Ex-5.1, (Part - 2) Factorization Of Algebraic Expressions, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q20 . x²−y²−4xz+4z²

SOLUTION :

On rearranging the terms

= x²−4xz+4z²−y²

= (x)²−2×x×2z+(2z)²−y²

Using the identity x²−2xy+y²=(x−y)²

= (x−2z)²−y²

Using the identity  p²−q²= ( p + q )( p – q )

= (x−2z+y)(x−2z−y)

∴ x²−y²−4xz+4z² = (x−2z+y)(x−2z−y)

Q21 .

SOLUTION :

Splitting the middle term ,

Q22 . 

SOLUTION :

Splitting the middle term,

=

Q23 . 

SOLUTION :

Splitting the middle term,

Q24 . 

SOLUTION :

Splitting the middle term,

Q25 .

SOLUTION :

Splitting the middle term,

Q26 . 

SOLUTION :

Splitting the middle term,

Q27 . 

SOLUTION :

Splitting the middle term,

Q28 .

 SOLUTION :

Using the identity  (x−y)²=x²+y²−2xy

Q29 .  

SOLUTION :

Splitting the middle term,

Q30 .

SOLUTION :

Splitting the middle term,

Q31 . 9(2a−b)²−4(2a−b)−13

SOLUTION :

Let  2a – b = x

=  9x²−4x−13

Splitting the middle term,

=  9x²−13x+9x−13

=  x(9x−13)+1(9x−13)

=  (9x−13)(x+1)

Substituting  x = 2a – b

=  [9(2a−b)−13](2a−b+1)

=  (18a−9b−13)(2a−b+1)

∴ 9(2a−b)²−4(2a−b)−13 = (18a−9b−13)(2a−b+1)

Q 32 . 7(x−2y)²−25(x−2y)+12

SOLUTION :

Let  x-2y = P

=  7P2−25P+12

Splitting the middle term,

=  7P²−21P−4P+12

=  7P(P−3)−4(P−3)

=  (P−3)(7P−4)

Substituting  P = x – 2y

=  (x−2y−3)(7(x−2y)−4)

=  (x−2y−3)(7x−14y−4)

∴ 7(x−2y)²−25(x−2y)+12 = (x−2y−3)(7x−14y−4)

Q33 . 2(x+y)2−9(x+y)−5

SOLUTION :

Let  x+y = z

=  2z2−9z−5

Splitting the middle term,

=  2z2−10z+z−5

=  2z(z−5)+1(z−5)

=  (z−5)(2z+1)

Substituting  z = x + y

=  (x+y−5)(2(x+y)+1)

=  (x+y−5)(2x+2y+1)

∴ 2(x+y)2−9(x+y)−5 = (x+y−5)(2x+2y+1)

Q34 . Give the possible expression for the  length  &  breadth of the rectangle having   35y²−13y−12 as its area.

SOLUTION :

Area is given as 35y²−13y−12

Splitting the middle term,

Area =  35y²+218y−15y−12

=  7y(5y+4)−3(5y+4)

=  (5y+4)(7y−3)

We also know that area of rectangle  = length ×breadth

∴ Possible length = (5y+4) and  breadth=(7y−3)

Or  possible length = (7y−3) and  breadth= (5y+4)

Q35 . What are  the possible expression for the cuboid having volume 3x²−12x.

SOLUTION :

Volume =  3x²−12x

= 3x(x−4)

= 3×x(x−4)

Also volume = Length ×Breadth×Height

∴ Possible expression for dimensions of cuboid are = 3 , x , (x−4)