Maharashtra Board Class 8 Maths part 1 Chapter 5 Expansion formulae PDF Download

Expansion Formulae

Let's Recall

We have studied the following expansion formulae in previous standard.

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

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

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

Use the above formulae to fill proper terms in the following boxes.

(i) \((x + 2y)^2 = x^2 + \underline{\quad} + 4y^2\)

(ii) \((2x - 5y)^2 = \underline{\quad} - 20xy + \underline{\quad}\)

(iii) \((101)^2 = (100 + 1)^2 = \underline{\quad} + \underline{\quad} + 1^2 = \underline{\quad}\)

(iv) \((98)^2 = (100 - 2)^2 = 10000 - \underline{\quad} + \underline{\quad} = \underline{\quad}\)

(v) \((5m + 3n)(5m - 3n) = \underline{\quad} - \underline{\quad} = \underline{\quad} - \underline{\quad}\)

Teacher's Note

These formulas help us calculate big numbers easily without using a calculator. For example, we can find 99 squared by using (100 - 1)² instead of multiplying 99 × 99.

Exam Trick

Remember the pattern: (a + b)² always gives three terms - square of first, square of second, and twice the product. Write it as a² + b² + 2ab to never forget any term.

Points to Remember

(a + b)² = a² + 2ab + b²
(a - b)² = a² - 2ab + b²
(a + b)(a - b) = a² - b²
Use these to solve problems faster
Practice with numbers like 101, 99, 98 to get better

Let's Learn

Activity: Expand (x + a)(x + b)

We can expand (x + a)(x + b) using formulae for areas of a square and a rectangle.

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

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

(I) Expansion of (x + a)(x + b)

(x + a) and (x + b) are binomials with one term in common. Let us multiply them.

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

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

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

Teacher's Note

This formula is very useful for multiplying numbers that are close to each other. For example, 23 × 25 can be written as (24-1)(24+1) and solved quickly.

Exam Trick

Remember: When you see (x + a)(x + b), the middle term is always the sum (a + b) multiplied by x, and the last term is the product of a and b.

Points to Remember

(x + a)(x + b) gives us x², then (a+b)x in the middle
The last term is always a times b
This works when both brackets have the same first term x
You can use this to multiply numbers quickly
Always arrange your answer in order: x², x, then constant

Expand

Ex. (1) \((x + 2)(x + 3) = x^2 + (2 + 3)x + (2 \times 3) = x^2 + 5x + 6\)

Ex. (2) \((y + 4)(y - 3) = y^2 + (4 - 3)y + (4) \times (-3) = y^2 + y - 12\)

Ex. (3) \((2a + 3b)(2a - 3b) = (2a)^2 + [(3b) + (-3b)]2a + [3b \times (-3b)]\)

\(= 4a^2 + 0 \times 2a - 9b^2 = 4a^2 - 9b^2\)

Ex. (4) \(\left(m + \frac{3}{2}\right)\left(m + \frac{1}{2}\right) = m^2 + \left(\frac{3}{2} + \frac{1}{2}\right)m + \frac{3}{2} \times \frac{1}{2}\)

\(= m^2 + 2m + \frac{3}{4}\)

Ex. (5) \((x - 3)(x - 7) = x^2 + (-3 - 7)x + (-3)(-7) = x^2 - 10x + 21\)

Practice Set 5.1

1. Expand.

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

(2) \((m - 4)(m + 6)\)

(3) \((p + 8)(p - 3)\)

(4) \((13 + x)(13 - x)\)

(5) \((3x + 4y)(3x + 5y)\)

(6) \((9x - 5t)(9x + 3t)\)

(7) \(\left(m + \frac{2}{3}\right)\left(m + \frac{7}{3}\right)\)

(8) \(\left(x + \frac{1}{x}\right)\left(x + \frac{1}{x}\right)\)

(9) \(\left(\frac{1}{y} + 4\right)\left(\frac{1}{y} - 9\right)\)

Teacher's Note

These practice problems help you master the (x + a)(x + b) formula. Try solving them step by step, just like the examples given above.

Exam Trick

Always write down what a and b are before you start. This prevents silly mistakes. For example, in (p + 8)(p - 3), write a = 8, b = -3 first.

Points to Remember

Check if both brackets have the same first term
Find a and b correctly, watch the signs
Use the formula: x² + (a+b)x + ab
Write your final answer in proper order
Double-check your arithmetic in the middle and last terms

Let's Learn

(II) Expansion of (a + b)³

\((a + b)^3 = (a + b)(a + b)(a + b) = (a + b)(a + b)^2\)

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

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

\(= a^3 + 2a^2b + ab^2 + ba^2 + 2ab^2 + b^3\)

\(= a^3 + 3a^2b + 3ab^2 + b^3\)

\((a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\)

Let us study some examples based on the above expansion formula.

Ex. (1) \((x + 3)^3\)

We know that \((a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\)

In the given example, a = x and b = 3

\((x + 3)^3 = (x)^3 + 3 \times x^2 \times 3 + 3 \times x \times (3)^2 + (3)^3\)

\(= x^3 + 9x^2 + 27x + 27\)

Teacher's Note

The cube formula has four terms, not three. Remember: first cube, then three times the square times second, then three times first times second squared, then second cube. Like a pyramid!

Exam Trick

Write the coefficients as 1, 3, 3, 1 to remember the pattern. This is from Pascal's triangle for cubes. The powers of the first term go down: 3, 2, 1, 0. The powers of the second term go up: 0, 1, 2, 3.

Points to Remember

(a + b)³ has four terms with coefficients 1, 3, 3, 1
The powers of a go from 3 to 0, and powers of b go from 0 to 3
Always multiply the numbers carefully: 3 times, not just 2 times
You can use this to find cubes of big numbers like 101³
Practice with small numbers first before trying big ones

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