Gujarat BoardĀ GSEB Textbook Solutions Class 8 Maths Chapter 14 Factorization Ex 14.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 8 Maths Chapter 14 Factorization Ex 14.2

Question 1.

Factorise the following expressions.

- a
^{2}+ 8a + 16 - p
^{2}– 10p + 25 - 25m
^{2}+ 30m + 9 - 49y
^{2}+ 84yz + 36z^{2} - 4x
^{2}– 8x + 4 - 121b
^{2}– 88bc + 16c^{2} - (1+ m)
^{2}– 41m - a
^{4}+ 2a^{2}b^{2}+ b^{4}[Hint: Expand (1 + m)^{2}first].

Solution:

1. We have a^{2} + 8a + 16

= (a)^{2} + 2(a)(4) + (4)^{2}

= (a + 4)^{2} = (a + 4)(a + 4)

ā“ a^{2} + 8a + 16 = (a + 4)^{2} = (a + 4)(a + 4)

2. We have p^{2} – 10p + 25

= p^{2} – 2(p)(5) + (5)^{2}

= (p – 5)^{2} = (p – 5)(p – 5)

3. We have 25m^{2} + 30m + 9

= (5m)^{2} + 2(5m)(3) + (3)^{2}

= (5m + 3)^{2} = (5m + 3)(5m + 3)

ā“ 25m^{2} + 30m + 9

= (5m + 3)^{2} = (5m + 3)(5m + 3)

4. We have 49y^{2} + 84yz + 36z

= (7y)^{2} + 2(7y)(6z) + (6z)^{2}

= (7y + 6z)^{2}

= (7y + 6z)(7y + 6z)

ā“ 49y^{2} + 84yx + 36z^{2}

= (7y + 6z)^{2}

= (7y + 6z)(7y + 6z)

5. We have 4x^{2} – 8x + 4

= (2x)^{2} – 2(2x)(2) + (2)^{2}

= (2x – 2)^{2} = (2x – 2)(2x – 2)

= 2(x – 1)2(x – 1)

= 4(x – 1)(x – 4)

ā“ 4x^{2} – 8x + 4 = 4(x – 1)(x – 1)

= 4(x – 1)^{2}

6. We have 121b^{2} – 88bc + 16c^{2}

= (11b)^{2} – 2(11b)(4c) + (4c)^{2}

= (11b – 4c)^{2} = (11b – 4c)(11b – 4c)

ā“ 121b^{2} – 88bc + 16c^{2}

7. We have (l + m)^{2} – 4lm

= (l^{2} + 2lm + m^{2}) – 4lm

[Collecting the like terms 2lm and -4lm]

= l^{2} + (2lm – 4lm) + m^{2}

= l^{2} + 2lm + m^{2}

= (l)^{2} – 2(l)(m) + (m)^{2}

= (l – m)^{2} = (l – m)(l – m)

ā“ (l + m)^{2} – 4lm = (l – m)^{2}

= (l – m)(l – m)

8. We have a^{4} + 2a^{2}b^{2} + b^{4}

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

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

ā“ a^{4} + 2a^{2}b^{2} + b^{2}

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

Question 2.

Factorise:

- 4p
^{2}– 9q^{2} - 63a
^{2}– 112b^{2} - 49x
^{2}– 36 - 16x
^{5}– 144x^{3} - (l + m)
^{2}– (l – m)^{2} - 9x
^{2}y^{2}– 16 - (x
^{2}– 2xy + y^{2}) – z^{2} - 25a
^{2}– 4b^{2}+ 28bc – 49c^{2}

Solution:

1. āµ 4p^{2} – 9q^{2} = (2p)^{2} – (3q)^{2}

= (2p – 3q)(2p + 3q)

[Using a^{2} – b^{2} (a + b)(a – b)

ā“ 4p^{2} – 9q^{2} = (2p – 3q)(2p + 3q)

2. We have

63a^{2} – 112b^{2} = 7 Ć 9a^{2} – 7 Ć 16b^{2}

= 7(9a^{2} – 16b^{2})

āµ 9a^{2} – 16b^{2} = (3a)^{2} – (4b)^{2}

= (3a + 4b)(3a – 4b)

[Using a^{2} – b^{2} = (a + b)(a – b)]

ā“ 63a^{2} – 112b^{2} = 7[(3a) + 4b)(3a – 4b)]

3. āµ 49x^{2} – 36 = (7x)^{2} – (6)^{2}

= (7x – 6)(7x + 6)

[(Using a^{2} – b^{2} = (a – b)(a + b)]

ā“ 49x^{2} – 36 = (7x – 6)(7x + 6)

