Difference of Squares Formula: Factorization & Solved Examples

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In Difference of Squares Formula, an expression of the form, a²- b², having the subtraction of two perfect squares is solved easily by splitting the expression into form of (a+b)(a-b), where a and b are Arbitrary Constants. This algebraic formula is made into use by the factorisation of the expression. Algebraic expressions contain variables and constants that are expressed along with algebraic operations.

Table of Content

Difference of Squares Formula
How to Factorize Differences of Squares?
Things to Remember
Sample Questions

Key Takeaways: Difference of Squares, Greatest Common Factor (GCF), Factorisation, Quadratic equation, Perfect Squares

Difference of Squares Formula

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To simplify the complex algebraic formula of the form, a² - b², having the expression as a difference of two perfect squares, the Difference of Squares Formula is used.

Difference of Squares Formula is given as:

a² - b² = (a+b)(a-b) OR a²- b² = (a-b)(a+b)

For example, when a=5 and b=2, then the difference of the perfect squares of a and b is given by,

LHS: a²- b²= 5²– 2²

5²– 2²= 25 - 4 = 21

RHS: (a+b)(a-b)=(5+2)(5-2)

(5+2)(5-2)=(7)(3)=21

Here, LHS = RHS

Therefore, it is proved that a²– b²= (a+b)(a-b)

Also Read: Difference between mean median and mode

How to Factorize Differences of Squares?

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When two perfect squares are to be subtracted, we factor out the terms using the Difference of Squares formula. The factorisation of the algebraic expression involving the subtraction of two perfect squares of the form, a²- b² is done the following way:

Firstly, check whether the given expression has terms having a greatest common factor (GCF) and then factor them out. The GCF must be included along with the simplified expression in the final answer.
Next, rewrite the expression in the form of a²– b², that is, the difference of two perfect squares for easy application of the formula to be used.
Now, apply the Difference of Squares formula and split the difference of perfect squares into the form, (a+b)(a-b). Again, check if there are any further terms that can be solved using the formula. This way the algebraic formula for difference of squares can be used for simplifying complex expressions.

Also Read: Trigonometry Values

Things to Remember

Using the Difference of Squares Formula, the difference of two perfect squares in a mathematical algebraic expression can be solved using the formula, a²– b²= (a+b)(a-b)
The factorisation of the difference of two perfect squares is done first by checking the greatest common factor (GCF) and then factoring the expression using the Difference of Squares formula. Make sure to add the GCF along the simplified terms in the final solved form.
The Sum of Squares Formula for algebraic expressions are given as, a²+ b²= (a+b)²– 2ab where a and b are arbitrary constants.
The Sum of Squares Formula for ‘n’ numbers is given as, 1²+ 2²+ 3²+....+ n²= \(\frac{n(n+1)(2n+1)}{6}\)6

Sample Questions

Ques. Factorise the expression x²y²– 64. (3 Marks)

Ans. Given the expression, x²y²– 64

Rewriting, we get x²y²– 64 = (xy)²- (8)²

Using the Difference of Squares Formula, we get

(xy)²– (8)²=(xy+8)(xy-8) [Since, a²– b²=(a+b)(a-b)]

Therefore, x²y²– 64= (xy+8)(xy-8)

Ques. Factorise the expression 1649 - 25m². (3 Marks)

Ans. Given the expression, 1649 - 25m²

Rewriting, we get 1649 - 25m²=(47)²- (5m)²

Using the Difference of Squares Formula, we get

(47)²– (5m)²= (47+5m)(47-5m) [Since, a²- b²=(a+b)(a-b)]

Therefore, 1649 - 25m²= (47+5m)(47-5m)

Ques. Factorise the expression as a difference between two squares -64p²+4q² (3 Marks)

