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# Exercise 3.4 (Factorisation)

8th Maths : Chapter 3 : Algebra : Factorisation, Factorisation using cubic identities : Exercise 3.4 : Text Book Back Exercises Questions with Answers, Solution

Exercise 3.4

1. Factorise the following by taking out the common factor

(i) 18xy 12 yz

(ii) 9x 5 y 3 + 6x 3 y2 18x 2 y

(iii) x (b 2c ) + y(b 2c)

(iv) (ax + ay) + (bx + by)

(v) 2x 2 (4x 1) 4x + 1

(vi) 3y (x 2)2 2(2 x)

(vii) 6xy 4 y 2 + 12xy 2 yzx

(viii) a 3 3a2 + a 3

(ix) 3y 3 48 y

(x) ab2 bc2 ab + c2

Solution:

(i) 18xy − 12yz = (2 × 3 × 3 × y × x) − (2 × 2 × 3 × y × z)

Taking out the common factors 2, 3, y, we get

= 2 × 3 × y (3x − 2z) = 6y (3x − 2z)

(ii) 9x5 y3 + 6x3y2 – 18x2y  = (3 × 3 × x2 × x3 × y × y2) + (2 × 3 × x2 × x × y × y)(2 × 3 × 3 × x2 × y)

Taking out the common factors 3, x2, y, we get

= 3 × x2 × y (3x3 y2 + 2xy − 6)

= 3x2y (3x3 y2 + 2xy − 6)

(iii) x(b − 2c) + y (b − 2c)

Taking out the binomial factor (b − 2c) from each term, we have

= (b − 2c) (x + y)

(iv) (ax + ay) + (bx + by)

Taking at ‘a’ from the first term and ‘b’ from the second term we have

(ax + ay) + (bx + by) =  a (x + y) + b (x + y)

Now taking out the binomial factor (x + y) from each term

= (x + y) (a + b)

(v) 2x2(4x − 1) − 4x + 1

Taking out −1 from last two terms

2x2(4x − 1) − 4x + 1  = 2x2 (4x − 1) − 1 (4x − 1)

Taking out the binomial factor 4x − 1, we get

= (4x − 1) (2x2 − 1)

(vi) 3y (x − 2)2 – 2 (2 − x)

3y (x − 2)2 – 2 (2 − x) = 3y (x − 2) (x − 2) – 2 (−l) (x − 2)

[ Taking out −1 from 2 − x]

= 3y (x − 2) (x − 2) + 2(x − 2)

Taking out the binomial factor x − 2 from each term, we get

= (x − 2) [3y (x − 2) + 2]

(vii) 6xy − 4y2 + 12xy − 2yzx

6xy + 12xy − 4y2 − 2yzx  [  Addition is commutative]

= (6 × x × y) + (2 × 6 × x × y) + ((−1) (2) (2) y + y) + ((−l) (2) (y) (z) (x))

Taking out 6 × x × y from first two terms and (−1) × 2 × y from last two terms we get

= 6 × x × y ( l + 2) + (−l) (2) y [2y + zx]

= 6xy (3) − 2y (2y + zx)

= (2 × 3 × 3 × x × y) − 2xy (2y + zx)

Taking out 2y from two terms

= 2y (9x − (2y + zx)) = 2y (9x −2yxz)

(viii) a3 − 3a2 + a – 3 = a2 (a − 3) + 1 (a − 3) [ Grouping the terms suitably]

= (a − 3) (a2 + 1)

(ix) 3y3 – 48y = 3 × y × y2 − 3 × 16 × y

Taking out 3 × y

= 3y (y2 − 16) = 3y (y2 − 42)

Comparing y2 − 42 with a2b2

a = y, b = 4

a2b2 = (a + b) (a − b)

y2 − 42 = (y + 4) (y − 4)

3y (y2 – 16 ) = 3y (y + 4) (y − 4)

(x) ab2bc2ab + c2

Grouping suitably

ab2bc2ab + c2 = b (abc2) − 1 (abc2)

Taking out the binomial factor abc2 = (abc2) (b − 1)

2. Factorise the following expressions

(i) x 2 + 14x + 49

(ii) y 2 10 y + 25

(iii) c 2 4c 12

(iv) m2 + m 72

(v) 4x 2 8x + 3

Solution:

(i) x2 + 14x + 49 = x2 + 14x + 72

Comparing with a2 + 2ab + b2  = (a + b)2 we have a = x, b = 7

x2 + 2(x)(7) + 72 = (x + 7)2

x2 + 14x + 49 =  (x + 7)2

(ii) y2 – 10y + 25 = y– 10y + 52

Comparing with a2 − 2ab + b2  = (a − b)2 we have a = y, b = 5

y2 + 2(y)(5) + 52 = (y − 5)2

y2 – 10y + 25  = (y − 5)2

(iii) c2  − 4c − 12

This is of the form ax2 + bx + c

Where a = l, b = −4  c = −12, x = c

Now the product ac = 1 × − 12 = −12 and the sum b = − 4

Product = − 72  :  Sum = 1

1 ×  (−12) = −12  :  1 + (−12) = −11

2 × (−6) = −12    :   2 + (−6) =  −4

The middle term − 4c can be written as 2c − 6c

c2 − 4c − 12 = c2 + 2c − 6c − 12

= c (c + 2) − 6 (c + 2)

Taking out (c + 2)

