myhelper: RS Aggarwal solution class 8 chapter 7 Factorisation Exercise 7C

Exercise 7C

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Q1 | Ex-7C | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 1:

Factorise:
x² + 8x + 16

Answer 1:

We have:
$x^{2} + 8 x + 16 = x^{2} + 2 \times x \times 4 + {(4)}^{2} = {(x + 4)}^{2}$

∴ $x^{2} + 8 x + 16 = {(x + 4)}^{2}$

Q2 | Ex-7C | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 2:

Factorise:
x² + 14x + 49

Answer 2:

We have:
$x^{2} + 14 x + 49 = x^{2} + 2 \times x \times 7 + {(7)}^{2} = {(x + 7)}^{2}$

∴ $x^{2} + 14 x + 49 = {(x + 7)}^{2}$

Q3 | Ex-7C | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 3:

Factorise:
1 + 2x + x²

Answer 3:

We have:
$1 + 2 x + x^{2} = x^{2} + 2 x + 1 = x^{2} + 2 \times x \times 1 + {(1)}^{2} = {(x + 1)}^{2}$

∴ $1 + 2 x + x^{2} = {(x + 1)}^{2}$

Q4 | Ex-7C | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 4:

Factorise:
9 + 6z + z²

Answer 4:

We have;
$9 + 6 z + z^{2} = z^{2} + 6 z + 9 = z^{2} + 2 \times x \times 3 + {(3)}^{2} = {(z + 3)}^{2}$

∴ $9 + 6 z + z^{2} = {(z + 3)}^{2}$

Q5 | Ex-7C | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 5:

Factorise:
x² + 6ax + 9a²

Answer 5:

We have:
$x^{2} + 6 a x + 9 a^{2} = x^{2} + 2 \times x \times 3 a + {(3 a)}^{2} = {(x + 3 a)}^{2}$

∴ $x^{2} + 6 a x + 9 a^{2} = {(x + 3 a)}^{2}$

Q6 | Ex-7C | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 6:

Factorise:
4y² + 20y + 25

Answer 6:

We have:
$4 y^{2} + 20 y + 25 = {(2 y)}^{2} + 2 \times 2 y \times 5 + {(5)}^{2} = {(2 y + 5)}^{2}$

∴ $4 y^{2} + 20 y + 25 = {(2 y + 5)}^{2}$

Q7 | Ex-7C | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 7:

Factorise:
36a² + 36a + 9

Answer 7:

We have:
$36 a^{2} + 36 a + 9 = {(6 a)}^{2} + 2 \times 6 a \times 3 + {(3)}^{2} = {(6 a + 3)}^{2}$

∴ $36 a^{2} + 36 a + 9 = {(6 a + 3)}^{2}$

Q8 | Ex-7C | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 8:

Factorise:
9m² + 24m + 16

Answer 8:

We have:
$9 m^{2} + 24 m + 16 = {(3 m)}^{2} + 2 \times 3 m \times 4 + {(4)}^{2}$
$= {(3 m + 4)}^{2}$

∴ $9 m^{2} + 24 m + 16 = {(3 m + 4)}^{2}$

Q9 | Ex-7C | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 9:

Factorise:
z² + z + $\frac{1}{4}$

Answer 9:

We have:
$z^{2} + z + \frac{1}{4} = z^{2} + 2 \times z \times \frac{1}{2} \times {(\frac{1}{2})}^{2}$
$= {(z + \frac{1}{2})}^{2}$

∴ $z^{2} + z + \frac{1}{4} = {(z + \frac{1}{2})}^{2}$

Q10 | Ex-7C | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 10:

Factorise:
49a² + 84ab + 36b²

Answer 10:

We have:
$49 a^{2} + 84 a b + 36 b^{2} = {(7 a)}^{2} + 2 \times 7 a \times 6 b + {(6 b)}^{2}$
$= {(7 a + 6 b)}^{2}$

