Exercise 3D RS Aggarwal Class 9 Solutions Factorisation of Polynomials Best

Exercise 1A Exercise 1B Exercise 1C Exercise 1D Exercise 1E Exercise 1F Exercise 1G Exercise 2A Exercise 2B Exercise 2C Exercise 2D Exercise 3A Exercise 3B Exercise 3C Exercise 3D Exercise 3E Exercise 3F

We have given Exercise 3D RS Aggarwal Class 9 Solutions of all the seven questions. The questions are based on the square of a trinomial.

Square of a Trinomial:
${(x + y + z)}^{2} = x^{2} + y^{2} + z^{2} + 2 x y + 2 y z + 2 z x$

Exercise 3D RS Aggarwal Class 9 Solutions

Q.1 Q.2 Q.3 Q.4 Q.5 Q.6 Q.7

Question 1 Exercise 3D RS Aggarwal Class 9 Solutions has been given below.

**1. Expand:**

(i). ${(a + 2 b + 5 c)}^{2}$

Answer

The given algebraic expression is
${(a + 2 b + 5 c)}^{2}$

We know that
${(x + y + z)}^{2} = x^{2} + y^{2} + z^{2} + 2 x y + 2 y z + 2 z x$

Putting $x = a, y = 2 b, z = 5 c$ , we get.

${(a + 2 b + 5 c)}^{2}$
$= a^{2} + {(2 b)}^{2} + {(5 c)}^{2} + 2 \cdot a \cdot 2 b + 2 \cdot 2 b \cdot 5 c + 2 \cdot 5 c \cdot a$
$= a^{2} + 2^{2} \cdot b^{2} + 5^{2} \cdot c^{2} + 4 a b + 20 b c + 10 c a$
$= a^{2} + 4 b^{2} + 25 c^{2} + 4 a b + 20 b c + 10 a c$

(ii). ${(2 a - b + c)}^{2}$

Answer

The given algebraic expression is
${(2 a - b + c)}^{2}$

We know that
${(x + y + z)}^{2} = x^{2} + y^{2} + z^{2} + 2 x y + 2 y z + 2 z x$

Putting $x = 2 a, y = - b, z = c$ , we get.

${(2 a - b + c)}^{2}$

$= {(2 a)}^{2} + {(- b)}^{2} + c^{2} + 2 \cdot 2 a \cdot (- b) + 2 \cdot (- b) \cdot c + 2 \cdot c \cdot 2 a$

$= 2^{2} a^{2} + b^{2} + c^{2} - 4 a b - 2 b c + 4 a c$

$= 4 a^{2} + b^{2} + c^{2} - 4 a b - 2 b c + 4 a c$

(iii). ${(a - 2 b - 3 c)}^{2}$

Answer

The given algebraic expression is
${(a - 2 b - 3 c)}^{2}$

We know that
${(x + y + z)}^{2} = x^{2} + y^{2} + z^{2} + 2 x y + 2 y z + 2 z x$

Putting $x = a, y = (- 2 b), z = (- 3 c)$ , we get.

${(a - 2 b - 3 c)}^{2}$

$= a^{2} + {(- 2 b)}^{2} + {(- 3 c)}^{2} + 2 \cdot a \cdot (- 2 b) + 2 \cdot (- 2 b) \cdot (- 3 c) + 2 \cdot (- 3 c) \cdot a$

$= a^{2} + {(- 2)}^{2} b^{2} + {(- 3)}^{2} c^{2} - 4 a b + 12 b c - 6 a c$

$= a^{2} + 4 b^{2} + 9 c^{2} - 4 a b + 12 b c - 6 a c$

Question 2 Exercise 3D RS Aggarwal Class 9 Solutions has been given below.

**2. Expand:**

(i). ${(2 a - 5 b - 7 c)}^{2}$

Answer

The given algebraic expression is
${(2 a - 5 b - 7 c)}^{2}$

We know that
${(x + y + z)}^{2} = x^{2} + y^{2} + z^{2} + 2 x y + 2 y z + 2 z x$

Putting $x = 2 a, y = (- 5 b), z = (- 7 c)$ , we get.

${(2 a - 5 b - 7 c)}^{2}$

$= {(2 a)}^{2} + {(- 5 b)}^{2} + {(- 7 c)}^{2} + 2 \cdot 2 a \cdot (- 5 b) + 2 \cdot (- 5 b) \cdot (- 7 c) + 2 \cdot (- 7 c) \cdot 2 a$

$= 2^{2} a^{2} + {(- 5)}^{2} b^{2} + {(- 7)}^{2} c^{2} - 20 a b + 70 b c - 28 a c$

$= 4 a^{2} + 25 b^{2} + 49 c^{2} - 20 a b + 70 b c - 28 a c$

(ii). ${(- 3 a + 4 b - 5 c)}^{2}$

Answer

The given algebraic expression is
${(- 3 a + 4 b - 5 c)}^{2}$

We know that
${(x + y + z)}^{2} = x^{2} + y^{2} + z^{2} + 2 x y + 2 y z + 2 z x$

Putting $x = - 3 a, y = 4 b, z = - 5 c$ , we get.

