# Multiplication of Polynomials Exercise 3.1.4-Class 9

1. Expand the following:

(i) (a + b + 2c)2

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

a = a, b = b and c = 2c

(a + b + 2c)2 = a2 + b2 + (2c)2 + 2ab + 2b (2c) + 2(2c)a

= a2 + b2 + 4c2 + 2ab + 4bc + 4ca

(ii) (x + y + 3z)2

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

a = x, b = y and c = 3z

(x + y + 3z)2 = x2 + y2 + (3z)2 + 2.x.y + 2y(3z) + 2.(3z)x

= x2 + y2 + 9z2 + 2xy + 6yz + 6zx

(iii) (p + q – 2r)2

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

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

(p + q – 2r)2 = p2 + q2 + (-2r)2 + 2.p.q + 2q(-2r) +2(-2r)p

= p2 + q2 + 4r2 + 2pq – 4pr – 4pr

(iv) ( a/2+b/2+c/2 )2

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

a = a/2, b = b/2 and c = c/2

(a/2+ b/2+ c/2  )2 =( a/2 )2 + (b/2 )2 + (c/2  )2 + 2 (a/2 ) (b/2) + 2 (b/2 ) (c/2) +

2 (c/2  ) (a/2 )

= a2/4+b2/4+c2/4 + ab/2+bc/2+ca/2

(v) (x2 + y2 + z)2

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

a = x2, b = y2 and c = z

(x2 + y2 + z)2 = (x2)2 + (y2)2 + (z)2 + 2x2y2 + 2y2z +2zx2

= x4 + y4 + z2 + 2x2y2 + 2y2z +2zx2

(vi) (m – 3 – 1/m )2

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

a = m, b = -3 and c = – 1/m

(m – 3 – 1/m )2 = m2 + (-3)2 + (1/m )2 + 2.m(-3) + 2(-3)(- 1/m ) + 2 (-1/m) m

= m2 + 9 + 1/m 2 – 6m + 6m – 2

= m2 + 1/m2 + 6/m – 6m + 7

(vii) (-a + b – c)2

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

a = –a b = b c = –c

(-a + b – c)2 = (–a)2 + b2 + (–c)2 + 2(–a)b + 2b(–c) +2(–c)a

= a2 + b2 + c2 – 2ab – 2bc +2ca

(viii) (x + 5 + 1/2x )2

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

a = x; b = 5; c = 1/2x

(x + 5 + 1/2x  )2 = x2 + 52 + (1/2x)2 + 2.x.5 + 2.5. 1/2x  + 2(1/2x)x

= x2 + 25 + 1/4x2 +10x + 5/x + 1

= x2 + 1/4x2 +10x + 5/x + 26

1. Simplify the following:

(i) (a – b + c)2 – (a – b – c)2

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

= [ a2 + (-b)2 + c2 + 2a(-b) + 2(-b)c + 2ca ] – [a + (-b)2 + (–c)2 + 2a(–b)

+ 2(–b)(–c) + 2(–c)a]

= a2 + b2 + c2 – 2ab – 2bc +2ca – [a2 + b2 + c2 – 2ab + 2bc –2ca]

= a2 + b2 + c2 – 2ab – 2bc +2ca – a2 – b2 – c2 + 2ab – 2bc +2ca

= 4ac – 4bc

= 4c (a – b)

(ii) (3x + 4y + 5)2 – (x + 5y – 4)2

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

(3x + 4y + 5)2 – (x + 5y – 4)2

= [(3x)2 + (4y)2 + 52 + 2.3x.4y + 2.4y.5 + 2.5(3x)] – [x2 + (5y)2 + (-4)2 +

1. X.5y + 2.5y (-4) + 2 (-4). x]

= 9x2 + 16y2 + 25 + 24xy + 40y + 30x – [x2 + 25y2 + 16 + 10xy – 40y-8x

= 9x2 + 16y2 + 25 + 24xy + 40y + 30x – x2 – 25y2 – 16 – 10xy + 40y + 8x

= 8x2 – 9y2 + 14xy + 80y + 38x + 9

(iii) (2m – n – 3p)2 + 4mn – 6np + 12pm

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

(2m – n – 3p)2 + 4mn – 6np + 12pm

= (2m)2 + (–n)2 + (-3p)2 + 2.2m(–n) + 2(n)(–3p) + 2 (-3p)(2m) + 4mn –

6np + 12pm

= 4m2 + n2 + 9p2 – 4mn – 6np – 12pm + 4mn – 6np + 12pm

= 4m2 + n2 + 9p2

(iv) (x + 2y + 3z + r)2 + (x + 2y + 3z – r)2

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

a = (x + 2y) b = 3z c = r

(x + 2y + 3z + r)2 + (x + 2y + 3z – r)2

= (x + 2y)2 + (3z)2 + r2 + 2 (x + 2y)3z + 2.3z.r + 2.r.(x + 2y) + (x + 2y)2 +

(3z)2 – (r)2 + 2 (x + 2y) 3z + 2.3z (-r) + 2 (-r)(x + 2y)

= 2(x + 2y)2 + 9z2 + r2 + 6 (x + 2y)z + 6zr + 2r (x + 2y) + 9z2 + r2 +

6 (x + 2y) z – 6zr – 2r (x + 2y)

= 2(x2 + 2.x.2y +4y2) + 18z2 + 2r2 + 12 (x + 2y)z

= 2x2 + 8xy + 8y2 + 18z2 + 2r2 +12xz + 24 yz

= 2x2 + 8y2 + 18z2 + 2r2 + 8xy +12xy + 24 yz

1. If a + b + c = 12 and a2 + b2 + c2 = 50, find ab + bc + ca.

Given a + b + c = 12 squaring both sides

(a + b + c)2 = 122

a2 + b2 + c2 + 2ab + 2bc + 2ca = 144

Given a2 + b2 + c2 = 50

50 + 2ab + 2bc + 2ca = 144

2(ab + bc + ca) = 144 – 50

2(ab + bc + ca) = 94

ab + bc + ca = 94/2

ab + bc + ca = 47

1. If a2 + b2 + c2 = 35 and ab + bc + ca = 23, find all possible values of a + b +c.

Given a2 + b2 + c2 = 35 and ab + bc + ca = 23,

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

= a2 + b2 + c2 + 2(ab + bc + ca)

= 35 + 2 (23)

= 35 + 46

(a + b + c)2 = 81

(a + b +c) = ± √81 = ±9

1. Express 4x + 9y + 16z + 12xy – 24yz – 16zx as the square of a trinomial

Using and comparing the coefficient of

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

4x + 9y + 16z + 12xy – 24yz – 16zx

= (2x)2 + (3y)2 + (–4z2) + 2.2x.3y.(–4z) + 2.(–4z) + 2.(–4z).2x

= (2x + 3y – 4z)2

1. If x and y are real numbers and satisfy the equation

(2x + 3y – 4z)2 + (5x – y – 4)2 = 0 find x, y.

[Hint: If a, b are real numbers such that a2 + b2 = 0, then a = b = 0.]

Given if a2 + b2 = 0, then a = b = 0

(2x + 3y – 4z)2 = 0 and (5x – y – 4)2 = 0

(2x + 3y = 5)x1 i.e., 2x + 3y = 5————-(1)

(5x – y = 4)x3 i.e., 15x – 3y = 12—————-(2)

Therefore,

2x + 3y = 5

15x – 3y = 12

—————

17x = 17

—————

x = 1

substitute the value of x in (1)

2(1)+3y = 5

2+ 3y = 5

3y = 5 – 2 = 3

y = 3/3 = 1

Thus, x = 1 and y = 1