KC Sinha Mathematics Solution Class 9 Algebraic identities exercise 4.4

Page 4.26

Exercise 4.4

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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)

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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)