Jharkhand Board JAC Class 9 Maths Solutions Chapter 2 Polynomials Ex 2.5 Textbook Exercise Questions and Answers.

## JAC Board Class 9th Maths Solutions Chapter 2 Polynomials Ex 2.5

Question 1.

Use suitable identities to find the following products:

(i) (x + 4) (x + 10)

(ii) (x + 8) (x – 10)

(iii) (3x + 4) (3x – 5)

(iv) (y^{2} + 3/2) (y^{2} – 3/2)

(v) (3 – 2x) (3 + 2x)

Answer:

(i) Using identity, (x + a) (x + b)

= x^{2} + (a + b) x + ab

In (x + 4) (x + 10), a = 4 and b = 10

Now,

(x + 4) (x + 10)

= x^{2} + (4 + 10)x + (4 × 10)

= x^{2} + 14x+ 40

(ii) (x + 8) (x – 10)

Using identity, (x + a) (x + b)

= x^{2} + (a + b) x + ab

Here, a = 8 and b = -10

(x + 8) (x – 10)

= x^{2} + {8 + (- 10)}x + {8 × (- 10)}

= x^{2} + (8 – 10)x – 80

= x^{2} – 2x – 80

(iii) (3x + 4) (3x – 5)

Using identity, (x + a) (x + b)

= x^{2} + (a + b) x + ab

Here, x as 3x, a = 4 and b = – 5

(3x + 4) (3x – 5)

= (3x)^{2} + {4 + (-5)}3x + {4 × (-5)}

= 9x^{2} + 3x(4 – 5) – 20

= 9x^{2} – 3x – 20

(iv) (y^{2} + 3/2) (y^{2} – 3/2)

Using identity, (x + y) (x – y) = x^{2} – y^{2}

Here, x = y^{2} and y = 3/2

(y^{2} + 3/2) (y^{2} – 3/2)

= (y^{2})^{2} – (3/2)^{2}

= y^{4} – 9/4

(v) (3 – 2x) (3 + 2x)

Using identity, (x + y) (x – y) = x^{2} – y^{2}

Here, x = 3 and y = 2x

(3 – 2x) (3 + 2x)

= 3^{2} – (2x)^{2}

= 9 – 4x^{2}

Question 2.

Evaluate the following products without multiplying directly:

(i) 103 × 107

(ii) 95 × 96

(iii) 104 × 96

Answer:

(i) 103 × 107

Answer:

(i) 103 × 107 = (100 + 3) (100 + 7)

Using identity, (x + a) (x + b)

= x^{2} + (a + b) x + ab

Here, x = 100, a = 3 and b = 7

103 × 107 = (100 + 3) (100 + 7)

= (100)^{2}+ (3 + 7)100 + (3 × 7)

= 10000 + 1000 + 21 = 11021

(ii) 95 × 96 = (90 + 5) (90 + 6)

Using identity, (x + a) (x + b)

= x^{2} + (a + b) x + ab

Here, x = 90, a = 5 and b = 6

95 × 96 = (90 + 5) (90 + 6)

= 90^{2} + 90(5 + 6) + (5 × 6)

= 8100 + (11 × 90) + 30 = 8100 + 990 + 30 = 9120

(iii) 104 × 96 = (100 + 4) (100 – 4)

Using identity, (x + y) (x – y) = x^{2} – y^{2}

Here, x = 100 and y = 4

104 × 96 = (100 + 4) (100 – 4)

= (100)^{2} – (4)^{2}

= 10000 – 16 = 9984

Question 3.

Factorise the following using appropriate identities:

(i) 9x^{2} + 6xy + y^{2}

(ii) 4y^{2} – 4y + 1

(iii) x^{2} – \(\frac{y^2}{100}\)

Answer:

(i) 9x^{2} + 6xy + y^{2}

= (3x)^{2} + (2 × 3x × y) + y^{2}

Using identity, (a + b)^{2} = a^{2} + 2ab + b^{2}

Here, a = 3x and b = y

9x^{2} + 6xy + y^{2}

= (3x)^{2} + (2 × 3x × y) + y^{2}

= (3x + y)^{2}

= (3x + y) (3x + y)

