Lesson Explainer: Difference of Two Squares | Nagwa Lesson Explainer: Difference of Two Squares | Nagwa

Lesson Explainer: Difference of Two Squares Mathematics • Second Year of Preparatory School

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In this explainer, we will learn how to factor the difference of two squares and apply the result to evaluate numerical expressions.

Dina and Nabil are both asked to calculate 5149. They each approach the problem in a different way: Nabil directly calculates each of the squares and then finds the difference, whereas Dina finds the sum of 51 and 49 and the difference between 51 and 49 and multiplies her two answers. Dina’s method is clearly quicker, but is it correct? Let us look at each of Nabil’s and Dina’s working out. Nabil used long multiplication as follows: 51×51512550260149×494411960240126012401200

Dina’s working is shown here: (51+49)(5149)=100×2=200.

Clearly, the methods produce the same result and, in fact, we can easily show that Dina’s method is correct. If we expand the two brackets in the first stage of Dina’s working, we get 51+(49)(51)(49)(51)49, which simplifies to 5149.

This result is called the difference of two squares. We can also express this idea more generally to obtain the following formula.

Formula: Difference of Two Squares

An expression of the form 𝑎𝑏 is called a difference of two squares; it can always be factored by using the formula 𝑎𝑏=(𝑎+𝑏)(𝑎𝑏).

This is a powerful result that can be used to simplify algebraic and numerical expressions.

Let us start with a numerical example similar to the one we discussed above, showing how we can reduce a calculation involving large square numbers to a much easier one by treating it as a difference of two squares.

Example 1: Evaluating a Numerical Expression Using the Difference of Squares

By factorizing or otherwise, evaluate (77)(23).

Answer

Recall that a difference of two squares is of the form 𝑎𝑏 and can be factored by using the formula 𝑎𝑏=(𝑎+𝑏)(𝑎𝑏).

Comparing the given calculation (77)(23) with the expression 𝑎𝑏, we have 𝑎=(77) and 𝑏=(23). Taking square roots in both cases, we get 𝑎=77 and 𝑏=23.

We are now ready to apply the formula and evaluate the result: (77)(23)=(77+23)(7723)=(100)(54)=5400.

Thus, we have worked out that (77)(23)=5400.

We now show how to apply the formula to a simple algebraic expression.

Example 2: Factoring a Simple Difference of Two Squares

Factor the expression 𝑥49.

Answer

If we consider the given expression, we can see that the first term, 𝑥, is the square of 𝑥, and the second term, 49, is the square of 7. This means that the expression is a difference of two squares.

Recall that a difference of two squares is of the form 𝑎𝑏, which can be factored by using the formula 𝑎𝑏=(𝑎+𝑏)(𝑎𝑏).

Therefore, taking 𝑎=𝑥 and 𝑏=7, we can rewrite the expression as 𝑥7 and factor it as follows: 𝑥49=𝑥7=(𝑥+7)(𝑥7).

Next, we look at a slightly more complicated example, where the coefficient of 𝑥 in the original expression is greater than 1.

Example 3: Factoring a Difference of Two Squares

Factorize fully 64𝑥81.

Answer

In the given expression, notice that 64=8 and 81=9 are perfect squares, and 𝑥 is the square of 𝑥. This suggests we can rewrite 64𝑥81 as a difference of two squares.

Recall that a difference of two squares is of the form 𝑎𝑏, which can be factored by using the formula 𝑎𝑏=(𝑎+𝑏)(𝑎𝑏).

Therefore, we start by comparing 64𝑥81 with the general form 𝑎𝑏 to work out the values of 𝑎 and 𝑏.

For the first term, taking the square root of 64𝑥, we get 64𝑥=64𝑥=8𝑥, so 𝑎=8𝑥. For the second term, taking the square root implies that 𝑏=81=9.

Rewriting the original expression as a difference of two squares and applying the formula, we get 64𝑥81=(8𝑥)9=(8𝑥+9)(8𝑥9).

Hence, factoring the expression 64𝑥81 gives (8𝑥+9)(8𝑥9).

