ICSE Class 8 Maths Chapter 13 Special Products Expansions

Chapter 13: Special Products And Expansions

Products of algebraic expressions can be obtained by using distributive laws:

\(a(b + c) = ab + ac\)

and \((a + b)c = ac + bc\)

In this chapter, we shall study some special products and expansions.

Some Special Products

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

Simplifying the left hand side, we get

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

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

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

Simplifying the left hand side, we get

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

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

Aliter: Using the product 1, we get

\((x + a)(x - b) = (x + a)[x + (-b)] = x^2 + [a + (-b)]x + a \times (-b)\)

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

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

Using the product 1, we get

\((x - a)(x + b) = [x + (-a)](x + b) = x^2 + [(-a) + b]x + (-a) \times b\)

\(= x^2 - (a - b)x - ab\)

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

Using the product 1, we get

\((x - a)(x - b) = [x + (-a)][x + (-b)] = x^2 + [(-a) + (-b)]x + (-a) \times (-b)\)

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

Example 1

Using the product \((x + a)(x + b) = x^2 + (a + b)x + ab\), simplify the following:

(i) \((x + 7)(x + 12)\)

(ii) \((y + 3)(y - 8)\)

(iii) \((s - 5)(s + 9)\)

(iv) \((p - 6)(p - 13)\)

Solution.

(i) \((x + 7)(x + 12) = x^2 + (7 + 12)x + 7 \times 12\)

\(= x^2 + 19x + 84\)

(ii) \((y + 3)(y - 8) = (y + 3)[y + (-8)]\)

\(= y^2 + [3 + (-8)]y + 3 \times (-8)\)

\(= y^2 - 5y - 24\)

(iii) \((s - 4)(s + 9) = [s + (-4)](s + 9)\)

\(= s^2 + [(-4) + 9]s + (-4) \times 9\)

\(= s^2 + 5s - 36\)

Example 2

Find the following products:

(i) \((3x + 5)(4x - 7)\)

(ii) \((4x - 3y)(5x + 6y)\)

(iii) \((5a^2 - 4b^2)(3a^2 - 7b^2)\)

(iv) \((x + 2)(2x - 3)(3x + 2)\)

Solution.

(i) \((3x + 5)(4x - 7) = 3x(4x - 7) + 5(4x - 7)\)

\(= 12x^2 - 21x + 20x - 35\)

\(= 12x^2 - x - 35\)

(ii) \((4x - 3y)(5x + 6y) = 4x(5x + 6y) - 3y(5x + 6y)\)

\(= 20x^2 + 24xy - 15xy - 18y^2\)

\(= 20x^2 + 9xy - 18y^2\)

(iii) \((5a^2 - 4b^2)(3a^2 - 7b^2) = 5a^2(3a^2 - 7b^2) - 4b^2(3a^2 - 7b^2)\)

\(= 15a^4 - 35a^2b^2 - 12a^2b^2 + 28b^4\)

\(= 15a^4 - 47a^2b^2 + 28b^4\)

(iv) \((x + 2)(2x - 3)(3x + 2) = (x + 2)[2x(3x + 2) - 3(3x + 2)]\)

\(= (x + 2)(6x^2 + 4x - 9x - 6)\)

\(= (x + 2)(6x^2 - 5x - 6)\)

\(= x(6x^2 - 5x - 6) + 2(6x^2 - 5x - 6)\)

\(= 6x^3 - 5x^2 - 6x + 12x^2 - 10x - 12\)

\(= 6x^3 + 7x^2 - 16x - 12\)

Exercise 13.1

Find the following (1 to 12) products:

1. (i) \((x + 3)(x + 5)\)

(ii) \((y + 2)(y - 5)\)

2. (i) \((a - 3)(a + 8)\)

(ii) \((t - 11)(t - 6)\)

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

(ii) \(\left(b + \frac{2}{5}\right)\left(b - \frac{2}{5}\right)\)

4. (i) \((x - 3)\left(x + \frac{2}{7}\right)\)

(ii) \((x + 0.4)(x - 0.7)\)

5. (i) \((8 - x)(5 + x)\)

(ii) \((3 - z)(11 - z)\)

6. (i) \((2x + 3)(2x + 7)\)

(ii) \((5y - 2)(5y + 9)\)

7. (i) \((7c - 11)(7c - 3)\)

(ii) \((p^2 + 3)(p^2 - 5)\)

8. (i) \((3x^2 - 7)(3x^2 + 5)\)

(ii) \((3 + xy)(7 - xy)\)

9. (i) \(\left(\frac{y}{3} - 2\right)\left(\frac{y}{3} - 7\right)\)

(ii) \((5x + 2y)(2x + 5y)\)

10. (i) \((3a - 5b)(7a + 2b)\)

(ii) \((3mn - 5)(4mn + 6)\)

11. (i) \((3x^2 + 2y^2)(4x^2 - 5y^2)\)

(ii) \((2c^2 - 3d^2)(7c^2 - 2d^2)\)

12. (i) \((ab - 2c)(3ab + 5c)\)

(ii) \((x + 1)(2x + 5)(3x - 1)\)

Teacher's Note

These algebraic products appear constantly in engineering and design - from calculating areas of rectangular plots to computing dimensions in construction projects.

