ICSE Class 8 Maths Chapter 14 Special Products and Expansions PDF Download

Chapter 14: Special Products and Expansions

14.1 Review

Special Products

The multiplications of certain types of expressions can be obtained by direct or short cut method. Such multiplications are known as special products.

For example (The product of two binomials):

1. \((x + a)(x + b) = x(x + b) + a(x + b) = x^2 + bx + ax + ab = x^2 + ax + bx + ab = x^2 + (a + b)x + ab\)

2. \((x + a)(x - b) = x(x - b) + a(x - b) = x^2 - bx + ax - ab = x^2 + ax - bx - ab = x^2 + (a - b)x - ab\)

3. \((x - a)(x + b) = x(x + b) - a(x + b) = x^2 + bx - ax - ab = x^2 - ax + bx - ab = x^2 - (a - b)x - ab\)

4. \((x - a)(x - b) = x(x - b) - a(x - b) = x^2 - bx - ax + ab = x^2 - ax - bx + ab = x^2 - (a + b)x + ab\)

Examples (Using direct method)

1. \((x + 5)(x + 3) = x^2 + (5 + 3)x + 5 \times 3 = x^2 + 8x + 15\)

2. \((x + 5)(x - 3) = x^2 + (5 - 3)x - 5 \times 3 = x^2 + 2x - 15\)

3. \((x - 5)(x + 3) = x^2 - (5 - 3)x - 5 \times 3 = x^2 - 2x - 15\)

4. \((x - 5)(x - 3) = x^2 - (5 + 3)x + 5 \times 3 = x^2 - 8x + 15\)

Test Yourself

1. \((x + 15)(x + 4) = \) _____________ \(= \) _____________

2. \((x + 15)(x - 4) = \) _____________ \(= \) _____________

3. \((x - 15)(x + 4) = \) _____________ \(= \) _____________

4. \((x - 15)(x - 4) = \) _____________ \(= \) _____________

14.2 Important

While using direct method, the product of two binomials gives three terms:

(i) The first term = Product of the first terms of the two binomials

(ii) The middle term = (First term of first binomial × second term of second binomial) + (second term of first binomial × first term of second binomial) = Product of outer terms + Product of inner terms

(iii) The third term = Product of the second terms of the two binomials.

Example 1

Evaluate:

(i) \((2x + 3y)(3x + 4y)\)

(ii) \((2a + 3)(5a - 7)\)

(iii) \((4a - 3b)(2a + 5b)\)

(iv) \((7x - 3)(2x - 9)\)

Solution

(i) \((2x + 3y)(3x + 4y) = (2x \times 3x) + (2x \times 4y + 3y \times 3x) + (3y \times 4y) = 6x^2 + (8xy + 9xy) + (12y^2) = 6x^2 + 17xy + 12y^2\)

(ii) \((2a + 3)(5a - 7) = (2a \times 5a) + (2a \times -7 + 3 \times 5a) + (3 \times -7) = 10a^2 + (-14a + 15a) + (-21) = 10a^2 + a - 21\)

(iii) \((4a - 3b)(2a + 5b) = (4a \times 2a) + (4a \times 5b + -3b \times 2a) + (-3b \times 5b) = 8a^2 + (20ab - 6ab) + (- 15b^2) = 8a^2 + 14ab - 15b^2\)

(iv) \((7x - 3)(2x - 9) = (7x \times 2x) + (7x \times -9 + -3 \times 2x) + (-3 \times -9) = 14x^2 + (-63x - 6x) + (27) = 14x^2 - 69x + 27\)

Teacher's Note

When you buy items from a store at different quantities and prices, you are essentially using binomial multiplication to calculate the total cost - the outer and inner terms represent different combinations of items and their prices.

14.3 Product of Sum and Difference of Two Terms

Consider the two terms 5x and 4y. the sum of these two terms = 5x + 4y and the difference of these terms = 5x - 4y. And, the product of their sum and their difference

\(= (5x + 4y)(5x - 4y) = 5x(5x - 4y) + 4y(5x - 4y) = 25x^2 - 20xy + 20xy - 16y^2 = 25x^2 - 16y^2 = (5x)^2 - (4y)^2 = \text{(First Term)}^2 - \text{(Second Term)}^2\)

Test Yourself

5. \((x + 3)(x - 3) = \) _____________ _____________ \(= \) _____________

6. \((3x + 4y)(3x - 4y) = \) _____________ _____________ \(= \) _____________

7. \((1.6x^2 - 5)(1.6x^2 + 5) = \) _____________ _____________ \(= \) _____________

8. \((5a^2 + 8b)(5a^2 - 8b) = \) _____________ _____________ \(= \) _____________