${(- 3 a + 4 b - 5 c)}^{2}$

$= {(- 3 a)}^{2} + {(4 b)}^{2} + {(- 5 c)}^{2} + 2 \cdot (- 3 a) \cdot 4 b + 2 \cdot (4 b) \cdot (- 5 c) + 2 \cdot (- 5 c) \cdot (- 3 a)$

$= {(- 3)}^{2} a^{2} + 4^{2} b^{2} + {(- 5)}^{2} c^{2} - 24 a b - 40 b c + 30 a c$

$= 9 a^{2} + 16 b^{2} + 25 c^{2} - 24 a b - 40 b c + 30 a c$

(iii). ${(\frac{1}{2} a - \frac{1}{4} b + 2)}^{2}$

Answer

The given algebraic expression is
${(\frac{1}{2} a - \frac{1}{4} b + 2)}^{2}$

We know that
${(x + y + z)}^{2} = x^{2} + y^{2} + z^{2} + 2 x y + 2 y z + 2 z x$

Putting $x = \frac{1}{2} a, y = - \frac{1}{4} b, z = 2$ , we get.

${(\frac{1}{2} a - \frac{1}{4} b + 2)}^{2}$

$= {(\frac{1}{2} a)}^{2} + {(- \frac{1}{4} b)}^{2} + 2^{2} + 2 \cdot \frac{1}{2} a \cdot (- \frac{1}{4} b) + 2 \cdot (- \frac{1}{4} b) \cdot 2 + 2 \cdot 2 \cdot (\frac{1}{2} a)$

$= {(\frac{1}{2})}^{2} \cdot a^{2} + {(- \frac{1}{4})}^{2} \cdot b^{2} + 4 - \frac{1}{4} a b - b + 2 a$

$= \frac{1}{4} a^{2} + \frac{1}{16} b^{2} + 4 - \frac{1}{4} a b - b + 2 a$

$= \frac{a^{2}}{4} + \frac{b^{2}}{16} + 4 - \frac{a b}{4} - b + 2 a$

Question 3 Exercise 3D RS Aggarwal Class 9 Solutions has been given below.

Factorise

3. $4 x^{2} + 9 y^{2} + 16 z^{2} + 12 x y - 24 y z - 16 z x$

Answer

Since the terms $- 24 y z$ and $- 16 z x$ are negative and $z$ is common in both, we write the term $16 z^{2} = {(- 4 z)}^{2}$ .

Now,

$4 x^{2} + 9 y^{2} + 16 z^{2} + 12 x y - 24 y z - 16 z x$

$= {(2 x)}^{2} + {(3 y)}^{2} + {(- 4 z)}^{2} + 2 \cdot 2 x \cdot 3 y + 2 \cdot 3 y \cdot (- 4 z) + 2 \cdot (- 4 z) \cdot 2 x$

$= {[2 x + 3 y + (- 4 z)]}^{2}$

$= {(2 x + 3 y - 4 z)}^{2}$

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

Question 4 Exercise 3D RS Aggarwal Class 9 Solutions has been given below.

4. $9 x^{2} + 16 y^{2} + 4 z^{2} - 24 x y + 16 y z - 12 x z$

Answer

Since the terms $- 24 x y$ and $- 12 x z$ are negative and $x$ is common in both, we write the term $9 x^{2} = {(- 3 x)}^{2}$ .

Now,

$9 x^{2} + 16 y^{2} + 4 z^{2} - 24 x y + 16 y z - 12 x z$

$= {(- 3 x)}^{2} + {(4 y)}^{2} + {(2 z)}^{2} + 2 \cdot (- 3 x) \cdot 4 y + 2 \cdot 4 y \cdot 2 z - 2 \cdot (- 3 x) \cdot 2 z$

$= {[(- 3 x) + 4 y + 2 z]}^{2}$

$= {(- 3 x + 4 y + 2 z)}^{2}$

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

Question 5 Exercise 3D RS Aggarwal Class 9 Solutions has been given below.

5. $25 x^{2} + 4 y^{2} + 9 z^{2} - 20 x y - 12 y z + 30 x z$

Answer

Since the terms $- 20 x y$ and $- 12 y z$ are negative and $y$ is common in both, we write the term $4 y^{2} = {(- 2 y)}^{2}$ .

