From Zero to Engineer

Factoring algebraic expressions

Taking out a common factor, factoring by grouping, the algebraic identities, and writing a quadratic expression as a product of two linear factors.

Factoring is the reverse of multiplying: instead of expanding brackets, we look for them again. There are three ways of doing it, and it pays to recognise them at once. If every term has something in common, take it outside the bracket. If not, but the terms can be paired up, look for a common factor in each pair. And if the expression matches one of the algebraic identities, the factorisation is immediate.

Factoring algebraic expressions

Common factors

\text{a)}\quad 8x^{4}y^{3} + 6x^{3}y^{2} = 2x^{3}y^{2}(4xy + 3)

\text{b)}\quad 15a^{3}b - 9a^{2}b^{2} = 3a^{2}b(5a - 3b)

Common factors obtained by grouping

\begin{aligned}\text{a)}\quad x^{3} - 4x^{2}y + xy^{2} - 4y^{3} &= x^{2}(x - 4y) + y^{2}(x - 4y) = \\&= (x^{2} + y^{2})(x - 4y)\end{aligned}

\begin{aligned}\text{b)}\quad 20x^{2} - 3y^{2} + 4xy^{2} - 15x &= 20x^{2} - 15x + 4xy^{2} - 3y^{2} = \\&= 5x(4x - 3) + y^{2}(4x - 3) = (5x + y^{2})(4x - 3)\end{aligned}

Algebraic identities (special products)

\text{1)}\quad (a + b)^{2} = a^{2} + 2ab + b^{2}

the square of a sum

(x + 5)^{2} = x^{2} + 10x + 25

\text{2)}\quad (a - b)^{2} = a^{2} - 2ab + b^{2}

the square of a difference

(2a - 3)^{2} = 4a^{2} - 12a + 9

\text{3)}\quad (a - b)(a + b) = a^{2} - b^{2}

the difference of two squares

(5x - 4y)(5x + 4y) = 25x^{2} - 16y^{2}

\text{a)}\quad 16x^{2} + 40xy + 25y^{2} = (4x + 5y)^{2}

\text{b)}\quad 9x^{2} - 12xy + 4y^{2} = (3x - 2y)^{2}

\begin{aligned}\text{c)}\quad (2x + 3y)^{2} - (x - 4y)^{2} &= [2x + 3y - (x - 4y)][2x + 3y + (x - 4y)] = \\&= (2x + 3y - x + 4y)(2x + 3y + x - 4y) = \\&= (x + 7y)(3x - y)\end{aligned}

Here a stands for 2x + 3y and b for x − 4y, so this is the difference of two squares.

Quadratic expressions written as a product of linear factors

\text{1)}\quad (x + a)(x + b) = x^{2} + (a + b)x + ab

a and b are natural numbers

x^{2} + 5x + 6 = (x + 2)(x + 3)

\text{2)}\quad (x - a)(x - b) = x^{2} - (a + b)x + ab

x^{2} - 9x + 20 = (x - 4)(x - 5)

\text{3)}\quad (x + a)(x - b) = x^{2} + (a - b)x - ab

x^{2} - 2x - 24 = (x + 4)(x - 6)

\text{a)}\quad x^{2} + 7x + 12 = (x + 3)(x + 4)

\text{d)}\quad x^{2} + 2x - 24 = (x + 6)(x - 4)

\text{b)}\quad x^{2} - 11x + 28 = (x - 4)(x - 7)

\text{e)}\quad x^{2} - 2x - 35 = (x + 5)(x - 7)

\text{c)}\quad x^{2} - 3x - 18 = (x + 3)(x - 6)

\text{f)}\quad x^{2} - 10x + 16 = (x - 2)(x - 8)

Review exercises

Factor the following expressions.

\text{a)}\quad 18xy^{3} - 8x^{3}y = 2xy(9y^{2} - 4x^{2}) = 2xy(3y - 2x)(3y + 2x)

\begin{aligned}\text{b)}\quad x^{3} - 6x^{2}y - 2xy + 12y^{2} &= x^{2}(x - 6y) - 2y(x - 6y) = \\&= (x^{2} - 2y)(x - 6y)\end{aligned}

\begin{aligned}\text{c)}\quad 16x^{2} - 24xy - 18x + 27y &= 8x(2x - 3y) - 9(2x - 3y) = \\&= (8x - 9)(2x - 3y)\end{aligned}

\begin{aligned}\text{d)}\quad (x - 2y)^{2} - (2x - y)^{2} &= [x - 2y - (2x - y)][x - 2y + (2x - y)] = \\&= (x - 2y - 2x + y)(x - 2y + 2x - y) = (-x - y)(3x - 3y) = \\&= -3(x + y)(x - y)\end{aligned}

\text{e)}\quad x^{2} + 7x - 30 = (x + 10)(x - 3)

\text{f)}\quad 4x^{2} - 36 = (2x - 6)(2x + 6) = 4(x - 3)(x + 3)

\text{g)}\quad x^{2} + 10x + 25 = (x + 5)^{2}

\begin{aligned}\text{h)}\quad 3x^{2} - 11x - 4 &= 3x^{2} - 12x + x - 4 = 3x(x - 4) + (x - 4) = \\&= (3x + 1)(x - 4)\end{aligned}

Frequently asked questions

Where do you start when factoring?

Always with a factor common to every term. In 18xy³ − 8x³y the common part is 2, x and y, which gives 2xy(9y² − 4x²), and only then does the difference of two squares appear inside the bracket.

What is factoring by grouping?

Split the terms into pairs so that each pair has a common factor and the brackets come out the same. Then take that bracket out once more: x³ − 4x²y + xy² − 4y³ gives x²(x − 4y) + y²(x − 4y), that is (x² + y²)(x − 4y).

How do you factor the quadratic x² + 7x + 12?

Look for two numbers that add up to 7 and multiply to 12. Those are 3 and 4, so the quadratic is (x + 3)(x + 4). When the constant term is negative, the two numbers have opposite signs.

What if the coefficient of x² is not 1?

Split the middle term so that grouping becomes possible. In 3x² − 11x − 4 write −11x as −12x + x, which gives 3x(x − 4) + (x − 4), that is (3x + 1)(x − 4).

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Algebra - review exercises