It is worth noting here that, for an expression of a difference of two squares in the form 𝑎𝑏, the terms 𝑎 and 𝑏 can contain both numbers and variables. In the previous two examples, 𝑏 was a number, but this is not always the case.

Example 4: Factoring a Difference of Two Squares with Two Variables

Factorize fully 100𝑥121𝑦.

Answer

In the given expression, notice that 100=10 and 121=11 are perfect squares, 𝑥 is the square of 𝑥, and 𝑦 is the square of 𝑦. This suggests that we can rewrite 100𝑥121𝑦 as a difference of two squares.

Recall that a difference of two squares is an expression of the form 𝑎𝑏, which can be factored by using the formula 𝑎𝑏=(𝑎+𝑏)(𝑎𝑏).

Comparing 100𝑥121𝑦 with 𝑎𝑏, by taking square roots, we get 𝑎=10𝑥 and 𝑏=11𝑦. We can, therefore, rewrite the expression as (10𝑥)(11𝑦), which factors as (10+11𝑦)(1011𝑦).

Another common assumption is that this method can only be applied to expressions that contain exponents of 2, for example, 9𝑥25 or 𝑎64𝑏. However, this is not true, as the next example shows.

Example 5: Factoring a Difference of Two Squares with Higher Exponents

Factorize fully 9𝑚64𝑛.

Answer

Though it is perhaps not immediately clear, the expression 9𝑚64𝑛 can be rewritten in the form 𝑎𝑏. Recall that an expression of the form 𝑎𝑏 is called a difference of two squares, which can be factored by using the formula 𝑎𝑏=(𝑎+𝑏)(𝑎𝑏).

In the first term, notice that 9=3 is a perfect square and 𝑚=𝑚, so 9𝑚=9𝑚=3𝑚, giving 𝑎=3𝑚. Similarly, in the second term, 64=8 is a perfect square and 𝑛=𝑛, so 64𝑛=64𝑛=8𝑛, giving 𝑏=8𝑛. This enables us to rewrite the original expression as the difference of two squares 3𝑚8𝑛. We can then factor it by applying the formula, as follows: 9𝑚64𝑛=3𝑚8𝑛=3𝑚+8𝑛3𝑚8𝑛.

Furthermore, we can still use this method even in cases where expressions feature more than two variables. Here is an example of this type.

Example 6: Factoring a Difference of Two Squares with Three Variables

Factorize fully 49𝑎64𝑏𝑐.

Answer

Notice that the numbers 49 and 64 are both perfect squares. This suggests that the given expression could be a difference of two squares. Recall that a difference of two squares is an expression of the form 𝑝𝑞, which can be factored by using the formula 𝑝𝑞=(𝑝+𝑞)(𝑝𝑞).

Comparing 49𝑎 with 𝑝 and 64𝑏𝑐 with 𝑞, we can work out the values of 𝑝 and 𝑞 by taking square roots. Therefore, 49𝑎=49𝑎=7𝑎, so 𝑝=7𝑎. Similarly, 64𝑏𝑐=64𝑏𝑐=8𝑏𝑐, giving 𝑞=8𝑏𝑐. Finally, we can apply the formula to get 49𝑎64𝑏𝑐=(7𝑎)8𝑏𝑐=7𝑎+8𝑏𝑐7𝑎8𝑏𝑐.

We conclude that completely factoring the expression 49𝑎64𝑏𝑐 gives 7𝑎+8𝑏𝑐7𝑎8𝑏𝑐.

In some questions, we can use this idea to rewrite a given expression as a difference of two squares and then use the formula to help us evaluate the result.

Example 7: Evaluating an Expression Using the Difference of Squares

If 𝑥𝑦=8, what is the value of (𝑥+3𝑦)(𝑥3𝑦)?

Answer

In this case, we are given the expression (𝑥+3𝑦)(𝑥3𝑦), which is in the form of a difference of two squares, 𝑎𝑏. Recall the formula for factoring a difference of two squares: 𝑎𝑏=(𝑎+𝑏)(𝑎𝑏).