(ii) 4y^{2} – 4y + 1

= (2y)^{2} – (2 × 2y × 1) + 1^{2}

Using identity, (a – b)^{2} = a^{2} – 2ab + b^{2}

Here, a = 2y and b = 1

4y^{2} – 4y + 1

= (2y)^{2} – (2 × 2y × 1) + 1^{2}

= (2y – 1)^{2}

= (2y – 1) (2y – 1)

(iii) x^{2} – \(\frac{y^2}{100}\) = x^{2}– (\(\frac{y}{10}\))^{2}

Using identity, a^{2} – b^{2}= (a + b) (a – b)

Here, a = x and b = (\(\frac{y}{10}\))

x^{2} – \(\frac{y^2}{100}\) = x^{2} – (\(\frac{y^2}{100}\))^{2}

= (x – \(\frac{y}{10}\)) (x + \(\frac{y}{10}\))

Page – 49

Question 4.

Expand each of the following, using suitable identities:

(i) (x + 2y + 4z)^{2}

(ii) (2x – y + z)^{2}

(iii) (-2x + 3y + 2z)^{2}

(iv) (3a – 7b – c)^{2}

(v) (-2x + 5y – 3z)^{2}

(vi) [\(\frac{1}{4}\) a – \(\frac{1}{2}\) b + 1]^{2}

Answer:

(i) (x + 2y + 4z)^{2}

Using identity, (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca

Here, a = x, b = 2y and c = 4z

(x + 2y + 4z)^{2}

= x^{2} + (2y)^{2} + (4z)^{2} + (2 × x × 2y) + (2 × 2y × 4z) + (2 × 4z × x)

= x^{2} + 4y^{2} + 16z^{2} + 4xy + 16yz + 8xz

(ii) (2x – y + z)^{2}

Using identity, (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca

Here, a = 2x, b = – y and c = z

(2x – y + z)^{2}

= (2x)^{2} + (-y)^{2} + z^{2} + (2 × 2x × (- y)) + (2 × (- y) × z) + (2 × z × 2x)

= 4x^{2} + y^{2} + z^{2} – 4xy – 2yz + 4xz

(iii) (-2x + 3y + 2z)^{2}

Using identity, (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca

Here, a = – 2x, b = 3y and c = 2z

(-2x + 3y + 2z)^{2}

=(-2x)^{2} + (3y)^{2} + (2z)^{2} + (2 × (- 2x) × 3y) + (2 × 3y × 2z) + (2 × 2z × (- 2x))

= 4x^{2} + 9y^{2} + 4z^{2} – 12xy + 12yz – 8xz

(iv) (3a – 7b – c)^{2}

Using identity, (a + b + c)^{2} = a^{2} + b^{2}+ c^{2} + 2ab + 2bc + 2ca

Here, a = 3a, b = – 7b and c = – c

(3a – 7b – c)^{2}

= (3a)^{2} + (-7b)^{2} + (-c)^{2} + (2 × 3a × (-7b)) + (2 × (- 7b) × (- c)) + (2 × (- c) × 3a)

= 9a^{2} + 49b^{2} + c^{2} – 42ab + 14bc – 6ac

(v) (-2x + 5y – 3z)^{2}

Using identity, (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca

Here, a = – 2x, b = 5y and c = – 3z

(-2x + 5y – 3z)^{2}

= (-2x)^{2} + (5y)^{2} + (-3z)^{2} + (2 × (-2x) × 5y) + (2 × 5y × (- 3z)) + (2 × (- 3z) × (- 2x))

= 4x^{2} + 25y^{2} + 9z^{2} – 20xy – 30yz + 12xz

(vi) [\(\frac{1}{4}\) a – \(\frac{1}{2}\) b + 1]^{2}

Using identity, (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca

Here, a = \(\frac{1}{4}\)a; b = \(\frac{1}{2}\) and c = 1

[\(\frac{1}{4}\) a – \(\frac{1}{2}\) b + 1]^{2}

= \(\left(\frac{1}{4} \mathrm{a}\right)^2+\left(-\frac{1}{2} \mathrm{~b}\right)^2+1^2+\left(2 \times \frac{1}{4} \mathrm{a} \times \frac{(-1)}{2} \mathrm{~b}\right)+\left(2 \times \frac{(-1)}{2} \mathrm{~b} \times 1\right)+\left(2 \times 1 \times \frac{1}{4} \mathrm{a}\right)\)

= \(\frac{1}{16} \mathrm{a}^2+\frac{1}{4} \mathrm{~b}^2+1-\frac{1}{4} \mathrm{ab}-\mathrm{b}+\frac{1}{2} \mathrm{a}\)

Question 5.