Ans. Given the expression, -64p²+ 4q²

Rewriting, we get -64p²+ 4q²= 4q²– 64p²= (2q)²- (8p)²

Using the Difference of Squares Formula, we get

(2q)²– (8p)²= (2q+8p)(2q-8p) [Since, a²– b²=(a+b)(a-b)]

Therefore, -64p²+ 4q²= (2q+8p)(2q-8p)

Ques. Factorise the expression 4x⁶- y⁸. (2 Marks)

Ans. Given the expression, 4x⁶- y⁸

Rewriting, we get 4x⁶- y⁸= (2x³)²- (y⁴)²

Using the Difference of Squares Formula, we get

(2x³)²- (y⁴)²= (2x³+y⁴)(2x³-y⁴) [Since, a²- b²=(a+b)(a-b)]

Therefore, 4x⁶- y⁸= (2x³+y⁴)(2x³-y⁴)

Ques. Factorise the expression 2x²- 32. (3 Marks)

Ans. Given the expression, 2x²– 32

Here, the greatest common factor (GCF) is 2

Factoring out the GCF, we get 2x²- 32=2(x²-16)

Rewriting, we get 2(x²- 16)=2(x²– 4²)

Using the Difference of Squares Formula, we get

2(x²– 4²)=2(x+4)(x-4) [Since, a²- b²=(a+b)(a-b)]

Therefore, 2x²- 32=2(x+4)(x-4)

Ques. Factorise the expression 1-0.09y². (3 Marks)

Ans. Given the expression, 1- 0.09y²

Rewriting, we get 1- 0.09y²=(1)²- (0.3y)²

Using the Difference of Squares Formula, we get

(1)²- (0.3y)²= (1+0.3y)(1-0.3y) [Since, a²-b²=(a+b)(a-b)]

Therefore, 1-0.09y²=(1+0.3y)(1-0.3y)

Ques. Factorise the expression 4x²-81. (3 Marks)

Ans. Given the expression, 4x²-81

Rewriting, we get 4x²- 81=(2x)²- (9)²

Using the Difference of Squares Formula, we get

(2x)²- (9)²=(2x+9)(2x-9) [Since, a²– b²=(a+b)(a-b)]

Therefore, 4x²– 81=(2x+9)(2x-9)

Ques. Factorise the expression as a difference between two squares 9xy²-x³ (3 Marks)

Ans. Given the expression, 9xy²– x³

Here, the greatest common factor (GCF) is x

Factoring out the GCF, we get 9xy²– x³= x[(3y)²– x²]

Rewriting, we get 9xy²- x³= x[(3y)²– x²]

Using the Difference of Squares Formula, we get

x[(3y)²- x²= x(3y+x)(3y-x) [Since, a²- b²=(a+b)(a-b)]

Therefore, 9xy²-x³= x(3y+x)(3y-x)

Ques. Factorise the expression y²– 144. (3 Marks)

Ans. Given the expression, y²- 144

Rewriting, we get y²- 144 = y²– (12)²

Using the Difference of Squares Formula, we get

y²– (12)²= (y+12)(y-12) [Since, a²– b²=(a+b)(a-b)]

Therefore, y²- 144=(y+12)(y-12)

Ques. Factorise the expression as a difference between two squares 1-121a² (3 Marks)

Ans. Given the expression, 1-121a²

Rewriting, we get 1-121a²=(1²– (11a)²)

Using the Difference of Squares Formula, we get

(1²– (11a)²)=(1+11a)(1-11a) [Since, a²- b²=(a+b)(a-b)]

Therefore, 1-121a²= (1+11a)(1-11a)

Ques. Factorise the expression x³-36x. (3 Marks)

Ans. Given the expression, x³-36x

Here, the greatest common factor (GCF) is x

Factoring out the GCF, we get x³-36x = x(x²-36)

Rewriting, we get x(x²- 36)=x(x²- 6²)

Using the Difference of Squares Formula, we get

x(x²- 6²)=x(x+6)(x-6) [Since, a²- b²=(a+b)(a-b)]