∴ $49 a^{2} + 84 a b + 36 b^{2} = {(7 a + 6 b)}^{2}$

Q11 | Ex-7C | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 11:

Factorise:
p² − 10p + 25

Answer 11:

We have:
$p^{2} - 10 p + 25 = p^{2} - 2 \times p \times 5 + {(5)}^{2}$
$= {(p - 5)}^{2}$

∴ $p^{2} - 10 p + 25 = {(p - 5)}^{2}$

Q12 | Ex-7C | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 12:

Factorise:
121a² − 88ab + 16b²

Answer 12:

We have:
$121 a^{2} - 88 a b + 16 b^{2} = {(11 a)}^{2} - 2 \times 11 a \times 4 b + {(4 b)}^{2}$
$= {(11 a - 4 b)}^{2}$

∴ $121 a^{2} - 88 a b + 16 b^{2} = {(11 a - 4 b)}^{2}$

Q13 | Ex-7C | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 13:

Factorise:
1 − 6x + 9x²

Answer 13:

We have:
$1 - 6 x + 9 x^{2} = 9 x^{2} - 6 x + 1$
$= {(3 x)}^{2} - 2 \times 3 x \times 1 + {(1)}^{2} = {(3 x - 1)}^{2}$

∴ $1 - 6 x + 9 x^{2} = {(3 x - 1)}^{2}$

Q14 | Ex-7C | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 14:

Factorise:
9y² − 12y + 4

Answer 14:

We have:
$9 y^{2} - 12 y + 4 = {(3 y)}^{2} - 2 \times 3 y \times 2 + {(2)}^{2}$
$= {(3 y - 2)}^{2}$

∴ $9 y^{2} - 12 y + 4 = {(3 y - 2)}^{2}$

Q15 | Ex-7C | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 15:

Factorise:
16x² − 24x + 9

Answer 15:

We have:
$16 x^{2} - 24 x + 9 = {(4 x)}^{2} - 2 \times 4 x \times 3 + {(3)}^{2}$
$= {(4 x - 3)}^{2}$

∴ $16 x^{2} - 24 x + 9 = {(4 x - 3)}^{2}$

Q16 | Ex-7C | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 16:

Factorise:
m² − 4mn + 4n²

Answer 16:

We have:
$m^{2} - 4 m n + 4 n^{2} = m^{2} - 2 \times m \times 2 n + {(2 n)}^{2}$
$= {(m - 2 n)}^{2}$

∴ $m^{2} - 4 m n + 4 n^{2} = {(m - 2 n)}^{2}$

Q17 | Ex-7C | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 17:

Factorise:
a²b² − 6abc + 9c²

Answer 17:

We have:
$a^{2} b^{2} - 6 a b c + 9 c^{2} = {(a b)}^{2} - 2 \times a b \times 3 c + {(3 c)}^{2}$
$= {(a b - 3 c)}^{2}$

Q18 | Ex-7C | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 18:

Factorise:
m⁴ + 2m²n² + n⁴

Answer 18:

We have:
$m^{4} + 2 m^{2} n^{2} + n^{4} = {(m^{2})}^{2} + 2 \times m^{2} \times n^{2} + {(n^{2})}^{2}$
$= {(m^{2} + n^{2})}^{2}$

∴ $m^{4} + 2 m^{2} n^{2} + n^{4} = {(m^{2} + n^{2})}^{2}$

Q19 | Ex-7C | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 19:

Factorise:
(l + m)² − 4lm

Answer 19:

We have:
${(l + m)}^{2} - 4 l m = (l^{2} + m^{2} + 2 l m) - 4 l m$
$= l^{2} + m^{2} + 2 l m - 4 l m = l^{2} + m^{2} - 2 l m = {(l)}^{2} + {(m)}^{2} - 2 \times l \times m = {(l - m)}^{2}$

∴ ${(l + m)}^{2} - 4 l m = {(l - m)}^{2}$

RS Aggarwal solution class 8 chapter 7 Factorisation Exercise 7C