Factoring the difference of two squares is a special case of factoring a polynomial, where you’ll be factoring a binomial which is a difference of two terms that are both perfect squares. For example, ???9x^2-16??? is the difference of two squares, because ???9x^2??? is the perfect square ???(3x)^2???, ???16??? is the perfect square ???4^2???, and ???16??? is being subtracted from ???9x^2???.
To factor a difference of two squares, you simply take the expressions that are squared (in the example the expressions ???3x??? and ???4???) and put both of them (in the given order) into the terms in both factors of the given binomial. In one factor, you’ll add the second expression to the first one; in the other factor, you’ll subtract the second expression from the first one. Therefore, ???9x^2-16??? is factored as ???(3x+4)(3x-4)???.
Example Factor the binomial. ???x^2-25??? Since ???x^2??? and ???25??? are both perfect squares (the squares of ???x??? and ???5???, respectively), ???x^2-25??? is factored as ???(x+5)(x-5)???
Let’s try another example of factoring the difference of two squares.
In one factor, you’ll add the second expression to the first one; in the other factor, you’ll subtract the second expression from the first one.
Example Factor the binomial. ???64x^4y^2-9z^6??? Notice that ???64x^4y^2=(8x^2y)^2???, and that ???9z^6=(3z^3)^2???. Therefore, ???64x^4y^2-9z^6??? is factored as ???(8x^2y+3z^3)(8x^2y-3z^3)???
Here we will learn how to factorise using the difference of two squares method for a quadratic in the form a2 – b2. There are also difference of two squares questions and an extensive worksheet based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.
The difference of two squares is a method of factorising used when an algebraic expression includes two squared terms, one subtracted from the other: a2 – b2 When we are subtracting a squared term from another squared term we can use the difference of two squares method. Square numbers are sometimes called perfect squares. An example of an expression we can factorise using the difference of two squares might be x2 – 4 or 4x2 – 25 To use quadratic factorisation on an expression in the form a2 – b2, we need double brackets.
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In order to factorise an algebraic expression using the difference of two squares:
Fully factorise x2 – 9 ( )( ) 2Square root the first term and write it on the left hand side of both brackets x2 – 9 √(x2) (x )(x ) 3Square root the last term and write it on the right hand side of both brackets. (x2 – 9) √9 (x 3)(x 3) 4Put a + in the middle of one bracket and a – in the middle of the other (the order doesn’t matter). (x + 3)(x – 3) We can check the final answer by multiplying out the brackets! (x + 3)(x – 3) = x2 – 9
Fully factorise 64 – y2
Square root the first term and write it on the left hand side of both brackets
Square root the last term and write it on the right hand side of both brackets
Put a + in the middle of one bracket and a – in the middle of the other (the order doesn’t matter.)
(8 + y)(8 – y) We can check the final answer by multiplying out the brackets! (8 – y)(8 + y) = 64 – y2
Fully factorise 25x2 – 16
Square root the first term and write it on the left hand side of both brackets
25x2 – 16 √(25x2) (5x )(5x )
Square root the last term and write it on the right hand side of both brackets
25x2 – 16 √16 (5x 4)(5x 4)
Put a + in the middle of one bracket and a – in the middle of the other (the order doesn’t matter.
(5x + 4)(5x – 4) We can check the final answer by multiplying out the brackets! (5x + 4)(5x – 4) = 25x2 – 16
Fully factorise 4x2 – 81y2
Square root the first term and write it on the left hand side of both brackets
4x2 – 81y2 √(4x2) (2x )(2x )
Square root the last term and write it on the right hand side of both brackets
4x2 – 81y2 √(81y2) (2x 9y)(2x 9y)
Put a + in the middle of one bracket and a – in the middle of the other (the order doesn’t matter.)
(2x + 9y)(2x – 9y) We can check the final answer by multiplying out the brackets! (2x + 9y)(2x – 9y) = 4x2 – 81y2
Fully factorise x3 – 64x Be careful, this one is not the difference of two squares! We first need to find the highest or greatest common factor (x) and write it outside of a single bracket. x(x2 – 64)
Write down two brackets with the x at the front
Square root the first term and write it on the left hand side of both brackets
x(x2 – 64) √(x2) x(x )(x )
Square root the last term and write it on the right hand side of both brackets
x(x2 – 64) √64 x(x 8)(x 8)
Put a + in the middle of one bracket and a – in the middle of the other (the order doesn’t matter.)
x(x + 8)(x – 8) We can check the final answer by multiplying out the brackets! x(x + 8)(x – 8) = x3 – 64x
There must simply be a + in one bracket, and a – in the other, the order doesn’t matter. ( + )( – ) or ( – )( + )
Remember to square root the entire term including the coefficients of the variables.E.g. 9x2 – 49 √(9x2) = 3x √49 = 7 The term factorising can sometimes be written as ‘factoring’ or factorization’ Practice difference of two squares questions
Recognising the expression as a difference of two squares means we can square root each term \begin{aligned} \sqrt{x^{2}}&=x\\ \sqrt{25}&=5\\ \end{aligned}and then rewrite as a product of two brackets (x+5)(x-5)
Recognising the expression as a difference of two squares means we can square root each term \begin{aligned} \sqrt{y^{2}}&=y\\ \sqrt{81}&=9\\ \end{aligned}and then rewrite as a product of two brackets (y+9)(y-9)
Recognising the expression as a difference of two squares means we can square root each term \begin{aligned} \sqrt{49}&=7\\ \sqrt{y^{2}}&=y\\ \end{aligned}and then rewrite as a product of two brackets (7+y)(7-y)
Recognising the expression as a difference of two squares means we can square root each term \begin{aligned} \sqrt{4}&=2\\ \sqrt{x^{2}}&=x\\ \end{aligned}and then rewrite as a product of two brackets (2+x)(2-x)
First, we can factor 4 out of each term. This gives 4(4x^{2}-25) Recognising the expression in the brackets as a difference of two squares means we can square root each term \begin{aligned} \sqrt{4x^{2}}&=2x\\ \sqrt{25}&=5\\ \end{aligned}and then rewrite as a product of two brackets muliplied by the 4. 4(2x+5)(2x-5)
Recognising the expression as a difference of two squares means we can square root each term \begin{aligned} \sqrt{49}&=7\\ \sqrt{9y^{2}}&=3y\\ \end{aligned}and then rewrite as a product of two brackets (7+3y)(7-3y)
Recognising the expression as a difference of two squares means we can square root each term \begin{aligned} \sqrt{81x^{2}}&=9x\\ \sqrt{16y^{2}}&=4y\\ \end{aligned}and then rewrite as a product of two brackets (9x+4y)(9x-4y)
The first thing we need to do is divide by the highest common factor of the two terms and rewrite the expression 4x^{3}-36x=4x(x^{2}-9)Recognising the expression in the bracket as a difference of two squares means we can square root each term \begin{aligned} \sqrt{x^{2}}&=x\\ \sqrt{9}&=3\\ \end{aligned}and then rewrite as a product of two brackets. Don’t forget to also multiply by the factor we initially divided by 4x(x+3)(x-3)Difference of two squares GCSE questions
(2 marks)
(2 marks)
= 2(x2 – 25) = 2(x + 5)(x – 5) (2 marks)
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How do you factor the difference of two squares
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