4. We have 16x^{5} = 16x^{3} Ć x^{2} and 144x^{3}

= 16x^{3} Ć 9

ā“ 16x^{5} – 144x^{3} = (16x^{3}) Ć x^{2} – (16x^{3}) Ć 9

= 16x^{3} Ć 9

= 16x^{3}[x^{2} – 9]

= 16x^{3}[(x)^{2} – (3)^{2}]

= 16x^{3}[(x + 3)(x – 3)]

[(Using a^{2} – b^{2} = (a – b)(a + b)]

ā“ 16x^{5} – 144x^{3} = 16x^{3}(x + 3)(x – 3)

5. Using the identiy

a^{2} – b^{2} = (a + b)(a – b)

we have (l + m)^{2} – (l – m)^{2}

= [(l + m) + (l – m)][(l + m) – (l – m)]

= [l + m + l – m][l + m – l + m]

= (2l)(2m) = 2 Ć 2(l Ć m) = 4lm

6. We have 9x^{2}y^{2} – 16 (3xy)^{2} – (4)^{2} = (3xy + 4)(3xy – 4)

[Using a^{2} – b^{2} = (a + b)(a – b)]

ā“ 9x^{2}y^{2} – 16 = (3xy + 4)(3xy – 4)

7. We have x^{2} – 2xy + y^{2} = (x – y)^{2}

ā“ (x^{2} – 2xy + y^{2}) – z^{2} = (x – y)^{2} – (z)^{2}

= [(x – y) + z][(x – y) – z]

[Using a^{2} – b^{2} = (a + b)(a – b)]

= (x – y + z)(x – y – z)

ā“ (x^{2} – 2xy + y^{2}) – z^{2} = (x – y + z)(x – y – z)

8. We have -4b^{2} + 28bc – 49c^{2}

= (-1)[4b^{2} – 28bc + 49c^{2}]

= -1[(2b)^{2} – 2(2b)(7c) + (7c)^{2}]

= -[2b – 7c]^{2}

ā“ 25a^{2} – 4b^{2} + 28bc – 49c^{2} = 25a^{2} – (2b – 7c)^{2}

Now, using a^{2} – b^{2} = (a + b)(a – b),

we have [5a]^{2} – [2b – 7c]^{2}

= (5a + 2b – 7c)(5a – 2b + 7c)

ā“ 25a^{2} – 4a^{2} + 28bc – 49c^{2}

= (5a + 2b – 7c)(5a – 2b + 7c)

Question 3.

Factorise the expressions?

- ax
^{2}+ bx - 7p
^{2}+ 21q^{2} - 2x
^{3}+ 2xy^{2}+ 2xz^{2} - am
^{2}+ bm^{2}+ bn^{2}+ an^{2} - (lm + l) + m + l
- y(y + z) + 9(y + z)
- 5y
^{2}– 20y – 8z + 2yz - 10ab + 4a + 5b + 2
- 6xy – 4y + 6 – 9x

Solution:

1. āµ ax^{2} = a Ć x Ć x = (x)[ax]

bx = b Ć x = (x)[b]

ā“ ax^{2} + bx = x[ax + b]

2. We have 7p^{2} + 21q^{2}

= 7 Ć p Ć p + 7 Ć 3 Ć q Ć q

= 7[p Ć p + 3 Ć q Ć q]

= 7[p^{2} + 3q^{2}]

3. Taking out 2x as common from each term, we have

2x^{3} + 2xy^{2} + 2xz^{2}= 2x[x^{2} + y^{2} + z^{2}]

4. We can take out m^{2} as common from the first two terms and n^{2} as common from the last two terms,

ā“ am^{2} + bm^{2} + bn^{2} + an^{2}

= m^{2}(a + b) + n^{2}(b + a)

= m^{2}(a + b) + n^{2}(a + b)

= (a + b)[m^{2} + n^{2}]

5. We have lm + l = l(m + 1)

ā“ (lm + l) + (m + 1)

= l(m + 1) + (m + 1)

= (m + 1)[l + 1]

6. We have (y + z) as a common factor to both terms.

ā“ y(y + z) + 9(y + z) = (y + z)(y + 9)

7. We have 5y^{2} – 20y = 5y(y – 4)

and -8z + 2yz = 2z(-4 + y)

= 2z(y – 4)

ā“ 5y^{2} – 20y – 8z + 2yz

= 5y(y – 4) + 2z(y – 4)

= (y – 4)[5y + 2z]

8. We have, 10ab + 4a = 2a(5b + 2)

and (5b + 2) = 1(5b + 2)

ā“ 10ab + 4a + 5b + 2

= 2a(5b + 2) + 1(5b + 2)

= (5b + 2)[2a + 1]

9. Regrouping the terms, we get

6xy – 4y + 6 – 9x = 6xy – 4y – 9x + 6

= 2y[3x – 2] – 3[3x – 2]

= (3x – 2)[2y – 3]

Question 4.

