THE BINOMIAL THEOREM, HOW DO I SOLVE BINOMIAL THEOREM ?

 THE BINOMIAL THEOREM

Binomial is an algebraic expression involving two terms operation, addition or subtraction, eg, (x + y), (x + y)^2 or (3a - 6y).

Monomial ; Is an Expression coulombs a simple term such as 3x, 4x, or 5x

Pascal  Theorem

(a + b)⁰ = 1

(a + b)² = a² + 2ab + b²

(a + b)³ = a³ + 3a¹b² + 3a²b¹ + b³

Note that ; 

1) Reading from either end of each, the coefficient is the same.

2) There are n + 1 terms.

3) Each term is of degree n

4) The coefficient are obtained from the rows in the Pascal triangle beginning with 1 (one) and ending with 1 (one) . 

5) The Pascal triangle coefficient of the second and second to the last term are always desame.

Examples ;

(2x - 6y)³ = (1)2x - (3)2x¹6y² - (3)2x²6y¹ - 6y(1)

2x - 6x18y² - 6x²18y - 6y

BINOMIAL AND SURD

Expand (1 + √2)³ - (1 - √2)³

Let's take it one by one

Firstly (1 + √2)³

(1)1³ + (3)1√2 + (3)1√2² + √2(1)

1 + 3√2 + (3)(2) + 2√2

1 + 3√2 + 6 + 2√2

7 + 5√2

Secondly (1 - √2)³

(1)1³ - (3)1²√2 + (3)1√2² - √2³(1)

1 - 3√2 + (3)1√4 - √2³

1 - 3√2 + (3)(2) - √2³ 

1 - 3√2 + 6 - 2√2 

 therefore

7 - 5√2

7+5√2 + 7 - 5√2 = 14

Suppose we wish to find (a+b)⁴

To find this we say the row begin with Pascal triangle.

a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴

Suppose we wish to find (2x + y)³

2x³+ 3(2x)²y + 3(2x)y² + y³

The Binomial n positive integers (a + 1b)^n = an + ⁿC₁ aⁿ⁻¹b¹ + ⁿC₂ anⁿ⁻²b₂.....bⁿ

This type of expansion is used usually when the power of the equation is higher, using the Pascal's triangle to solve a given problem may consume a lot of time especially during exam. The sequence above will be applied to solve the following term.

Examples

Find the coefficient of x¹⁰ in the expansion of (2x - 3)¹⁴

Solution

The term x¹⁰ can be simplified as (2x)¹⁰ (-3)⁴. the only term are needed and binomial theorem its ¹⁴C₄ where n = 14 but the x power is 10 

now 14 - 10 = 4

¹⁴C₄ (2x)¹⁰ (-3)⁴ = the coefficient of x.

Now we Compute

1001* 2¹⁰ * x¹⁰ * 81

1001 (1024)x¹⁰ × 81

83026944(x)¹⁰ hence proved

Example 2

Write down the term indicated in the expansion of the following terms.

1. (x + 2)⁸ at term x⁵

2. (3x - 2)⁵ at term x²

Solution to number one1

(x)⁵(2)³

8 - 5 = 3

⁸C₃ x⁵ 8 = 56 × 8 × x⁵

448x⁵

take number 2 as exercise