# Polynomials

## Exercise 2.5 Part 5

Question: 4 – Expand each of the following using suitable identities:

(iv) (3a – 7b – c)^2

Answer: Given, (3a – 7b – c)^2

Let, x = 3a, y = - 7b and z = - c

Using the identity (x + y + z)^2= x^2 + y^2 + z^2 + 2xy + 2yz + 2xz, we get;

(3a)^2+(-7b)^2+(- c)^2+2xx(3a)xx(-7b)+

2xx(-7b)xx(- c)+2xx(3a)xx(- c)

= 9a^2 + 49b^2 + c^2 + 2( - 21ab) + 2(7bc) + 2( - 3ac)

= 9a^2 + 49b^2 + c^2 - 42ab + 14bc – 6ac

(v) ( - 2x + 5y – 3z)^2

Answer: Given, ( - 2x + 5y – 3z) ^2

Let, a = - 2x, b = 5y and c = - 3z

Using the identity (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac, we get;

(-2x)^2+(5y)^2+(-3z)^2+2(-2x)xx(5y)+

2(5y)xx(-3z)+2(-2x)xx(-3z)

= 4x^2 + 25y^2 + 9z^2 + 2( - 10xy) + 2( - 15yz) + 2(6xz)

= 4ax^2 + 25y^2 + 9z^2 - 20xy – 30yz + 12xz

(vi) (1/4\a-1/2\b+1)^2

Solution: Given, (1/4\a-1/2\b+1)^2

Let, x = 1/4a, y = -1/2b and c = 1
[Using identity, (x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2xz]
Given expression can be written as follows:
(1/4\a)^2+(-1/2\b)^2+1^2+2(1/4\a)xx(-1/2\b)+

2(-1/2\bxx1)+2(1/4\axx1)

=1/16\a^2+1/4\b^2+1-1/4\ab-b+1/2\a