Factorise:

- a
^{4}– b^{4} - p
^{4}– 81 - x
^{4}– (y + z)^{4} - x
^{4}– (x – z)^{4} - a
^{4}– 2a^{2}b^{2}+ b^{4}

Solution:

1. Using a^{2} – b^{2} = (a – b)(a + b),

we have a^{4} – b^{4} = (a^{2})^{2} – (b^{2})^{2}

= (a^{2} + b^{2})(a^{2} – b^{2})

= (a^{2} + b^{2})[(a + b)(a – b)]

= (a^{2} + b^{2})(a + b)(a – b)

2. We have p^{4} – 81 = (p^{2})^{2} – (9)^{2}

Now using a^{2} – b^{2} = (a + b)(a – b),

we have (p^{2})^{2} – (9)^{2} = (p^{2} + 9) (p^{2} – 9)

We can factorise p^{2} – 9 further as

p^{2} – 9 = (p)^{2} – (3)^{2}

= (p + 3)(p – 3)

ā“ p^{4} – 81 = (p + 3)(p – 3)(p^{2} + 9)

3. āµ x^{4} – (y + z)^{4} = [x^{2}]^{2} – [(y + z)^{2}]^{2}

= [(x)^{2} + (y + z)^{2}][(x^{2}) – (y + z)^{2}]

Using a^{2} – b^{2} = (a + b)(a – b)]

We can factorise [x^{2} – (y + z)^{2}] further as

(x)^{2} – (y + z)^{2}

= [(x) + (y + z)][(x) – (y + z)]

= (x + y + z)(x – y + z)

ā“ x^{4} – (y + z)^{4}

= (x + y + z)(x – y – z)[x^{2} + (y + z)^{2}]

4. We have x^{4} – (x – z)^{4} = [x^{2}]^{2} – [(x – z)^{2}]^{2}

= [x^{2} + (x – z)^{2}][x^{2} – (x – z)^{2}]

Now, factorising x^{2} – (x – z)^{2} further,

we have

x^{2} – (x – z)^{2} = [x + (x – z)[x – (x – z)]

= (x + x – z)(x – x + z)

= (2x – z)(z)

ā“ x^{4} – (y – z)^{4} = z(2x – 2)[x^{2} + (x – 2)^{2}]

= z(2x – 2)[x^{2} + (x^{2} – 2xz + z^{2})]

= z(2x – z)[x^{2} + x^{2} – 2xz + z^{2}]

= z(2x – z)(2x^{2} – 2xz + z^{2})

5. āµ a^{4} – 2a^{2}b^{2} + b^{4}

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

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

= [(a – b)(a + b)(a^{2} + b^{2})]

ā“ a^{4} – 2a^{2}b^{2} + b^{4} = (a – b)(a + b)(a^{2} + b^{2})

Question 5.

Factorise the following expressions.

- p
^{2}+ 6p + 8 - q
^{2}– 10q + 21 - p
^{2}+ 6p – 16

Solution:

1. We have p^{2} + 6p + 8 = p^{2} + 6p + 9 – 1 [āµ 8 = 9 – 1]

= [(p)^{2} + 2(p)(3) + 3^{2}] – 1 = (p + 3)^{2} – 1^{2}

[Using a^{2} + 2ab + b^{2} = (a + b)^{2}]

= [(p + 3) + 1][(p + 3) – 1]

= (p + 4)(p + 2)

ā“ p^{2} + 6p + 8 = (p + 4)(p + 2)

2. We have q^{2} – 10q + 21

= q^{2} – 10q + 25 – 4 [āµ 21 = 25 – 4]

= [(q)^{2} – 2(q)(5) + (5)^{2}] – (2)^{2}

= [q – 5]^{2} – [2]^{2}

= [(q – 5) + 2][(q – 5) – 2]

[Using a^{2} – b^{2} = (a + b)(a – b)] = (q – 3)(q – 7)

3. We have p^{2} + 6p – 16 = p^{2} + 6p + 9 – 25 [āµ -16 = 9 – 25]

= [(p)^{2} + 2(p)(3) + (3)^{2}] – (5)^{2}

= [p + 3]^{2} – [5]^{2}

[Using a^{2} + 2ab + b^{2} = (a + b)^{2}]

= [(p + 3) + 5][(p + 3) – 5]

[Using a^{2} – b^{2} = (a + b)(a – b)]

= (p + 8)(p – 2)