Polynomials: 9 Math
NCERT Exercise 2.5: 9th math
NCERT Exercise 2.5 Polynomials class ninth math Question (1) Use suitable identities to find the following products:
(i) (x +4) (x +10)
Solution
Given, (x +4) (x + 10)
We know that, (x + a)(x + b) = x2 + (a + b)x + ab
Here, we have a = 4 and b = 10
Thus, by replacing a = 4 and b = 10, we get
(x + 4)(x + 10)
= x2 + (4 + 10)x + 4 ร 10
= x2 + 14x + 40 Answer
(ii) (x + 8)(x โ 10)
Solution
Given, (x + 8)(x โ 10)
Then product using suitable identity = ?
We know that, (x + a)(x + b) = x2 + (a + b)x + ab
Now, (x +8)(x โ 10)
= (x + 8)(x + (โ10)]
Now, using (x + a)(x + b) = x2 + (a + b)x + ab, we get
x2 + [8 + (โ10)]x + 8(โ10)
= x2 + (8 โ10) x โ 80
= x2 + (โ2)x โ 80
= x2 โ 2x โ 80 Answer
(iii) (3x + 4)(3x โ 5)
Solution
Given, (3x + 4)(3x โ 5)
Then product using suitable identity = ?
(3x + 4)(3x โ 5)
= (3x + 4)[3x + (โ5)]
Using (x + a)(x + b) = x2 + (a + b)x + ab, we get
(3x)2 + [4 + (โ5)]3x + [4 ร (โ5)]
= 9x2 + (4 โ 5) 3x + (โ20)
= 9x2 + (โ1)3x โ 20
= 9x2 โ 3x โ 20 Answer
(iv) `(y^2+3/2)(y^2-3/2)`
Solution
Given, `(y^2+3/2)(y^2-3/2)`
Then product using suitable identity = ?
We know that, (a + b)(a โ b) = a2 โ b2
Here we have a = y2 and b = `3/2`
Thus, `(y^2+3/2)(y^2-3/2)`
`=y^2-(3/2)^2`
`=y^2-9/4` Answer
(v) (3 โ 2x)(3 + 2x)
Solution
Given, (3 โ 2x)(3 + 2x)
We know that, (a + b)(a โ b) = a2 โ b2
Here we have a = 3 and b = 2x
Thus, (3 โ 2x)(3 + 2x)
= (3)2 โ (2x)2
= 9 โ 4x2 Answer
NCERT Exercise 2.5 Polynomials class ninth math Question (2) Evaluate the following products without multiplying directly:
(i) 103 ร 107
Solution
Given, 103 ร 107
= (100 + 3)(100 + 7)
Now, we know that, (x + a)(x + b) = x2 + (a + b)x + ab
Here, we have x = 100, a = 3 and b = 7
Thus, (100 + 3)(100 + 7)
= (100)2 + (3 + 7)100 + 3 ร 7
= 10000 + 10 ร 100 + 21
= 10000 + 1000 + 21
= 11021 Answer
(ii) 95 ร 96
Solution
First Method
Given, 95 ร 96
= (90 + 5)(90 +6)
Now, we know that, (x + a)(x + b) = x2 + (a + b)x + ab
Here, we have x = 90, a = 5 and b = 6
Thus, (90 + 5)(90 +6)
= (90)2 + (5 + 6)90 + 5 ร 6
= 8100 + (11 ร 90) + 30
= 8100 + 990 + 30
= 9120 Answer
Second Method
Given, 95 ร 96
= (100 โ 5)(100 โ4)
= [100 +(โ5)][100 + (โ4)]
Now, we know that, (x + a)(x + b) = x2 + (a + b)x + ab
Here, we have x = 100, a = โ5 and b = โ4
Thus, [100 +(โ5)][100 + (โ4)]
= (100)2 + [โ5 +(โ4)] 100 + [โ5 (โ4)]
= 10000 + (โ5 โ4)100 + 20
= 10000 + (โ9)100 + 20
= 10000 โ 900 + 20
= 10000 โ 880
= 9120 Answer
(iii) 104 ร 96
Solution
Given, 104 ร 96
= (100 + 4)(100 โ 4)
We know that, (a + b)(a โb) = a2 โ b2
Here, we have a = 100 and b = 4
Thus, by replacing a = 100 and b = 4, we get
= (100)2 โ (4)2
= 10000 โ 16
= 9984 Answer
NCERT Exercise 2.5 Polynomials class ninth math Question (3) Factorise the following using appropriate identities:
(i) 9x2 + 6xy + y2
