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Polynomials: 9 Math


mathematics Class Nine

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 + (โ€“10)]

Now, using (x + a)(x + b) = x2 + (a + b)x + ab, we get

x2 + [8 + (โ€“10)]x + 8(โ€“10)

= x2 + (8 โ€“10) x โ€“ 80

= x2 + (โ€“2)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 + (โ€“5)]

Using (x + a)(x + b) = x2 + (a + b)x + ab, we get

(3x)2 + [4 + (โ€“5)]3x + [4 ร— (โ€“5)]

= 9x2 + (4 โ€“ 5) 3x + (โ€“20)

= 9x2 + (โ€“1)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 โ€“4)

= [100 +(โ€“5)][100 + (โ€“4)]

Now, we know that, (x + a)(x + b) = x2 + (a + b)x + ab

Here, we have x = 100, a = โ€“5 and b = โ€“4

Thus, [100 +(โ€“5)][100 + (โ€“4)]

= (100)2 + [โ€“5 +(โ€“4)] 100 + [โ€“5 (โ€“4)]

= 10000 + (โ€“5 โ€“4)100 + 20

= 10000 + (โ€“9)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 โ€“b) = 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 + (โ€“y) + z]2

We know that, (x + y + z)2 = x2 + y2 + z2 + 2xy +2yz + 2zx

We have x = 2x, y = โ€“y and z = z

Thus, after replacing x = 2x, y = โ€“y and z = z, we get

(2x)2 + (โ€“y)2 + z2 + 2 ร— 2x ร— (โ€“y) + 2ร— (โ€“y) ร— z + 2 z ร— (2x)

= 4x2 + y2 + z2 โ€“ 4xy โ€“ 2yz + 4zx Answer

(iii) (โ€“2x + 3y + 2z)2

Solution

Given, (โ€“2x + 3y + 2z)2

= [(โ€“2x) + 3y + 2z2

We know that, (x + y + z)2 = x2 + y2 + z2 + 2xy +2yz + 2zx

Here we have, x = โ€“2x, y = 3y and z = 2z

Thus, after replacing x = โ€“2x, y = 3y and z = 2z, we get

(โ€“2x)2 + (3y)2 + (2z)2 + 2(โ€“2x)(3y) + 2(3y)(2z) + 2(2z)(โ€“2x)

= 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

solution of ncert exercise 2.5  polynomials class ninth

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