Therefore, x³- 36x=x(x+6)(x-6)

Ques. Factorise the expression as a difference between two squares m⁴- n⁴. (3 Marks)

Ans. Given the expression, m⁴- n⁴

Rewriting, we get m⁴- n⁴=((m²)²- (n²)²)

Using the Difference of Squares Formula, we get

((m²)²-(n²)²)=(m²+n²)(m²- n²) —------(1) [Since, a²- b²= (a+b)(a-b)]

Now, factorising (m²– n²)in (1) using the Difference of Squares Formula, (1) becomes

(m²)²– (n²)²) = (m²+n²)(m+n)(m-n)

Therefore, m⁴- n⁴=(m²+n²)(m+n)(m-n)

Ques. Factorise the expression 169p²-1 (3 Marks)

Ans. Given the expression, 169p²-1

Rewriting, we get 169p²-1=(13p)²-(1)²

Using the Difference of Squares Formula, we get

(13p)²– (1)²= (13p+1)(13p-1) [Since, a²– b²= (a+b)(a-b)]

Therefore, 169p²– 1=(13p+1)(13p-1)

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Difference of Squares Formula: Factorization & Solved Examples

Difference of Squares Formula

How to Factorize Differences of Squares?

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.
A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is 3 cm more than its height. Twice of its length, thrice of its breadth and its height add up to 10 cm. Its breadth added to 7 times its height is 1 cm less than 3 times its length.

3.
Find:

If \[ (3\hat{i}-2\hat{j}+5\hat{k})\times(4\hat{i}+p\hat{j}+q\hat{k})=\vec{0} \] then find the values of \(p\) and \(q\).

4.
Using integration, find the area of the region bounded by the curve \( y = x|x| \), the x-axis, and the vertical lines \( x = -2 \) and \( x = 2 \).

5.
Find:

The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]

6.
At a birthday party, children are being served orange juice in conical cups, as shown in the figure.

Each cup is 15 cm deep and has a radius 5 cm. The juice is being poured into this cup at a rate of 0·1 cm3/s.
On the basis of the above information, answer the following questions :

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Difference of Squares Formula: Factorization & Solved Examples

Difference of Squares Formula

How to Factorize Differences of Squares?

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is 3 cm more than its height. Twice of its length, thrice of its breadth and its height add up to 10 cm. Its breadth added to 7 times its height is 1 cm less than 3 times its length.

3. Find: If \[ (3\hat{i}-2\hat{j}+5\hat{k})\times(4\hat{i}+p\hat{j}+q\hat{k})=\vec{0} \] then find the values of \(p\) and \(q\).

4.Using integration, find the area of the region bounded by the curve \( y = x|x| \), the x-axis, and the vertical lines \( x = -2 \) and \( x = 2 \).

5. Find: The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]

6.At a birthday party, children are being served orange juice in conical cups, as shown in the figure. Each cup is 15 cm deep and has a radius 5 cm. The juice is being poured into this cup at a rate of 0·1 cm3/s.On the basis of the above information, answer the following questions :

Comments

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1.
A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is 3 cm more than its height. Twice of its length, thrice of its breadth and its height add up to 10 cm. Its breadth added to 7 times its height is 1 cm less than 3 times its length.

3.
Find:

If \[ (3\hat{i}-2\hat{j}+5\hat{k})\times(4\hat{i}+p\hat{j}+q\hat{k})=\vec{0} \] then find the values of \(p\) and \(q\).

4.
Using integration, find the area of the region bounded by the curve \( y = x|x| \), the x-axis, and the vertical lines \( x = -2 \) and \( x = 2 \).

5.
Find:

The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]

6.
At a birthday party, children are being served orange juice in conical cups, as shown in the figure.

Each cup is 15 cm deep and has a radius 5 cm. The juice is being poured into this cup at a rate of 0·1 cm3/s.
On the basis of the above information, answer the following questions :