Solution
Given, 9x2 + 6xy + y2
= (3x)2 + 2 ร 3x ร y + y2
Here, a = 3x and b = y
Thus, by using (a +b)2 = a2 + 2ab + b2, we get
(3x + y)2
= (3x +y)(3x +y) Answer
(ii) 4y2 โ 4y + 1
Solution
Given, 4y2 โ 4y + 1
= (2y)2 โ 2 ร 2y ร 1 + 12
Here, a = 2y and b = 1
Thus, by using (a โ b)2 = a2 โ 2ab + b2, we get
(2y โ 1)2
= (2y โ 1)(2y โ 1) Answer
(iii) `x^2-y^2/100`
Solution
Given, `x^2-y^2/100`
Here a = x and y `=y^2/10`
Thus, by using a2 โ b2 = (a + b)(a โ b)
`x^2-y^2/100`
`= (x + y/10)(x โ y/10)` Answer
NCERT Exercise 2.5 Polynomials class ninth math Question (4) Expand each of the following, using suitable identities:
(i) (9x + 2y + 4z)2
Solution
(9x + 2y + 4z)2
We know that, (x + y + z)2 = x2 + y2 + z2 + 2xy +2yz + 2zx
We have x = 9x, y = 2y and z = 4z
Thus, after replacing x = 9x, y = 2y and z = 4z, we get
(9x)2 + (2y)2 + (4z)2 + 2ร 9x ร 2y + 2 ร 2y ร 4z + 2 ร 4z ร 9x
= 81x2 + 4y2 + 16z2 + 36 xy + 16 yz + 72 zx Answer
(ii) (2x โ y + z)2
Solution
Given, (2x โ y + z)2
= [2x + (โy) + z]2
We know that, (x + y + z)2 = x2 + y2 + z2 + 2xy +2yz + 2zx
We have x = 2x, y = โy and z = z
Thus, after replacing x = 2x, y = โy and z = z, we get
(2x)2 + (โy)2 + z2 + 2 ร 2x ร (โy) + 2ร (โy) ร z + 2 z ร (2x)
= 4x2 + y2 + z2 โ 4xy โ 2yz + 4zx Answer
(iii) (โ2x + 3y + 2z)2
Solution
Given, (โ2x + 3y + 2z)2
= [(โ2x) + 3y + 2z2
We know that, (x + y + z)2 = x2 + y2 + z2 + 2xy +2yz + 2zx
Here we have, x = โ2x, y = 3y and z = 2z
Thus, after replacing x = โ2x, y = 3y and z = 2z, we get
(โ2x)2 + (3y)2 + (2z)2 + 2(โ2x)(3y) + 2(3y)(2z) + 2(2z)(โ2x)
= 4x2 + 9y2 + 4z2 โ 12xy + 12yz โ 8zx Answer
(iv) (3a โ 7b โ c)2
Solution
Given, (3a โ 7b โ c)2
= [3a + (โ 7b) + (โ c)]2
We know that, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Here, we have x = 3a, y = โ 7b and z = โ c
Thus, after replacing x = 3a, y = โ 7b and z = โ c, we get
[3a + (โ 7b) + (โ c)]2
= (3a)2 + (โ 7b)2 + (โ c)2 + 2(3a)(โ 7b) + 2 (โ 7b)(โ c) + 2(โ c)(3a)
= 9a2 + 49b2 + c2 โ 42ab + 14bc โ 6ca Answer
(v) (โ 2x + 5y โ 3z)2
Solution
Given, (โ 2x + 5y โ 3z)2
= [(โ 2x) + 5y + (โ 3z)]2
We know that, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Here, we have x = โ 2x, y = 5y and z = โ 3z
Thus, after replacing x = โ 2x, y = 5y and z = โ 3z, we get
[(โ 2x) + 5y + (โ 3z)]2
= (โ 2x)2 + (5y)2 + (โ 3z)2 + 2(โ 2x)(5y) + 2(5y)(โ 3z) + 2(โ 3z)(โ 2x)
= 4x2 + 25y2 + 9z2 โ 20xy โ 30yz + 12zx Answer
(vi) `[1/4a-1/2b+1]^2`
Solution
Given, `[1/4a-1/2b+1]^2`
We know that, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Here, we have x = `1/4a`, `y = -1/2b` and z = 1
After replacing values of x, y and z, given equation can be written using above identity
`(1/4a)^2 + (-1/2b)^2+1^2` `+2(1/4a)(-1/2b)` `+2(-1/2b)xx1` `+2(1)(1/4a)`
`=1/16a^2+1/4b^2+1` `-1/4ab-b+1/2a` Answer
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