=> (c + 2) (c − 6)

c2 − 4c − 12 = (c + 2) (c − 6)

(iv) m2 + m − 72

This is of the form ax2 + bx + c

where a = 1, b = l, c = −72

Product a × c = l × −72 = −72

Sum b = 1

The middle term m can be written as 9m − 8m

m2 + m – 72 =  m2 + 9m – 8m – 72

= m (m + 9) – 8 (m + 9)

Taking out (m + 9)

= (m + 9) (m − 8)

m2 + m – 72 = (m + 9) (m − 8)

(v) 4x2 − 8x + 3

This is of the form ax2 + bx + c with a =  4 b = − 8  c = 3

Product ac =  4 × 3 = 12

sum b = − 8

The middle term can be written as − 8x = −2x − 6x

4x2 − 8x + 3 = 4x2 − 2x − 6x + 3

= 2x (2x − 1) − 3 (2x − 1)

= (2x − l) (2x − 3)

4x2 – 8x + 3 = (2x − l) (2x − 3)

3. Factorise the following expressions using (a + b)3 = a 3 + 3a2b + 3ab 2 + b3 identity

(i) 64x3+144x2 + 108x+27

(ii) 27p3+54p2q+36pq2+8q3

Solution:

(i) 64x3 + 144x2 + 108x + 27

= (4x)3 + 3(4x)2 (3) + 3(4) (3)2 + 33

= (4x + 3)3

(ii) 27 p3 + 54p2q + 36pq2 + 8q3

= (3p)3 + 3(3p)2 (2q) + 3(3p) (2q)2 + (2q)3

= (3p + 2q)3

4. Factorise the following expressions using (a b)3 = a 3 3a2b + 3ab 2 b3 identity

(i) y3–18y2+108y–216

(ii) 8m3–60m2n+150mn2–125n3

(i) y3 – 18y2 + 108y − 216  = y3 – 3y2(6) + 3(6)2y − 63

= (y − 6)3

(ii) 8m3 – 60 m2n + 150mn2 – 125 n3

= (2m)3 − 3(2m)2 (5) + 3(2m) (5n)2 − (5n)3

= (2m – 5n)3

Objective type Questions

5. Factors of 9x2+6xy are

(A) 3y, (x+2)

(B) 3x, (3x+3y)

(C) 6x, (3x+2y)

(D) 3x, (3x+2y)

[Answer: (D) 3x, (3x + 2y)]

Solution:

9x2 + 6xy =  3x (3x + 2y)

6. Factors of 4–m2 are

(A) (2+m)(2+m)

(B) (2–m)(2–m)

(C) (2+m)(2–m)

(D) (4+m)(4–m)

[Answer: (C) (2 + m) (2 − m)]

Solution: 4 – m2 = 22m2

= (2 + m) (2 − m

7. (x+4) and (x–5) are the factors of ___________

(A) x2x+20

(B) x2–9x–20

(C) x2+x–20

(D) x2x–20

[Answer: (D) x2 – x – 20]

Solution:

(x + 4) (x − 4) = x (x) + 4 (−5) + x (−5) + 4(x)

= x2 − 20 − 5x + 4x

= x2 − x − 20

8. The factors of x2–5x + 6 are (x–2)(x–p) then the value of p is _________

(A) –3

(B) 3

(C) 2

(D)–2

Solution:

x2 − 5x + 6 = x2 −3x − 2x + 6

= x (x − 3) – 2 (x − 3)

= (x − 3) (x − 2)

= (xp) (x − 2)

p = 3

9. The factors of 1–m3

(A) (1+m), (1+m+m2)

(B) (1–m), (1–mm2)

(C) (1–m), (1+m+m2)

(D) (1+m), (1–m+m2)

[Answer: (C) (1 − m), (1 + m + m2)]

Solution:

a3b3 = (ab) (a2 + ab + b2)

1 – m2 = (1 − m) (1+ m + m2)

10. One factor of x3+y3 is

(A) (xy)

(B) (x + y)

(C) (x + y)3

(D) (xy)3

a3 + b3 = (a + b) (a2ab + b2)

x3 + y3 = (x + y) (x2xy + y2

Exercise 3.4

1. (i ) 6 y (3 x − 2z) (ii) 3 x2y(3x3 y2 + 2xy − 6) (iii) (b − 2c )( x + y)

(iv) ( x + y)( a + b) ( v ) ( 4 x − 1)( 2 x2 −1) ( vi) ( x − 2)[3 y(x − 2) + 2]

(vii ) 2y(9x − 2 y zx) ( viii) ( a − 3) ( a2 +1) (ix) 3y ( y + 4)( y − 4) (x) (b − 1)(ab − c2 )

2. (i ) (x + 7)2 (ii) ( y − 5)2 (iii ) ( c + 2)( c − 6) (iv) ( m + 9)( m − 8) (v) ( 2 x − 3)( 2 x −1)

3. (i) (4x + 3)3 (ii) (3p + 2q)3

4. (i) ( y − 6)3 (ii) ( 2 m − 5n)3

5. (D) 3x, (3x+2y)

6. (C) (2+m)(2–m)

7. (D) x2x–20

8. (B) 3

9. (C) (1–m)(1+m+m2)

10. (B) (x+y)

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8th Maths : Chapter 3 : Algebra : Exercise 3.4 (Factorisation) | Questions with Answers, Solution | Algebra | Chapter 3 | 8th Maths