EXPANSION OF TRINOMIAL WITH POWER 2

In this section, you will learn how to expand a trinomial with power 2. 

Expansion of (a + b + c) Whole Square

(a + b + c)² = (a + b + c)(a + b + c)

(a + b + c)² = a² + ab + ac + ab + b² + bc + ac + bc + c²

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac

Expansion of (a + b - c) Whole Square 

To get expansion for (a + b - c)², let us consider the expansion of (a + b + c)². 

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac

In (a + b + c)², if c is negative, then we have 

(a + b - c)²

In the terms of the expansion for (a + b + c)², consider the terms in which we find 'c'.

They are c², bc, ca.

Even if we take negative sign for 'c' in c², the sign of c² will be positive.  Because it has even power 2. 

The terms bc, ac will be negative. Because both 'b' and 'a' are multiplied by 'c' that is negative.  

Finally, we have 

(a + b - c)² = a² + b² + c² + 2ab - 2bc - 2ac

Expansion of (a - b + c) Whole Square

To get expansion for (a - b + c)², let us consider the expansion of (a + b + c)². 

The expansion of (a + b + c)² is

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

In (a + b + c)², if b is negative, then we have 

(a - b + c)²

In the terms of the expansion for (a + b + c)², consider the terms in which we find "b".

They are b², ab, bc.

Even if we take negative sign for 'b' in b², the sign of b² will be positive.  Because it has even power 2. 

The terms ab, bc will be negative. Because both 'a' and 'c' are multiplied by 'b' that is negative.  

Finally, we have 

(a - b + c)² = a² + b² + c² - 2ab - 2bc + 2ac

Expansion of (a - b - c) Whole Square 

To get the expansion of (a - b - c)², let us consider the expansion of (a + b + c)². 

The expansion of (a + b + c)² is

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

In (a + b + c)², if b and c are negative, then we have 

(a - b - c)²

In the terms of the expansion for (a + b + c)², consider the terms in which we find 'b' and 'c'.

They are b², c², ab, bc, ac.

Even if we take negative sign for 'b' in b² and negative sign for 'c' in c², the sign of both b² and c² will be positive.  Because they have even power 2. 

The terms 'ab' and 'ac' will be negative.

Because, in 'ab', 'a' is multiplied by "b" that is negative. 

Because, in 'ac', 'a' is multiplied by "c" that is negative.  

The term 'bc' will be positive.

Because, in 'bc', both 'b' and 'c' are negative.    

That is,

negative ⋅ negative  =  positive  

Finally, we have 

(a - b - c)² = a² + b² + c² - 2ab + 2bc - 2ac

Summary

Example 1 :

Expand :

(2x + 3y + 4z)²

Solution :

(a + b + c)²  =  a² + b² + c² + 2ab + 2bc + 2ca

a = 2x, b = 3y and c = 4z.

  = (2x)² + (3y)² + (4z)² + 2(2x)(3y) + 2(3y)(4z) + 2(4z)(2x)

  = 4x² + 9y² + 16z² + 12xy + 24yz + 16zx

Example 2 :

Expand :

(-p +2q + 3r)²

Solution :

a = -p, b = 2q and c = 3r.

  = (-p)² + (2q)² + (3r)² + 2(-p)(2q) + 2(2q)(3r) + 2(3r)(-p)

= p² + 4q² + 9r² - 4pq + 12qr - 6rp

Product of Three Binomials

Example 3 :

Expand :

(2p + 3)(2p −4)(2p −5)

Solution :

x = 2p, a = 3, b = -4 and c = -5.

a + b + c = 3 + (-4) + (-5) = -6

ab + bc + ca = 3(-4) + (-4)(-5) + (-5)(3)

= -12 + 20 - 15

= -27 + 20

= -7

abc = 3(-4)(-5) = 60.

= x³ + (a + b + c)x² + (ab + bc + ca)x + abc

= (2p)³ + (-6)(2p)² + (-7)(2p) + 60

= 8p³ -24p² -14p + 60

Example 4 :

Expand :

(3a +1)(3a −2)(3a + 4)

Solution : 

x = 3a, a = 1, b = -2 and c = 4.

a + b + c = 1 + (-2) + 4 = 3

ab + bc + ca = 1(-2) + (-2)(4) + 4(1)

= -2 - 8 + 4

 = -10 + 4

= -6

abc = 1(-2)(4) = -8.

= x³ + (a + b + c) x² + (ab + bc + ca)x + abc

= (3a)³ + 3(3a)² + (-6)(3a) - 8

= 27a³ + 27a² -18a - 8

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.