Factorise:

(i) 4x^{2} + 9y^{2} + 16z^{2} + 12xy – 24yz – 16xz

(ii) 2x^{2} + y^{2} + 8z^{2} – 2\( \sqrt{2} \)xy + 4 \( \sqrt{2} \) yz – 8xz

Answer:

(i) 4x^{2} + 9y^{2} + 16z^{2} + 12xy – 24yz -16xz

Using identity, (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca

4x^{2} + 9y^{2} + 16z^{2} + 12xy – 24yz – 16xz

= (2x)^{2} + (3y)^{2} + (-4z)^{2} + (2 × 2x × 3y) + (2 × 3y × (- 4z)) + (2 × (- 4z) × 2x)

= (2x + 3y – 4z)^{2} = (2x + 3y – 4z) (2x + 3y – 4z)

(ii) 2x^{2} + y^{2} + 8z^{2} – 2 \( \sqrt{2} \) xy + 4 \( \sqrt{2} \) yz – 8xz

Using identity, (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca

2×2 + y2 + 8z2 – 2 \( \sqrt{2} \) xy + 4 \( \sqrt{2} \) yz – 8xz

= (-\( \sqrt{2} \) x)^{2} + (y)^{2} + (2\( \sqrt{2} \)z)^{2} + (2 × (- \( \sqrt{2} \) x) × y) + (2 × y × 2 \( \sqrt{2} \) z) + (2 × 2\( \sqrt{2} \) z × (- \( \sqrt{2} \)x))

= (-V/2x + y+ 2>/2z)^{2} = (-V2x + y + 2 \( \sqrt{2} \) z) (- \( \sqrt{2} \) x + y + 2 \( \sqrt{2} \) z)

Question 6.

Write the following cubes in expanded form:

(i) (2x + 1)^{3}

(ii) (2a – 3b)^{3}

(iii) \(\left[\frac{3}{2} x+1\right]^3\)

(iv) \(\left[x-\frac{2}{3} y\right]^3\)

Answer:

(i) (2x + 1)^{3}

Using identity, (a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)

(2x+1)^{3} = (2x)^{3} + 1^{3} + (3 × 2x × 1)(2x + 1) = 8x^{3} + 1 + 6x(2x + 1)

= 8x^{3} + 12x^{2} + 6x + 1

(ii) (2a – 3b)^{3}

Using identity, (a – b)^{3} = a^{3} – b^{3} – 3ab(a – b)

(2a – 3b)^{3}

= (2a)^{3} – (3b)^{3} – (3 × 2a × 3b)(2a – 3b) = 8a^{3} – 27b^{3} – 18ab(2a – 3b)

= 8a^{3} – 27b^{3} – 36a^{2}b + 54ab^{2}

(iii) \(\left[\frac{3}{2} x+1\right]^3\)

Using identity, (a – b)^{3} = a^{3} – b^{3} – 3ab(a – b)

(iv) \(\left[x-\frac{2}{3} y\right]^3\)

Using identity, (a – b)^{3} = a^{3} – b^{3} – 3ab(a – b)

Question 7.

Evaluate the following using suitable identities:

(i) (99)^{3}

(ii) (102)^{3}

(iii) (998)^{3}

Answer:

(i) (99)^{3} = (100 – 1)^{3}

Using identity, (a – b)^{3} = a^{3} – b^{3} – 3ab(a – b)

(99)^{3} = (100 – 1)^{3}

= (100)^{3} – (1)^{3} – (3 × 100 × 1) (100 – 1)

= 1000000 – 1 – 30000 + 300

= 970299

(ii) (102)^{3} = (100 + 2)^{3}

Using identity, (a – b)^{3} = a^{3} – b^{3} – 3ab(a – b)

(100 + 2)^{3}

= (100)^{3} + (2)^{3} + (3 × 100 × 2) (100 + 2)

= 1000000 + 8 + 600 (100 + 2)

= 1000000 + 8 + 60000 + 1200

= 1061208

(iii) (998)^{3} = (1000 – 2)^{3}

Using identity, (a – b)^{3} = a^{3} – b^{3} – 3ab(a – b)

= (1000)^{3} – (2)^{3} – (3 × 1000 × 2) (1000 – 2)

= 1000000000 – 8 – 6000 × (1000 – 2)

= 1000000000 – 8 – 6000000 + 12000

= 1000012000 – 6000008

= 994011992

Question 8.

