ICSE Class 8 Maths Algebra Chapter 04 Special Products and Expansions

Special Products and Expansions

Special Products

You can find the products of certain types of algebraic expressions directly without actually carrying out the multiplication. These products called special products, come in handy while simplifying expressions or solving equations. Some of the special products are as follows.

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

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

Examples:

(i) \((x + 2)(x + 3) = x^2 + (2 + 3)x + 2 \times 3 = x^2 + 5x + 6\)

(ii) \((m + 3n)(m + 7n) = m^2 + (3n + 7n)m + 3n \times 7n = m^2 + 10mn + 21n^2\)

(iii) \((4x^2 + 5y^2)(4x^2 + 9y^2) = (4x^2)^2 + (5y^2 + 9y^2) \times 4x^2 + 5y^2 \times 9y^2 = 16x^4 + 56x^2y^2 + 45y^4\)

(iv) \((3ab + 2xy)(3ab + 7xy) = (3ab)^2 + (2xy + 7xy) \times 3ab + 2xy \times 7xy = 9a^2b^2 + 27abxy + 14x^2y^2\)

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

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

Alternative method:

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

Substituting -b for b, \((x + a)[x + (-b)] = x^2[a + (-b)]x + a \times (-b)\).

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

Examples:

(i) \((x + 9)(x - 7) = x^2 + (9 - 7)x - 9 \times 7 = x^2 + 2x - 63\)

(ii) \((l + 3m)(l - 5m) = l^2 + (3m - 5m)l - 3m \times 5m = l^2 - 2lm - 15m^2\)

(iii) \((2c^2 + 4d^2)(2c^2 - d^2) = (2c^2)^2 + (4d^2 - d^2) \times 2c^2 - 4d^2 \times d^2 = 4c^4 + 6c^2d^2 - 4d^4\)

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

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

Alternative method:

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

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

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

Examples:

(i) \((x - 9)(x + 7) = x^2 - (9 - 7)x - 9 \times 7 = x^2 - 2x - 63\)

(ii) \((l^2 - 6mn)(l^2 + 9mn) = l^2 - (6mn - 9mn)l^2 - 6mn \times 9mn = l^2 + 3l^2mn - 54m^2n^2\)

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

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

Alternative method:

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

Substituting -a for a and -b for b.

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

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

Examples:

(i) \((a - 5)(a - 7) = a^2 - (5 + 7)a + 5 \times 7 = a^2 - 12a + 35\)

(ii) \((m - n)(m - 6n) = m^2 - (n + 6n)m + n \times 6n = m^2 - 7mn + 6n^2\)

(iii) \((4ab - 3cd)(4ab - 5cd) = (4ab)^2 - (3cd + 5cd) \times 4ab + 3cd \times 5cd = 16a^2b^2 - 8cd \times 4ab + 15c^2d^2 = 16a^2b^2 - 32abcd + 15c^2d^2\)

Solved Examples

Example 1: Multiply each of the following using a special product.

(i) \((a + 0.1)(a + 0.2)\)

(ii) \(\left(\frac{x}{2} + 3\right)\left(\frac{x}{2} + 6\right)\)

(iii) \(\left(\frac{m}{5} - \frac{n}{3}\right)\left(\frac{m}{5} + \frac{n}{6}\right)\)

(iv) \((2p^2 + 0.2qr)(2p^2 + 0.3qr)\)

Solution: In each case, we can use \((x + a)(x + b) = x^2 + (a + b)x + ab\).

(i) \((a + 0.1)(a + 0.2) = a^2 + (0.1 + 0.2)a + 0.1 \times 0.2 = a^2 + 0.3a + 0.02\)

(ii) \(\left(\frac{x}{2} + 3\right)\left(\frac{x}{2} + 6\right) = \left(\frac{x}{2}\right)^2 + (3 + 6)x + 3 \times 6 = \frac{x^2}{4} + \frac{9x}{2} + 18\)

(iii) \(\left(\frac{m}{5} - \frac{n}{3}\right)\left(\frac{m}{5} + \frac{n}{6}\right) = \left(\frac{m}{5}\right)^2 + \left(\frac{n}{3} + \frac{n}{6}\right) \cdot \frac{m}{5} - \frac{n}{3} \times \frac{n}{6} - \frac{m^2}{25} + \left(\frac{1}{3} + \frac{1}{6}\right) \frac{mn}{5} - \frac{n^2}{18} = \frac{m^2}{25} + \frac{1}{2} \times \frac{mn}{5} - \frac{n^2}{18} = \frac{m^2}{25} + \frac{mn}{10} - \frac{n^2}{18}\)

(iv) \((2p^2 + 0.2qr)(2p^2 + 0.3qr) = (2p^2)^2 + (0.2qr + 0.3qr) \times 2p^2 + 0.2qr \times 0.3qr = 4p^4 + 0.5qr \times 2p^2 + 0.06q^2r^2 = 4p^4 + 2 \times 0.5p^2qr + 0.06q^2r^2 = 4p^4 + p^2qr + 0.06q^2r^2\)

Teacher's Note

Understanding special products helps students recognize patterns in algebra, much like recognizing shortcuts in real-world calculations such as computing areas or volumes quickly without detailed formulas every time.

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