topic badge

2.03 Factoring polynomials completely


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):

  1. GCF factorization: $AB+AC+\dots=A\left(B+C+\dots\right)$AB+AC+=A(B+C+), can be with any number of terms, not just two. Just a case of finding the GCF. 
  2. Difference of two squares: $A^2-B^2=\left(A+B\right)\left(A-B\right)$A2B2=(A+B)(AB), so look for the difference of two terms which are both perfect squares. 
  3. Grouping in pairs: $2x^2+2x+3x+3=2x^2+2x+3x+3$2x2+2x+3x+3=2x2+2x+3x+3 Look for four terms where you can split them up into two pairs and factorize separately, then finally factorize using basic factorization afterwards $2x\left(x+1\right)+3\left(x+1\right)=\left(2x+3\right)\left(x+1\right)$2x(x+1)+3(x+1)=(2x+3)(x+1).
  4. Perfect squares: $A^2+2AB+B^2=\left(A+B\right)^2$A2+2AB+B2=(A+B)2, $A^2-2AB+B^2=\left(A-B\right)^2$A22AB+B2=(AB)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. 
  5. Quadratics with leading coefficient 1: 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. 
  6. Quadratics with leading coefficient not 1: Like quadratics with leading coefficient of $1$1, but the coefficient of $x^2$x2 is not one, e.g. $Px^2+Qx+R$Px2+Qx+R. Here we need four numbers, $A$A, $B$B, $C$C and $D$D where $A\times B=P$A×B=P, $C\times D=R$C×D=R and $A\times D+B\times C=Q$A×D+B×C=Q. This gives us the factorization $\left(Ax+C\right)\left(Bx+D\right)$(Ax+C)(Bx+D).
  7. Sums and differences of cubes: An expression of the form $A^3+B^3$A3+B3 has the factorization $\left(A+B\right)\left(A^2-AB+B^2\right)$(A+B)(A2AB+B2). Similarly, an expression of the form $A^3-B^3$A3B3 has the factorization $\left(A-B\right)\left(A^2+AB+B^2\right)$(AB)(A2+AB+B2).

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:

Worked examples

Question 1

Factor $\left(y+11y^2\right)^2-y^2$(y+11y2)2y2 completely

Think: We can treat the expression in the parentheses as one term.


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


$\left(y+11y^2\right)^2-y^2$(y+11y2)2y2 $=$= $\left(\left(y+11y^2\right)+y\right)\left(\left(y+11y^2\right)-y\right)$((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)
Question 2

Factor and simplify $\left(3m-n\right)\left(4n+7m\right)-2n+6m$(3mn)(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.


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(n3m). If we take out $2$2 then we get $2\left(-n+3m\right)=2\left(3m-n\right)$2(n+3m)=2(3mn).

Since we have $\left(3m-n\right)$(3mn) in the first term, we should take $2$2 out.

$\left(3m-n\right)\left(4n+7m\right)-2\left(n-3m\right)$(3mn)(4n+7m)2(n3m) $=$= $\left(3m-n\right)\left(4n+7m\right)+2\left(3m-n\right)$(3mn)(4n+7m)+2(3mn)
  $=$= $\left(3m-n\right)\left(4n+7m+2\right)$(3mn)(4n+7m+2)
Question 3

Factor and simplify completely: $j^3-27j^2k+50jk^2$j327j2k+50jk2

Think: What kind of expression do we have after factoring the GCF $j$j?


$j^3-27j^2k+50jk^2$j327j2k+50jk2 $=$= $j\left(j^2-27jk+50k^2\right)$j(j227jk+50k2)

Notice that $\left(j^2-26jk+50k^2\right)$(j226jk+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(j227jk+50k2) $=$= $j\left(j-25k\right)\left(j-2k\right)$j(j25k)(j2k)


Practice questions

Question 4

Factor the following expression:



Question 5

Factor the expression $20x^2-25x-30$20x225x30.


Question 6

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).

What is Mathspace

About Mathspace