Product Of Sum And Difference Of Two Terms

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

Simplifying the left hand side, we get

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

In words, this result can be stated as:

\((1\text{st term} + 2\text{nd term}) \times (1\text{st term} - 2\text{nd term}) = (1\text{st term})^2 - (2\text{nd term})^2\)

Example 1

Using the product \((a + b)(a - b) = a^2 - b^2\), simplify the following:

(i) \((5x + 7y)(5x - 7y)\)

(ii) \(\left(\frac{2}{3}a + \frac{5}{4}b\right)\left(\frac{2}{3}a - \frac{5}{4}b\right)\)

(iii) \((7pq + 11)(7pq - 11)\)

(iv) \(\left(6c^2 - \frac{5}{7}d^2\right)\left(6c^2 + \frac{5}{7}d^2\right)\)

Solution.

(i) \((5x + 7y)(5x - 7y) = (5x)^2 - (7y)^2 = 25x^2 - 49y^2\)

(ii) \(\left(\frac{2}{3}a + \frac{5}{4}b\right)\left(\frac{2}{3}a - \frac{5}{4}b\right) = \left(\frac{2}{3}a\right)^2 - \left(\frac{5}{4}b\right)^2 = \frac{4}{9}a^2 - \frac{25}{16}b^2\)

(iii) \((7pq + 11)(7pq - 11) = (7pq)^2 - (11)^2 = 49p^2q^2 - 121\)

(iv) \(\left(6c^2 - \frac{5}{7}d^2\right)\left(6c^2 + \frac{5}{7}d^2\right) = (6c^2)^2 - \left(\frac{5}{7}d^2\right)^2 = 36c^4 - \frac{25}{49}d^4\)

Example 2

Find the product of:

(i) \((x + 3)(x - 3)(x^2 + 9)\)

(ii) \((3p - 2q)(3p + 2q)(9p^2 + 4q^2)\)

Solution.

(i) \((x + 3)(x - 3)(x^2 + 9) = [(x + 3)(x - 3)](x^2 + 9)\)

\(= (x^2 - 3^2)(x^2 + 9) = (x^2 - 9)(x^2 + 9)\)

\(= (x^2)^2 - (9)^2 = x^4 - 81\)

(ii) \((3p - 2q)(3p + 2q)(9p^2 + 4q^2) = [(3p - 2q)(3p + 2q)](9p^2 + 4q^2)\)

\(= [(3p)^2 - (2q)^2](9p^2 + 4q^2)\)

\(= (9p^2 - 4q^2)(9p^2 + 4q^2)\)

\(= (9p^2)^2 - (4q^2)^2 = 81p^4 - 16q^4\)

Example 3

Using the product \((a + b)(a - b) = a^2 - b^2\), find the value of:

(i) \(507 \times 493\)

(ii) \(25.3 \times 24.7\)

Solution.

(i) \(507 \times 493 = (500 + 7)(500 - 7)\)

\(= (500)^2 - 7^2 = 250000 - 49 = 249951\)

(ii) \(25.3 \times 24.7 = (25 + 0.3)(25 - 0.3)\)

\(= (25)^2 - (0.3)^2 = 625 - 0.09 = 624.91\)

Exercise 13.2

Find the following (1 to 8) products:

1. (i) \((x + 7)(x - 7)\)

(ii) \((5x + 9)(5x - 9)\)

2. (i) \(\left(y + \frac{2}{3}\right)\left(y - \frac{2}{3}\right)\)

(ii) \((4 + 3x)(4 - 3x)\)

3. (i) \((4x + 11y)(4x - 11y)\)

(ii) \(\left(\frac{2}{3}p - \frac{4}{5}q\right)\left(\frac{2}{3}p + \frac{4}{5}q\right)\)

4. (i) \((3 - ab)(3 + ab)\)

(ii) \(\left(p + \frac{1}{q}\right)\left(p - \frac{1}{q}\right)\)

5. (i) \(\left(\frac{2}{a} + \frac{5}{b}\right)\left(\frac{2}{a} - \frac{5}{b}\right)\)

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

6. (i) \(\left(3x^2 - \frac{2}{5}y^2\right)\left(3x^2 + \frac{2}{5}y^2\right)\)

(ii) \((1.4a - 0.3b)(1.4a + 0.3b)\)

7. (i) \((y + 2)(y - 2)(y^2 + 4)\)

(ii) \((2p + 3)(2p - 3)(4p^2 + 9)\)

8. (i) \((x + a)(x - a)(x^2 + a^2)\)

(ii) \((x + yz)(x - yz)(x^2 + y^2z^2)\)

9. Using the product \((a + b)(a - b) = a^2 - b^2\), find the value of:

(i) \(108 \times 92\)

(ii) \(306 \times 294\)

(iii) \(10.4 \times 9.6\)

(iv) \(14.7 \times 15.3\)

Teacher's Note

The difference of squares formula is fundamental in physics and engineering, helping calculate areas, forces, and energy differences in real-world applications.

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