Example 2

Evaluate:

(i) \((x - 2)(x + 2)(x^2 + 4)\)

(ii) \((2a - 5b)(2a + 5b)(4a^2 + 25b^2)\)

Solution

(i) \((x - 2)(x + 2)(x^2 + 4) = [(x - 2)(x + 2)](x^2 + 4) = (x^2 - 2^2)(x^2 + 4) = (x^2 - 4)(x^2 + 4) = (x^2)^2 - (4)^2 = x^4 - 16\)

(ii) \((2a - 5b)(2a + 5b)(4a^2 + 25b^2) = [(2a - 5b)(2a + 5b)](4a^2 + 25b^2) = [(2a)^2 - (5b)^2](4a^2 + 25b^2) = (4a^2 - 25b^2)(4a^2 + 25b^2) = (4a^2)^2 - (25b^2)^2 = 16a^4 - 625b^4\)

Example 3

Use of the formula \((a + b)(a - b) = a^2 - b^2\) to find the value of:

(i) \(107 \times 93\)

(ii) \(30.8 \times 29.2\)

Solution

(i) \(107 \times 93 = (100 + 7)(100 - 7) = (100)^2 - (7)^2 = 10000 - 49 = 9951\)

(ii) \(30.8 \times 29.2 = (30 + 0.8)(30 - 0.8) = (30)^2 - (0.8)^2 = 900 - 0.64 = 899.36\)

Teacher's Note

When calculating areas of rectangular yards or land parcels where dimensions vary slightly, using the difference of squares formula allows quick mental math without needing a calculator.

Exercise 14 (A)

1. Use direct method to evaluate the following products:

(i) \((x + 3)(x + 3)\)

(ii) \((y + 5)(y - 3)\)

(iii) \((a - 8)(a + 2)\)

(iv) \((b - 3)(b - 5)\)

(v) \((3x - 2y)(2x + y)\)

(vi) \((5a + 16)(3a - 7)\)

(vii) \((8 - b)(3 + b)\)

2. Use direct method to evaluate:

(i) \((x + 1)(x - 1)\)

(ii) \((2 + a)(2 - a)\)

(iii) \((3b - 1)(3b + 1)\)

(iv) \((4 + 5x)(4 - 5x)\)

(v) \((2a + 3)(2a - 3)\)

(vi) \((xy + 4)(xy - 4)\)

(vii) \((ab + x^2)(ab - x^2)\)

(viii) \((3x^2 + 5y^2)(3x^2 - 5y^2)\)

(ix) \(\left(z - \frac{2}{3}\right)\left(z + \frac{2}{3}\right)\)

(x) \(\left(\frac{3}{5}a + \frac{1}{2}\right)\left(\frac{3}{5}a - \frac{1}{2}\right)\)

(xi) \((0.5 - 2a)(0.5 + 2a)\)

(xii) \(\left(\frac{a}{2} - \frac{b}{3}\right)\left(\frac{a}{2} + \frac{b}{3}\right)\)

3. Evaluate:

(i) \((a + 1)(a - 1)(a^2 + 1)\)

(ii) \((a + b)(a - b)(a^2 + b^2)\)

(iii) \((2a - b)(2a + b)(4a^2 + b^2)\)

(iv) \((3 - 2x)(3 + 2x)(9 + 4x^2)\)

(v) \((3x - 4y)(3x + 4y)(9x^2 + 16y^2)\)

4. Use the product \((a + b)(a - b) = a^2 - b^2\) to evaluate:

(i) \(21 \times 19\)

(ii) \(33 \times 27\)

(iii) \(103 \times 97\)

(iv) \(9.8 \times 10.2\)

(v) \(7.7 \times 8.3\)

(vi) \(4.6 \times 5.4\)

5. Evaluate:

(i) \((6 - xy)(6 + xy)\)

(ii) \(\left(7x + \frac{2}{3}y\right)\left(7x - \frac{2}{3}y\right)\)

(iii) \(\left(\frac{a}{2b} + \frac{2b}{a}\right)\left(\frac{a}{2b} - \frac{2b}{a}\right)\)

(iv) \(\left(3x - \frac{1}{2y}\right)\left(3x + \frac{1}{2y}\right)\)

(v) \((2a + 3)(2a - 3)(4a^2 + 9)\)

(vi) \((a + bc)(a - bc)(a^2 + b^2c^2)\)

(vii) \((5x + 8y)(3x + 5y)\)

(viii) \((7x + 15y)(5x - 4y)\)

(ix) \((2a - 3b)(3a + 4b)\)

(x) \((9a - 7b)(3a - b)\)

14.4 Expansions

In expansion, we study the multiplication of an expression by itself to obtain its second, third or higher power.

