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CLASS 9 MATHS CHAPTER 2 POLYNOMIALS
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Answers:-
I.
1)Given product is (x+4)(x+10)
It is in the form of (x+a)(x+b)
Where , x = x ,a = 4 and b = 10
We know that
(x+a)(x+b)=x^2+(a+b)x+ab
(x+4)(x+10) = x^2+(4+10)x+(4)(10)
(x+4)(x+10) = x^2+14x+40
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2)Given product is (x+8)(x-10)
=(x+8)(x+(-10))
It is in the form of (x+a)(x+b)
Where , x = x ,a = 8 and b = -10
We know that
(x+a)(x+b)=x^2+(a+b)x+ab
(x+8)(x-10) = x^2+(8-10)x+(8)(-10)
(x+8)(x-10) = x^2+(-2)x+(-80)
(x+8)(x-10) = x^2-2x-80
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3))Given product is (3x+4)(3x-5)
=(3x+4)(3x+(-5))
It is in the form of (x+a)(x+b)
Where , x = 3x ,a = 4 and b = -5
We know that
(x+a)(x+b)=x^2+(a+b)x+ab
(3x+4)(3x-5) =(3x)^2+(4-5)(3x)+(4)(-5)
(3x+4)(3x-5) = 9x^2+(-1)(3x)+(-20)
(3x+4)(3x-5) = 9x^2-3x-20
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4)Given product is (y^2 +3/2)(y^2- 3/2)
It is in the form of (a+b)(a-b)
Where a = y^2 , b= 3/2
We know that
(a+b)(a-b)=a^2-b^2
(y^2 +3/2)(y^2- 3/2) = [(y^2)^2]-(3/2)^2
(y^2 +3/2)(y^2- 3/2) = y^4 -(9/4)
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5)Given product is (3-2x)(3+2x)
It is in the form of (a+b)(a-b)
Where a = 3 , b= 2x
We know that
(a+b)(a-b)=a^2-b^2
(3-2x)(3+2x) = (3)^2-(2x)^2
(3-2x)(3+2x) = 9-4x^2
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II.
1)Given product is 103×107
=> (100+3)(100+7)
It is in the form of (x+a)(x+b)
Where , x = 100 ,a = 3 and b = 7
We know that
(x+a)(x+b)=x^2+(a+b)x+ab
103×107= (100)^2+(3+7)(100)+(3×7)
103×107= 10000+10(100)+21
103×107=10000+1000+21
103×107 = 11021
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2)Given product is 95×96
=> (100-5)(100-4)
=> (100+(-5))(100+(-4))
It is in the form of (x+a)(x+b)
Where , x = 100 ,a = -5and b =-4
We know that
(x+a)(x+b)=x^2+(a+b)x+ab
95×96= (100)^2+(-5-4)(100)+(-5)(-4)
95×96= 10000+(-9)(100)+20
95×96=10000-900+20
95×96 = 9120
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3)Given product is 104×96
=>(100+4)(100-4)
It is in the form of (a+b)(a-b)
Where a = 100 , b= 4
We know that
(a+b)(a-b)=a^2-b^2
104×96 = (100+4)(100-4)
104×96 = 100^2-4^2
104×96 = 10000-16
104×96 = 9984
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III.
1)Given expression is 9x^2+6xy+y^2
=> (3x)^2+2(3x)(y)+y^2
It is in the form of a^2+2ab+b^2
Where a = 3x and b = y
We know that
(a+b)^2=a^2+2ab+b^2
(3x)^2+2(3x)(y)+y^2= (3x+y)^2
(3x)^2+2(3x)(y)+y^2= (3x+y)(3x+y)
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2)Given expression is 4y^2-4y+1
=> (2y)^2-2(2y)(1)+(1)^2
It is in the form of a^2-2ab+b^2
Where a = 2y and b = 1
We know that
(a-b)^2=a^2-2ab+b^2
(2y)^2-2(2y)(1)+(1)^2 = (2y-1)^2
4y^2-4y+1 = (2y-1)(2y-1)
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3)Given expression is x^2-(y^2/100)
=> (x)^2 - (y/10)^2
It is in the form of a^2-b^2
Where a = x and b = y/10
We know that
a^2-b^2 = (a+b)(a-b)
(x)^2 - (y/10)^2 = [x+(y/10)][x-(y/10)]
x^2-(y^2/100) = [x+(y/10)][x-(y/10)]
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Used Identities:-
- (a+b)^2 =a^2+2ab+b^2
- (a-b)^2 =a^2-2ab+b^2
- (a+b)(a-b)=a^2-b^2
- (x+a)(x+b)=x^2+(a+b)x+ab