**Expand the following:**

**(i) (a + b + 2c) ^{2} **

Using (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc +2ca we get

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

(a + b + 2c)^{2} = a^{2} + b^{2} + (2c)^{2} + 2ab + 2b (2c) + 2(2c)a

= a^{2} + b^{2} + 4c^{2} + 2ab + 4bc + 4ca

**(ii) (x + y + 3z) ^{2} **

Using (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc +2ca

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

(x + y + 3z)^{2} = x^{2} + y^{2} + (3z)^{2} + 2.x.y + 2y(3z) + 2.(3z)x

= x^{2} + y^{2} + 9z^{2} + 2xy + 6yz + 6zx

**(iii) (p + q – 2r) ^{2} **

Using (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc +2ca

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

(p + q – 2r)^{2} = p^{2} + q^{2} + (-2r)^{2} + 2.p.q + 2q(-2r) +2(-2r)p

= p^{2} + q^{2} + 4r^{2} + 2pq – 4pr – 4pr

**(iv) ( **^{a}**/ _{2}+^{b}/_{2}+^{c}/_{2} **

**)**

^{2}Using (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 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} )

= a^{2}/_{4}+b^{2}/_{4}+c^{2}/_{4} + ^{ab}/_{2}+^{bc}/_{2}+^{ca}/_{2}

**(v) (x ^{2} + y^{2} + z)^{2} **

Using (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc +2ca

a = x^{2}, b = y^{2} and c = z

(x^{2} + y^{2} + z)^{2} = (x^{2})^{2} + (y^{2})^{2} + (z)^{2} + 2x^{2}y^{2} + 2y^{2}z +2zx^{2}

= x^{4} + y^{4} + z^{2} + 2x^{2}y^{2} + 2y^{2}z +2zx^{2}

**(vi) (m – 3 – **^{1}**/ _{m} **

**)**

^{2}Using (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc +2ca

a = m, b = -3 and c = – ^{1}/_{m}

(m – 3 – ^{1}/_{m} )^{2} = m^{2} + (-3)^{2} + (^{1}/_{m} )^{2} + 2.m(-3) + 2(-3)(- ^{1}/_{m} ) + 2 (-^{1}/_{m}) m

= m^{2} + 9 + ^{1}/_{m} ^{2} – 6m + 6m – 2

= m^{2} + ^{1}/_{m}^{2} + ^{6}/_{m} – 6m + 7

**(vii) (-a + b – c) ^{2} **

Using (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc +2ca

a = –a b = b c = –c

(-a + b – c)^{2} = (–a)^{2} + b^{2} + (–c)^{2} + 2(–a)b + 2b(–c) +2(–c)a

= a^{2} + b^{2} + c^{2} – 2ab – 2bc +2ca

**(viii) (x + 5 + **^{1}**/ _{2x} **

**)**

^{2}Using (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc +2ca

a = x; b = 5; c = ^{1}/_{2x }

(x + 5 + ^{1}/_{2x} )^{2} = x^{2} + 5^{2} + (^{1}/_{2x})^{2} + 2.x.5 + 2.5. ^{1}/_{2x} + 2(^{1}/_{2x})x

= x^{2} + 25 + ^{1}/_{4x}^{2} +10x + ^{5}/_{x} + 1

= x^{2} + ^{1}/_{4x}^{2} +10x + ^{5}/_{x} + 26

**Simplify the following:**

**(i) (a – b + c) ^{2} – (a – b – c)^{2} **

Using (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc +2ca (a – b + c)2 – (a – b – c)2

= [ a^{2} + (-b)^{2} + c^{2} + 2a(-b) + 2(-b)c + 2ca ] – [a + (-b)2 + (–c)2 + 2a(–b)

+ 2(–b)(–c) + 2(–c)a]

= a^{2} + b^{2} + c^{2} – 2ab – 2bc +2ca – [a2 + b2 + c2 – 2ab + 2bc –2ca]

= a^{2} + b^{2} + c^{2} – 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} = a^{2} + b^{2} + c^{2} + 2ab + 2bc +2ca

(3x + 4y + 5)^{2} – (x + 5y – 4)^{2}

= [(3x)^{2} + (4y)^{2} + 5^{2} + 2.3x.4y + 2.4y.5 + 2.5(3x)] – [x^{2} + (5y)^{2} + (-4)^{2} +

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

= 9x^{2} + 16y^{2} + 25 + 24xy + 40y + 30x – [x^{2} + 25y^{2 }+ 16 + 10xy – 40y-8x

= 9x^{2} + 16y^{2} + 25 + 24xy + 40y + 30x – x^{2} – 25y^{2} – 16 – 10xy + 40y + 8x

= 8x^{2} – 9y^{2} + 14xy + 80y + 38x + 9

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

Using (a + b + c)^{2} = a^{2 }+ b^{2} + c^{2 }+ 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

= 4m^{2} + n^{2} + 9p^{2} – 4mn – 6np – 12pm + 4mn – 6np + 12pm

= 4m^{2} + n^{2} + 9p^{2}

** **

**(iv) (x + 2y + 3z + r) ^{2} + (x + 2y + 3z – r)^{2} **

Using (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 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} + r^{2 }+ 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} + 9z^{2} + r^{2} + 6 (x + 2y)z + 6zr + 2r (x + 2y) + 9z2 + r2 +

6 (x + 2y) z – 6zr – 2r (x + 2y)

= 2(x^{2} + 2.x.2y +4y^{2}) + 18z^{2} + 2r^{2} + 12 (x + 2y)z

= 2x^{2} + 8xy + 8y^{2} + 18z^{2} + 2r^{2} +12xz + 24 yz

= 2x^{2} + 8y^{2} + 18z^{2} + 2r^{2} + 8xy +12xy + 24 yz

**If a + b + c = 12 and a**^{2}+ b^{2}+ c^{2}= 50, find ab + bc + ca.

Given a + b + c = 12 squaring both sides

(a + b + c)^{2} = 12^{2}

a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca = 144

Given a^{2} + b^{2} + c^{2} = 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

**If****a**^{2}+ b^{2}+ c^{2}= 35 and ab + bc + ca = 23, find all possible values of a + b +c.

Given a^{2} + b^{2} + c^{2} = 35 and ab + bc + ca = 23,

(a + b +c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca

= a^{2} + b^{2} + c^{2} + 2(ab + bc + ca)

= 35 + 2 (23)

= 35 + 46

(a + b + c)^{2} = 81

(a + b +c) = ± √81 = ±9

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

Using and comparing the coefficient of

(a +b+c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc +2ca we get

4x + 9y + 16z + 12xy – 24yz – 16zx

= (2x)^{2} + (3y)^{2 }+ (–4z^{2}) + 2.2x.3y.(–4z) + 2.(–4z) + 2.(–4z).2x

= (2x + 3y – 4z)^{2 }

**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 a^{2} + b^{2} = 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

## 1 thought on “Multiplication of Polynomials Exercise 3.1.4-Class 9”

Comments are closed.