Now,

$25 x^{2} + 4 y^{2} + 9 z^{2} - 20 x y - 12 y z + 30 x z$

$= {(5 x)}^{2} + {(- 2 y)}^{2} + {(3 z)}^{2} + 2 \cdot 5 x \cdot (- 2 y) + 2 \cdot (- 2 y) \cdot 3 z + 2 \cdot 3 z \cdot 5 x$

$= {[5 x + (- 2 y) + 3 z]}^{2}$

$= {(5 x - 2 y + 3 z)}^{2}$

$= (5 x - 2 y + 3 z) (5 x - 2 y + 3 z)$

Question 6 Exercise 3D RS Aggarwal Class 9 Solutions has been given below.

6. $16 x^{2} + 4 y^{2} + 9 z^{2} - 16 x y - 12 y z + 24 x z$

Answer

Since the terms $- 16 x y$ and $- 12 y z$ are negative and $y$ is common in both, we write the term $4 y^{2} = {(- 2 y)}^{2}$ .

Now,

$16 x^{2} + 4 y^{2} + 9 z^{2} - 16 x y - 12 y z + 24 x z$

$= {(4 x)}^{2} + {(- 2 y)}^{2} + {(3 z)}^{2} + 2 \cdot 4 x \cdot (- 2 y) + 2 \cdot (- 2 y) \cdot 3 z + 2 \cdot 3 z \cdot 4 x$

$= {[4 x + (- 2 y) + 3 z]}^{2}$

$= {(4 x - 2 y + 3 z)}^{2}$

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

Question 7 Exercise 3D RS Aggarwal Class 9 Solutions has been given below.

7. Evaluate

(i). ${(99)}^{2}$

Answer

We have
${(99)}^{2}$

$= {(100 - 1)}^{2}$ ... $[∵ 99 = 100 - 1]$

$= {(100)}^{2} + 1^{2} - 2 \times 100 \times 1$ ... $[∵ {(a - b)}^{2} = a^{2} + b^{2} - 2 a b]$

$= 10000 + 1 - 200$

$= 10001 - 200$

$= 9801$

(ii). ${(995)}^{2}$

Answer

We have
${(995)}^{2}$

$= {(1000 - 5)}^{2}$ ... $[∵ 995 = 1000 - 5]$

$= {(1000)}^{2} + 5^{2} - 2 \times 1000 \times 5$ ... $[∵ {(a - b)}^{2} = a^{2} + b^{2} - 2 a b]$

$= 1000000 + 25 - 10000$

$= 1000025 - 10000$

$= 990025$

(iii). ${(107)}^{2}$

Answer

We have
${(107)}^{2}$

$= {(100 + 7)}^{2}$

$= {(100)}^{2} + 7^{2} + 2 \times 100 \times 7$ ... $[∵ {(a + b)}^{2} = a^{2} + b^{2} + 2 a b]$

$= 10000 + 49 + 1400$

$= 11449$

Exercise 3D RS Aggarwal Class 9 Solutions

1. Expand:

(i). (a+2b+5c)2

Answer

(ii). (2a−b+c)2

Answer

(iii). (a−2b−3c)2

Answer

2. Expand:

(i). (2a−5b−7c)2

Answer

(ii). (−3a+4b−5c)2

Answer

(iii). (12a−14b+2)2

Answer

Factorise

3. 4x2+9y2+16z2+12xy−24yz−16zx

Answer

4. 9x2+16y2+4z2−24xy+16yz−12xz

Answer

5. 25x2+4y2+9z2−20xy−12yz+30xz

Answer

6. 16x2+4y2+9z2−16xy−12yz+24xz

Answer

7. Evaluate

(i). (99)2

Answer

(ii). (995)2

Answer

(iii). (107)2

Answer

Important Links

**1. Expand:**

(i). ${(a + 2 b + 5 c)}^{2}$

(ii). ${(2 a - b + c)}^{2}$

(iii). ${(a - 2 b - 3 c)}^{2}$

**2. Expand:**

(i). ${(2 a - 5 b - 7 c)}^{2}$

(ii). ${(- 3 a + 4 b - 5 c)}^{2}$

(iii). ${(\frac{1}{2} a - \frac{1}{4} b + 2)}^{2}$

3. $4 x^{2} + 9 y^{2} + 16 z^{2} + 12 x y - 24 y z - 16 z x$

4. $9 x^{2} + 16 y^{2} + 4 z^{2} - 24 x y + 16 y z - 12 x z$

5. $25 x^{2} + 4 y^{2} + 9 z^{2} - 20 x y - 12 y z + 30 x z$

6. $16 x^{2} + 4 y^{2} + 9 z^{2} - 16 x y - 12 y z + 24 x z$

(i). ${(99)}^{2}$

(ii). ${(995)}^{2}$

(iii). ${(107)}^{2}$