Our strategy will be to work out the values of 𝑎 and 𝑏 and then apply the formula. Once we have done this, we can use information from the question to work out the numerical answer.

If 𝑎=(𝑥+3𝑦) and 𝑏=(𝑥3𝑦), then taking square roots gives 𝑎=𝑥+3𝑦 and 𝑏=𝑥3𝑦. Hence, applying the formula, we have (𝑥+3𝑦)(𝑥3𝑦)=((𝑥+3𝑦)+(𝑥3𝑦))((𝑥+3𝑦)(𝑥3𝑦)).

Simplifying the right-hand side, we get 𝑥+3𝑦+𝑥3𝑦𝑥+3𝑦𝑥+3𝑦=(2𝑥)(6𝑦)=12𝑥𝑦.

Therefore, we have used the formula to show that the original expression simplifies to 12𝑥𝑦. Moreover, the question tells us that 𝑥𝑦=8, so substituting for 𝑥𝑦 in the expression 12𝑥𝑦 gives the answer 12×8=96.

We conclude that if 𝑥𝑦=8, the value of (𝑥+3𝑦)(𝑥3𝑦) is 96.

In our final example, we show how the difference of two squares method can be applied in a geometric context to solve problems about right triangles. As we will show, this is not really surprising, because the relationship between side lengths in right triangles is given by the Pythagorean theorem, which is expressed in terms of the squares of the side lengths.

Example 8: Using the Difference of Squares to Find an Unknown Value given a Right Triangle

The length of the hypotenuse of a right triangle is 64 cm and the length of one of the other sides is 59 cm. Find the area of the square drawn on that unknown side.

Answer

In this question, we are given the lengths of the hypotenuse and one other side of a right triangle, and we must use this information to find the area of the square drawn on the unknown side. It helps to draw a sketch of the situation, as shown below. Note that if we label the unknown side length as 𝑐 cm long, then the square drawn on the unknown side will have an area of 𝑐 cm2.

We recall the Pythagorean theorem, which states that in any right triangle, the square of the hypotenuse is equal to the sum of the squares of the two shorter sides. Writing 𝑎 for the length of the hypotenuse and 𝑏 and 𝑐 for the lengths of the two shorter sides, this can be expressed algebraically as 𝑏+𝑐=𝑎.

We will first use the Pythagorean theorem to find an expression for 𝑐, the square of the unknown side length. Importantly, this will also be the area of the square drawn on the unknown side, which is what we have been asked to find.

Starting with the Pythagorean theorem 𝑏+𝑐=𝑎, we make 𝑐 the subject by subtracting 𝑏 from both sides: 𝑐=𝑎𝑏.

Notice now that the right-hand side, 𝑎𝑏, is in the form of a difference of two squares. Recall that a difference of two squares can be factored by using the formula 𝑎𝑏=(𝑎+𝑏)(𝑎𝑏).

Applying the formula, we get 𝑐=(𝑎+𝑏)(𝑎𝑏).

Finally, we have that 𝑎=64 and 𝑏=59, so substituting these values gives 𝑐=(64+59)(6459)=(123)(5)=615.

Since the side lengths were given in centimetres, then this area will be in square centimetres. Therefore, we have found that the area of the square drawn on the unknown side is 615 cm2.

Let us finish by recapping some key concepts from this explainer.

Key Points

  • An expression of the form 𝑎𝑏 is called a difference of two squares; it can always be factored by using the formula 𝑎𝑏=(𝑎+𝑏)(𝑎𝑏).
  • For an expression of a difference of two squares in the form 𝑎𝑏, the terms 𝑎 and 𝑏 can contain both numbers and multiple variables, including variables with exponents greater than 1.
  • We can rewrite numerical expressions as differences of two squares and then use the formula to simplify and evaluate them.
  • The difference of two squares method can also be applied to solve problems about right triangles. This is a natural context because the relationship between side lengths in right triangles is given by the Pythagorean theorem, which is expressed in terms of the squares of the side lengths.

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