# KC Sinha Mathematics Solution Class 9 Algebraic identities exercise 4.4

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Page 4.26

### Exercise 4.4

Type 6

#### Question 1

Factorize the following if :

(i) (x+1) is a factor of 2x3 – 3x2 – 8x – 3

Sol :

Step-1⇒2x3-3x2-8x-3

Step-2⇒2x3+2x2-5x2-5x-3x-3

Step-3⇒2x2(x+1)-5x(x+1)-3(x+1)

[After writing step 1 , jumped to 3 and written 3 times (x+1) and multiply it with suitable algebric expression or number so as to get the expression of step 1 . Now step 2 can be written down from step 3]

[Step 4 = Now taking common (x+1) ]

Step-4⇒(x+1)(2x2-5x-3)

Step-5⇒(x+1)(2x2+x-6x-3) [middle term split]

⇒(x+1)[x(2x+1)-3(2x+1)]

⇒(x+1)(x-3)(2x+1)

(ii) (2x+3) is a factor of 4x3+20x2+33x+18

Sol :

⇒4x3+20x2+33x+18

⇒4x3+6x2+14x2+21x+12x+18

⇒2x2(2x+3)+7x(2x+3)+6(2x+3)

[taking common (2x+3)]

⇒(2x+3)(2x2+7x+6)

[mid term split]

⇒(2x+3)(2x2+4x+3x+6)

⇒(2x+3)[2x(x+2)+3(x+2)]

⇒(x+2)(2x+3)(2x+3)

(iii)  (x+9) is a factor of x3+13x2+31x-45

Sol :

⇒x3+13x2+31x-45

⇒x3+9x2+4x2+36x-5x-45

⇒x2(x+9)+4x(x+9)-5(x+9)

[taking common (x+9)]

⇒(x+9)[x2+4x-5]

⇒(x+9)[x2+4x-5]

⇒(x+9)[x2-x+5x-5]

⇒(x+9)[x(x-1)+5(x-1)]

⇒(x+9)(x+5)(x-1)

(iv) (x-1) is a factor of 2x3-5x2+x+2

Sol :

⇒2x3-5x2+x+2

⇒2x3-2x2-3x2+3x-2x+2

⇒2x2(x-1)-3x(x-1)-2(x-1)

[taking common (x-1)]

⇒(x-1)(2x2-3x-2)

[mid-term split]

⇒(x-1)(2x2-4x+x-2)

⇒(x-1)[2x(x-2)+1(x-2)]

⇒(x-1)(2x+1)(x-2)

Page 4.27

#### Question 2

Find factors of the following :

(i) x3+2x2-6x+3

Sol :

Here c=3 and factors are ±1 , ±3

Also , p(1)=13+2×12-6×1+3

=1+2-6+3 = 0

∴(x-1) is factor of p(x)

⇒x3+2x2-6x+3

⇒x3-x2+3x2-3x-3x+3

⇒x2(x-1)+3x(x-1)-3(x-1)

[Taking common (x-1)]

⇒(x-1)(x2+3x-3)

(ii) x3-7x+6

Sol :

Here c=6 and factors are ±1 , ±2 and ±3

Also , p(1)=13-7×1+6=0

∴(x-1) is factor of p(x)

⇒x3-7x+6

⇒x3-x2+x2-x-6x+6

⇒x2(x-1)+x(x-1)-6(x-1)

[Taking common (x-1)]

⇒(x-1)(x2+x-6)

[mid term split]

⇒(x-1)(x2-2x+3x-6)

⇒(x-1)[x(x-2)+3(x-2)]

⇒(x-1)(x+3)(x-2)

(iii) x3-2x2-x+2

Sol :

Here c=2

Also , the factors are ±1 , ±2

p(1)=13-2×12-1+2

=1-2-1+2=0

∴(x-1) is factor of p(x)

⇒x3-2x2-x+2

⇒x3-x2-x2+x-2x+2

⇒x2(x-1)-x(x-1)-2(x-1)

[Taking common (x-1)]

⇒(x-1)[x2-x-2]

⇒(x-1)[x2+x-2x-2]

⇒(x-1)[x(x+1)-2(x+1)]

⇒(x-1)(x-2)(x+1)

(iv) x3+3x2-4

Sol :

Here c=-4 and numerical value is 4

Also , factors are ±1,±2

p(1)=13+3×12-4

=1+3-4=0

∴(x-1) is factor of p(x)

⇒x3+3x2-4

⇒x3-x2+4x2-4x+4x-4

⇒x2(x-1)+4x(x-1)+4(x-1)

[Taking common (x-1)]

⇒(x-1)[x2+4x+4]

[mid term split]

⇒(x-1)[x2+2x+2x+4]

⇒(x-1)[x(x+2)+2(x+2)]

⇒(x-1)(x+2)(x+2)

(v) x3-6x+4

Sol :

Here c=4 and factors are ±1 , ±2

Also , p(2)=23-6×2+4

=8-12+4=0

∴(x-2) is factor of p(x)

⇒x3-6x+4

⇒x3-2x2+2x2-4x-2x+4

⇒x2(x-2)+2x(x-2)-2(x-2)

[Taking common (x-2)]

⇒(x-2)(x2+2x-2)

(vi) x3-13x-12

Sol :

Here c=-12 and numerical value is 12

factors are ±1,±2,±3,±4,±6,±12

Also , p(-1)=13-13×-1-12

=-1+13-12=0

∴(x+1) factor of p(x)

⇒x3-13x-12

⇒x3+x2-x2-x-12x-12

⇒x2(x+1)-x(x+1)-12(x+1)

[Taking common (x-1)]

⇒(x+1)(x2-x-12)

[mid term split]

⇒(x+1)(x2+3x-4x-12)

⇒(x+1)[x(x+3)-4(x+3)]

⇒(x+1)(x+3)(x-4)

(vii) x3-8x2+17x-10

Sol :

Here c=-10 and numerical value is 10 and factors are ±1 , ±2 and ±5

Clearly , p(1)=13-8×12+17×1-10=0

∴(x-1) is a factor of p(x)

Also ,

⇒x3-8x2+17x-10

⇒x3-x2-7x2+7x+10x-10

⇒x2(x-1)-7x(x-1)+10(x-1)

[Taking common (x-1)]

⇒(x-1)(x2-7x+10)

[mid term split]

⇒(x-1)(x2-5x-2x+10)

⇒(x-1)[x(x-5)-2(x-5)]

⇒(x-1)(x-2)(x-5)

(viii) x3+13x2+31x-45

Sol :

Here c=-45 and numerical value is 45 and factors are ±1 , ±5 and ±3

Clearly , p(1)=13+13×12+31×1-45=0

∴(x-1) is a factor of p(x)

⇒x3+13x2+31x-45

⇒x3-x2+14x2-14x+45x-45

⇒x2(x-1)+14x(x-1)+45(x-1)

[Taking common (x-1)]

⇒(x-1)(x2+14x+45)

[mid term split]

⇒(x-1)(x2+9x+5x+45)

⇒(x-1)[x(x+9)+5(x+9)]

⇒(x-1)(x+5)(x+9)

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