-y^2$x2−6x+9−y2.
Factor the expression $20x^2-25x-30$20x2−25x−30.
Factor the polynomial $\left(k+3\right)^3+8$(k+3)3+8 by using the substitution $u=k+3$u=k+3.
Use the structure of an expression to rewrite it in different equivalent forms. For example,see x^4 - y^4 as (x^2)^2 - (y^2)^2, thus recognizing it as a difference of squares that can be factored as (x^2 - y^2)(x^2 + y^2).