Algebra II/Factoring Rules

Depending on the type of polynomial given in the problem, the method used to factor will differ. The standard form of polynomial we often see is the trinomial $ax^{2}+bx+c$ where our goals are to find constants that sum up to $b$ and multiply to $c$ .

However, what if our polynomial is too complex? How do we factor $a^{2}-b^{2}$ ? What about $a^{2}+2ab+b^{2}$ ? Below are various forms of factoring that are commonly used.

1. GCF

Always take out the greatest common factor first!

For $6x^{2}y^{3}-10xy^{5}$ , greatest common factor is $2xy^{3}$ and so, we will factor out that whole term from $6x^{2}y^{3}$ and $10xy^{5}$ .

$6x^{2}y^{3}-10xy^{5}=2xy^{3}(3x-5y^{2})$

Firstly, the common term is $(x+y)$ and so, we must factor out that term and combine the remaining terms. After, we are left with another polynomial that is able to be factored using GCF.

$(x+y)a^{4}b^{9}c+(x+y)a^{2}bc^{5}=(x+y)(a^{4}b^{9}c+a^{2}bc^{5})$

$(x+y)(a^{4}b^{9}c+a^{2}bc^{5})=(x+y)a^{2}bc(a^{2}b^{8}+c^{4})$

2. Difference of two squares: a² - b² = (a+b)(a-b)

In the $a^{2}-b^{2}$ format, our primary goal is to separate both terms into "square" forms, meaning that it is a term multiplied by itself. The main feature of the difference of two squares is that both terms will be the same, but the signs in the factored form will differ: $(a+b)(a-b)$

$9x^{2}-49y^{2}=(3x+7y)(3x-7y)$

$121x^{2}-64=(11x-8)(11x+8)$

If there is a coefficient that can be factored out, do that step first before finding the difference of squares. Below, 3 is a factor of both $3x^{2}$ and $12$ as well as $3h^{2}$ and $-48$ .

$3x^{2}-12=3(x^{2}-4)$

$3(x^{2}-4)=3(x+2)(x-2)$

$3h^{2}-48=3(h^{2}-16)$

$3(h^{2}-16)=3(h+4)(h-4)$

3. Trinomial whose leading coefficient is one

When factoring in the format of $ax^{2}+bx+c$ , think about what two numbers will sum up to the $b$ term and what two numbers with multiply to the $c$ term.

$3x^{2}-18x+24=3(x^{2}-6x+8)$

$3(x^{2}-6x+8)=3(x-4)(x-2)$

Above, we are able to factor 3 out of the polynomial. When we have $x^{2}-6x+8$ , we have to find two numbers to sum to -6 and multiply to 8. We are able to accomplish this with -4 and -2 as our terms. $-4+(-2)=-6$ (our $b$ term) and $-4(-2)=8$ (our $c$ term).

We can treat $a^{2}b^{2}$ the same way we treat $ax^{2}$ in $ax^{2}+bx+c$ . $-8+12=4$ (our $b$ term) and $-8(12)=-96$ (our $c$ term).

$a^{2}b^{2}+4ab-96=(ab-8)(ab+12)$

4. Sum of two cubes: a³ + b³ = (a+b)(a² - ab + b²)

When we have a problem in the form of $a^{3}+b^{3}$ , we must break down our cubed variable and our cubed constant. The factored form will follow the format of $(a+b)(a^{2}-ab+b^{2})$ .

$x^{3}+125=(x+5)(x^{2}-5x+25)$

$27x^{3}-8=(3x+2)(9x^{2}-6x+4)$

$1000c^{3}-343b^{3}=(10c-7b)(100c^{2}+70bc+49b^{2})$

$8x^{3}-729=(2x-9)(4x^{2}+18x+81)$

5. Perfect Square Trinomials: a² + 2ab + b² = (a+b)²

Here, we have a trinomial in the format $a^{2}+2ab+b^{2}$ and we want an answer in the format of $(a+b)^{2}$ . These problems may seem tricky to start, but the key is to find the square that gives us the $a^{2}$ and $b^{2}$ terms and work from there; the $2ab$ term will complete itself.

Below, we know that $(3x)^{2}$ will result in our $a^{2}$ term of $9x^{2}$ and that $5^{2}$ will be our $b^{2}$ term of $25$ .

$9x^{2}+30x+25=(3x+5)(3x+5)$

Similarly, $(6c)^{2}$ will result in our $a^{2}$ term of $36c^{2}$ and that $(11d)^{2}$ will be our $b^{2}$ term of $121d^{2}$ .

$36c^{2}+132cd+121d^{2}=(6c+11d)(6c+11d)$

6. Factor by Grouping

If there are four or more terms, factor by grouping is the method you want to use. Find a common factor between the two terms of the problem as well as the other two terms. The factored term from both sets should be common.

$cm^{3}-dm^{3}-8c+8d=m^{3}(c-d)-8(c-d)$