Polynomials: 9 Math

NCERT Exercise 2.5 Part-2: 9th math

NCERT Exercise 2.5 Polynomials class ninth math Question (5) Factorize:

(i) 4x² + 9y² + 16z² + 12xy – 24yz – 16xz

Solution

Given, 4x² + 9y² + 16z² + 12xy – 24yz – 16xz

= (– 2x)² +(– 3y)² + (4z)² + 2(– 2x)(– 3y) + 2(– 3y)(4z) + 2(– 2x)(4z)

Now, using (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx, we get

(– 2z – 3y +4z)²

= (– 2z – 3y +4z) (– 2z – 3y +4z) Answer

(ii) 2x² + y² + 8z² – 2√2 xy + 4√2 yz – 8xz

Solution

Given, 2x² + y² + 8z² – 2√2 xy + 4√2 yz – 8xz

= (– x√2)² + y² + (2z √2)² + 2(– x√2)(y) + 2(y)(2z √2) – 2(– x√2)(2z √2)

Now, using (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx, we get

(– x√2 + y + 2z√2)²

= (– x√2 + y + 2z√2) (– x√2 + y + 2z√2) Answer

NCERT Exercise 2.5 Polynomials class ninth math Question (6) Write the following cubes in expanded from:

(i) (2x + 1)³

Solution

Given, (2x + 1)³

Here, we have x = 2x and y = 1

We know that, (x + y)³ = x³ + y³ + 3xy(x + y)

Thus, by replacing x = 2x and y = 1

(2x + 1)³

= (2x)³ + 1³ + 3(2x)(1)(2x + 1)

= 8x³ + 1 + 6x(2x + 1)

= 8x³ + 1 + 12x² + 6x

= 8x³ + 12x² + 6x + 1 Answer

(ii) (2a – 3b)³

Solution

Given, (2a – 3b)³

Here, we have x = 2a and y = 3b

We know that, (x – y)³ = x³ – 3x²y + 3xy² – y³

Thus, by replacing x = 2a and y = 3b

(2a – 3b)³

= (2a)³ – 3(2a)²(3b) + 3(2a)(3b)² – (3b)³

= 8a³ – 18a²b + 18ab² – 9b³ Answer

(iii) `[3/2x+1]^3`

Solution

Given, `[3/2x+1]^3`

Here we have x = 3/2x and y = 1

We know that, (x + y)³ = x³ + y³ + 3xy(x + y)

Thus, after replacing x = 3/2x and y = 1 we get

Thus, `27/8x^3+27/4x^2+9/2x+1` Answer

(iv) `[x-2/3y]^3`

Solution

Given, `[x-2/3y]^3`

Here, x = x and y = 2/3y

We know that, (x – y)³ = x³ – 3x²y + 3xy² – y³

Thus, by replacing x = x and y = 2/3y we get

Thus, `x^3-2x^2y-4/3xy^2-8/27y^3` Answer

NCERT Exercise 2.5 Polynomials class ninth math Question(7) Evaluate the following using suitable identities:

(i) (99)³

Solution

Given, (99)³

= (100 – 1)³

Using (x – y)³ = x³ – y³ – 3xy(x – y), we get

(100 – 1)³

= (100)³ – (1)³ – 3(100)(1)(100 – 1)

=1000000 – 1 – 300(99)

= 1000000 – 1 – 29700

= 970299 Answer

(ii) (102)³

Solution

Given, (102)³

= (100 + 2)³

Using identity, (x + y)³ = x³ + y³ + 3xy(x + y), we get

(100 + 2)³

= (100)³ + (2)³ + 3(100)(2)(100 + 2)

= 1000000 + 8 + 600 (102)

= 1000008 + 61200

= 1061208 Answer

(iii) (998)³

Solution

Given, (998)³

= (1000 – 2)³

Using identity, (x – y)³ = x³ – y³ + 3xy(x – y), we get

(1000 – 2)³

= (1000)³ – (2)³ – 3(1000)(2)(1000 – 2)

= 1000000000 – 8 – 6000(998)

= 1000000000 – 8 – 5988000

= 994011992 Answer

NCERT Exercise 2.5 Polynomials class ninth math Question (8) Factorize each of the following:

(i) 8a³ + b³ + 12a²b + 6ab²

Solution

Given, 8a³ + b³ + 12a²b + 6ab²

= (2a)³ + b³ + 3(2a)²b + 3(2a)b²

Now, using identity (x + y)³ = x³+ y³ 3x²y + 3xy², we get

(2a)³ + b³ + 3(2a)²b + 3(2a)b²

= (2a + b)³

= (2a +b)(2a + b)(2a + b) Answer

(ii) 8a³ – b³ – 12a²b + 6ab²

Solution

Given, 8a³ – b³ – 12a²b + 6ab²

= (2a)³ + (– b)³ + 3(2a)²(– b) + 3(2a)(– b)²

Now, using identity (x + y)³ = x³+ y³ 3x²y + 3xy², we get

[2a +(– b)]³

= (2a –b)³

= (2a – b)(2a – b)(2a – b) Answer

(iii) 27 – 125a³ – 135a + 225a²

Solution

Given, 27 – 125a³ – 135a + 225a²

=(3)³ + (– 5a)³ + 3(3)²(– 5a) + 3(3)(– 5a)²

Here, x = 3 and y = – 5a

Thus, by using identity (x + y)³ = x³+ y³ 3x²y + 3xy², we get

[3 + (– 5a)]³

= (3 – 5a)³

= (3 – 5a)(3 – 5a)(3 – 5a) Answer

(iv) 64a³ – 27b³ – 144a²b + 108ab²

Solution

Given, 64a³ – 27b³ – 144a²b + 108ab²

= (4a)³ + (– 3b)³ + 3(4a)²(– 3b) + 3(4a)(– 3b)²

Here, x = 4a and y = – 3b

Thus, by using identity (x + y)³ = x³+ y³ 3x²y + 3xy², we get

[4a + (– 3b)]³

= (4a – 3b)³

= (4a – 3b)(4a – 3b)(4a – 3b) Answer

(v) `27p^3-1/216-9/2p^2+1/4p`

Solution

Given,

`27p^3-1/216-9/2p^2+1/4p`

`=(3p)^3 + (-1/6)^3` `+ 3(3p)^2(-1/6) + 3(3p)(-1/6)^2`

Here, x = 3p and y = – 1/6

Thus, by using identity (x + y)³ = x³+ y³ 3x²y + 3xy², we get

[3p + (– 1/6)]³

= (3p – 1/6)³

= (3p – 1/6)(3p – 1/6)(3p – 1/6) Answer

NCERT Exercise 2.5 Polynomials class ninth math Question (9) Verify:

(i) x³ + y³ = (x + y)(x² – xy + y²)

Solution

Given, x³ + y³ = (x + y)(x² – xy + y²)

LHS = (x + y)(x² – xy + y²)

= x(x² – xy + y²) + y(x² – xy + y²)

= x³ – x²y + xy² + yx² – xy² + y³

= x³ + y³

= RHS Proved

(ii) x³ – y³ = (x – y)(x² + xy + y²)

Solution

Given, x³ – y³ = (x – y)(x² + xy + y²)

RHS = (x – y)(x² + xy + y²)

= x(x² + xy + y²) – y(x² + xy + y²)

= x³ + x²y + xy² – yx² – xy² – y³

= x³ – y³

= RHS Proved

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