1. \((a + b)^2 = (a + b)(a + b) = a^2 + ab + ab + b^2 = a^2 + 2ab + b^2\)

(Sum of two terms)^2 = (1st term)^2 + 2 × 1st term × 2nd term + (2nd term)^2

2. \((a - b)^2 = (a - b)(a - b) = a^2 - ab - ab + b^2 = a^2 - 2ab + b^2\)

(Difference of two terms)^2 = (1st term)^2 - 2 × 1st term × 2nd term + (2nd term)^2

Examples

1. \((3x + 4y)^2 = \text{(1st term)}^2 + 2 \times \text{1st term} \times \text{2nd term} + \text{(2nd term)}^2 = (3x)^2 + 2 \times 3x \times 4y + (4y)^2 = 9x^2 + 24xy + 16y^2\)

2. \(\left(\frac{3x}{2y} - \frac{2y}{3x}\right)^2 = \text{(1st term)}^2 - 2 \times \text{1st term} \times \text{2nd term} + \text{(2nd term)}^2 = \left(\frac{3x}{2y}\right)^2 - 2 \times \frac{3x}{2y} \times \frac{2y}{3x} + \left(\frac{2y}{3x}\right)^2 = \frac{9x^2}{4y^2} - 2 + \frac{4y^2}{9x^2}\)

3. \((208)^2 = (200 + 8)^2 = (200)^2 + 2 \times 200 \times 8 + (8)^2 = 40000 + 3200 + 64 = 43264\)

4. \((9.7)^2 = (10 - 0.3)^2 = (10)^2 - 2 \times 10 \times 0.3 + (0.3)^2 = 100 - 6 + 0.09 = 94.09\)

Test Yourself

Using expansions, evaluate:

9. \(\left(2a - \frac{3}{2}\right)^2 = \) _________________________ \(= \) _______________

10. \(\left(x + \frac{1}{2x}\right)^2 = \) _________________________ \(= \) _______________

11. \((2x^2 - 3y)^2 = \) _________________________ \(= \) ___________________________

12. \((107)^2 = \) _____________ \(= \) ____________________________ \(= \) _____________

13. \((97)^2 = \) _____________ \(= \) ____________________________ \(= \) _____________

14. \((10.6)^2 = \) _____________ \(= \) ____________________________ \(= \) _____________

15. \((19.8)^2 = \) _____________ \(= \) ____________________________ \(= \) _____________

14.5 Important Formulae to be Memorised

1. \((a + b)^2 = a^2 + b^2 + 2ab\)

2. \((a - b)^2 = a^2 + b^2 - 2ab\)

3. \(\left(a + \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} + 2\)

4. \(\left(a - \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} - 2\)

5. \((a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = a^2 + b^2 + c^2 + 2(ab + bc + ca)\)

6. \((a + b - c)^2 = a^2 + b^2 + (-c)^2 + 2(a \times b) + 2(b \times -c) + 2(-c \times a) = a^2 + b^2 + c^2 + 2ab - 2bc - 2ca\)

Example 4

Expand:

(i) \(\left(2x + \frac{1}{2x}\right)^2\)

(ii) \(\left(3a - \frac{1}{a}\right)^2\)

(iii) \((a + 2b - 5c)^2\)

(iv) \((a - 2b - 5c)^2\)

Solution

(i) \(\left(2x + \frac{1}{2x}\right)^2 = (2x)^2 + \left(\frac{1}{2x}\right)^2 + 2 \times 2x \times \frac{1}{2x} = 4x^2 + \frac{1}{4x^2} + 2\)

(ii) \(\left(3a - \frac{1}{a}\right)^2 = (3a)^2 + \left(\frac{1}{a}\right)^2 - 2 \times 3a \times \frac{1}{a} = 9a^2 + \frac{1}{a^2} - 6\)

(iii) \((a + 2b - 5c)^2 = (a)^2 + (2b)^2 + (-5c)^2 + 2(a \times 2b) + 2(2b \times -5c) + 2(-5c \times a) = a^2 + 4b^2 + 25c^2 + 4ab - 20bc - 10ca\)

(iv) \((a - 2b - 5c)^2 = (a)^2 + (-2b)^2 + (-5c)^2 + 2(a \times -2b) + 2(-2b \times -5c) + 2(-5c \times a) = a^2 + 4b^2 + 25c^2 - 4ab + 20bc - 10ca\)

Teacher's Note

Understanding polynomial expansions is essential in engineering and architecture, where calculating areas and volumes of composite shapes requires breaking them down into simpler polynomial terms.

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