S.chand books class 8 maths solution chapter 14 Factorisation exercise 14 F

EXERCISE 14 F

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Q1 | Ex-14F | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 1

(a+b)² - 1

Sol :

=(a+b)² - 1²

[ Using : a²-b²=(a+b)(a-b) ]

=(a+b+1)(a+b-1)

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Q2 | Ex-14F | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 2

(3x-2y)² - 9z²

Sol :

=(3x-2y)² - (3z)²

[ Using : a²-b²=(a+b)(a-b) ]

=(3x-2y+3z)(3x-2y-3z)

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Q3 | Ex-14F | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 3

1-(x-y)²

Sol :

=(1)²-(x-y)²

[ Using : a²-b²=(a+b)(a-b) ]

=(1+x-y)[1-(x-y)]

=(1+x-y)(1-x+y)

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Q4 | Ex-14F | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 4

4x² -9(2x-y)²

Sol :

=(2x)²-[3(2x-y)]²

[ Using : a²-b²=(a+b)(a-b) ]

=[2x-3(2x-y)][2x+3(2x-y)]

=(2x-6x+3y)(2x+6x+3y)

=(-4x+3y)(8x+3y)

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Q5 | Ex-14F | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 5

16(x+y)² - 25(x-y)²

Sol :

=[4(x+y)]² - [5(x-y)]²

[ Using : a²-b²=(a+b)(a-b) ]

=[4(x+y)+5(x-y)][4(x+y)-5(x-y)]

=(4x+4y+5x-5y)(4x+4y-5x+5y)

=(9x-y)(-x+9y)

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Q6 | Ex-14F | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 6

ab² - ac²

Sol :

=a(b² - c²) taking "a" common 

[ Using : a²-b²=(a+b)(a-b) ]

=a[(b+c)(b-c)]

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Q7 | Ex-14F | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 7

36x³ -x
Sol :

=x(36x²-1) taking "x" common

=x[(6x)²-(1)²]

=x(6x+1)(6x-1)

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Q8 | Ex-14F | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 8

3 - 12b²
Sol :

=3(1-4b²) taking "3" common

=3[(1²-(2b²)] 

[ Using : a²-b²=(a+b)(a-b) ]

=3[(1+2b)(1-2b)]

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Q9 | Ex-14F | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 9

18ax² - 98ay²

Sol :

=2a(9x² - 49y²)

[ Using : a²-b²=(a+b)(a-b) ]

=2a[(3x)² - (7y²)]

=2a(3x+7y)(3x-7y)

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Q10 | Ex-14F | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 10

2x² -$\dfrac{1}{2}$   [Hint: Given Exp$=2\left(x²-\dfrac{1}{4}\right)$]

Sol :

$=\frac{4x^2-1}{2}$

$=\frac{1}{2}\left[(2x)^2-(1)^2\right]$ 

[ Using : a²-b²=(a+b)(a-b) ]

$=\frac{1}{2}]\left[(2x-1)(2x+1)\right]$ taking 2 common 

$=2 \times \frac{1}{2}\left[\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)\right]$

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Q11 | Ex-14F | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 11

x⁴ - y⁴

Sol :

=(x²)²-(y²)²

[ Using : a²-b²=(a+b)(a-b) ]

=(x²+y²)(x²-y²)

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Q12 | Ex-14F | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 12

x⁴ - 625

Sol :

=(x²)²-(25)²

[ Using : a²-b²=(a+b)(a-b) ]

=(x²+25)(x²-25)

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Q13 | Ex-14F | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 13

16y⁴ - 81

Sol :

=(4y)²-(9)²

[ Using : a²-b²=(a+b)(a-b) ]

=(4y²+9)(4y²-9)

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Q14 | Ex-14F | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 14

xy⁵ - yx⁵

Sol :

=xy(y⁴ - x⁴) taking common xy

=xy[(y²)²-(x²)²]

[ Using : a²-b²=(a+b)(a-b) ]

=xy[(y²-x²)(y²+x²)]

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Q15 | Ex-14F | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 15

(a+b)² -(b-a)²

Sol :

[ Using : a²-b²=(a+b)(a-b) ]

=[(a+b)+(b-a)][(a+b)-(b-a)]

=[a+b+b-a][a+b-b+a]

=(2b)(2a)

=4ab

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Q16 | Ex-14F | Class 8 | S.Chand | Composite maths | chapter 14 |Factorisation | myhelper

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Question 16

(x+2y-5z)² - (x-2y+5z)²

Sol :

[ Using : a²-b²=(a+b)(a-b) ]

=[(x+2y-5z)+(x-2y+5z)][(x+2y+5z)-(x-2y+5z)]

=[x+2y-5z+x-2y-5z][x+2y-5z-x+2y-5z]

=(2x-10z)(4y-10z)