# Binomial Expansion - Mathematics Form 3 Notes

## Introduction

• A binomial is an expression of two terms

Examples

(a + y), a + 3, 2a + b

• It easy to expand expressions with lower power but when the power becomes larger, the expansion or multiplication becomes tedious. We therefore use pascal triangle to expand the expression without multiplication.
• We can use Pascal triangle to obtain coefficients of expansions of the form( a + b) n

## Pascal's Triangle

Note;

• Each row starts with 1
• Each of the numbers in the next row is obtained by adding the two numbers on either side of it in the preceding row
• The power of first term (a ) decreases as you move to right while the powers of the second term (b) increases as you move to the right

Example

Expand (p + q)5

Solution

The terms without coefficients are;
p5, p4q, p3q2, p2q3, pq4, q5

From Pascal triangle, the coefficients when n =5 are; 1, 5, 10, 10, 5, 1
Therefore (p +
q)5
p5 +5p4q + 10p3q2 + 10q2q3 + 5pq4 + q5

Example

Expand (x − y)7

Solution

(x−y) 7 = (x + (−y)7

The terms without the coefficient are;
x7x6(−y), x5(−y)2x4(−y)3x3(−y)4, x2(−y)5x(−y)6y7

From Pascal triangle, the coefficients when n =7 are; 1, 7, 21, 35, 35, 21, 7, 1

Therefore (x−y)7x7 7x6y + 21x5y2 35x4y3 + 35x3y4 21x2y5 + 7xy6 y7

Note;

When dealing with negative signs, the signs alternate with the positive sign but first start with the negative sign.

### Applications to Numeric Cases

Example

Use binomial expansion to evaluate (1.02)6 to 4 S.F

Solution

(1.02) = (1+0.02)
Therefore (1 .02
)6 = (1 + 0.02)6

The terms without coefficients are
1615(0.02)114(0.02)213(0.02)312(0.02)411(0.02)5(0.02)6

From Pascal triangle, the coefficients when n = 6 are; 1, 6, 15, 20, 15, 6, 1

Therefore;
(1.02)6 = 1 + 6 (0.02) + 15 (0.02)2 + 20(0.02)3 + 15(0.02)4 + 6(0.02)5 + (0.02)6
=1 + 0.12 + 0.0060 + 0.00016 + 0.0000024 + 0.0000000192 + 0.000000000064
=1.1261624
=1.126 (4 S.F)

Note;

• To get the answer just consider addition of up to the 4th term of the expansion. The other terms are too small to affect the answer.

Example

Expand (1 + x)9 up to the term x3. Use the expansion to estimate (0.98) 9 correct to 3 decimal places.

Solution

(1 + x)9

The terms without the coefficient are;
1918(x), 17x216(x)315(x)4

From Pascal triangle, the coefficients when n =9 are; 1, 9, 36, 84, 126, 126, 84, 36, 9, 1

Therefore (1 + x)9= 1 + 9x + 36x2 + 84x3………………..

(0.98)9= 1 + 9(-0.02) + 36 + (-0.02)2 + 84 (-0.02)3
= 1 – 0.1 8 + 0.0144 – 0.000672
= 0.833728
= 0.834 (3 d.p)

Example

Expand (1 + ½x)10 upto the term in in ascending powers of hence find the value of (correct to four decimal places.)

Solution

= 1 + 10(1/2x) + 45(1/2 x)2 + 120(1/2x)3
= 1 + 10 × 1/2x + 45 × 1/2x2 + 120 × 1/8x3
1 + 5x + 45/4x2 + 15x3
(1.005)10 = (1+0.005)10
Here 1/2 x = 0.005
x = 0.010

Substituting for x = 0.01 in the expansion
[1+ 1/2 (0.01)]10 = 1 + 5 ×0.01 + 45/4 ×(0.01)2 + 15(0.01)3
= 1 + 0.05 +0.001125 +0.00001 5
= 1.051140
= 1.0511 (4 decimal places)

## Past KCSE Questions on the Topic.

1. Write down the simplest expansion (1 + x)6
Use the expansion up to the fourth term to find the value of (1.03)to the nearest one thousandth.
2. Use binomial expression to evaluate (0.96)5 correct to 4 significant figures.
3. Expand and simplify (3x – y)4 hence use the first three terms of the expansion to proximate the value of (6 – 0.2)4
4. Use binomial expression to evaluate
(2 + 1/√2)5+ (2 − 1/√2)5
5. Expand the expression (1 + 1/2x)5 in ascending powers of x, leaving the coefficients as fractions in their simplest form.
6.
1. Expand (a − b)6
2. Use the first three terms of the expansion in (a) above to find the approximate value of (1 .98)6
7. Expand (2 + x)5 in ascending powers of x up to the term in x3 hence approximate the value of (2.03)5 to 4 s.f
8.
1. Expand (1 + x)5
Hence use the expansion to estimate (1 .04)5 correct to 4 decimal places
2. Use the expansion up to the fourth term to find the value of (1 .03)6 to the nearest one thousandth.
9. Expand and Simplify (1 − 3x)5 up to the term in x3
Hence use your expansion to estimate (0.97)5 correct to decimal places.
10. Expand (1 + a)5
Use your expansion to evaluate (0.8)5 correct to four places of decimal
11.
1. Expand (1 + x)5
2. Use the first three terms of the expansion in (a) above to find the approximate value of (0.98)5

• ✔ To read offline at any time.
• ✔ To Print at your convenience
• ✔ Share Easily with Friends / Students

### Related items

.
Subscribe now

access all the content at an affordable rate
or
Buy any individual paper or notes as a pdf via MPESA
and get it sent to you via WhatsApp