Factorise each of the following:

(i) 8a^{3} + b^{3} + 12a^{2}b + 6ab^{2}

(ii) 8a^{3} – b^{3} – 12a^{2}b + 6ab^{2}

(iii) 27 – 125a^{3} – 135a + 225a^{2}

(iv) 64a^{3} – 27b^{3} – 144a^{2}b + 108ab^{2}

(v) 27p^{3} – \(\frac{1}{216}-\frac{9}{2}\)p^{2} + \(\frac {1}{4}\)p

Answer:

(i) 8a^{3} + b^{3} + 12a^{2}b + 6ab^{2}

Using identity, (a + b)^{3} = a^{3} + b^{3} + 3a^{2}b + 3ab^{2}

= (2a)^{3} + (b)^{3} + 3 (2a) (b) (2a + b)

= (2a + b)^{3}

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

(ii) 8a^{3} – b^{3} – 12a^{2}b + 6ab^{2}

Using identity, (a + b)^{3} = a^{3} + b^{3} + 3a^{2}b + 3ab^{2}

= (2a)^{3} – (b)^{3} – 3 (2a)(b)(2a – b)

= (2a – b)^{3} = (2a – b)(2a – b)(2a – b)

(iii) 27 – 125a^{3} – 135a + 225a^{2}

Using identity, (a + b)^{3} = a^{3} + b^{3} + 3a^{2}b + 3ab^{2}

= (3)^{3} – (5a)^{3} – 3(3)(5a) (3 – 5a)

= (3 – 5a)^{3} = (3 – 5a) (3 – 5a) (3 – 5a)

(iv) 64a^{3} – 27b^{3} – 144a^{2}b + 108aba^{2}

Using identity, (a + b)^{3} = a^{3} + b^{3} + 3a^{2}b + 3ab^{2}

= (4a)^{3} – (3b)^{3} – 3 (4a) (3b) (4a – 3b)

= (4a – 3b)^{3}

= (4a – 3b) (4a – 3b) (4a – 3b)

(v) 27p^{3} – \(\frac{1}{216}-\frac{9}{2}\)p^{2} + \(\frac {1}{4}\)p

Using identity, (a + b)^{3} = a^{3} + b^{3} + 3a^{2}b + 3ab^{2}

Question 9.

Verify:

(i) x^{3} + y^{3} = (x + y) (x^{2} – xy + y^{2})

(ii) x^{3} – y^{3} = (x – y) (x^{2} + xy + y^{2})

Answer:

(i) (x + y) (x^{2} – xy + y^{2})

We know that,

(x + y)^{3} = x^{3} + y^{3} + 3xy(x + y)

⇒ x^{3} + y^{3} = (x + y)^{3} – 3xy(x + y)

⇒ x^{3} + y^{3} = (x + y)[(x + y)^{2} – 3xy] {Taking (x + y) common}

⇒ x^{3} + y^{3} = (x + y) (x^{2} + y^{2} – xy)

⇒ x^{3} + y^{3} = (x + y) (x^{2} + xy + y^{2})

(ii) x^{3} – y^{3} = (x – y) (x^{2} + xy + y^{2})

We know that,

(x – y)^{3} = x^{3} + y^{3} – 3xy(x – y)

⇒ x^{3} – y^{3} = (x – y)^{3} + 3xy(x – y)

⇒ x^{3} + y^{3} = (x – y)[(x – y)^{2} + 3xy] {Taking (x – y) common}

⇒ x^{3} + y^{3} = (x – y) [(x^{2} + y^{2} – 2xy) + 3xy]

⇒ x^{3} + y^{3} = (x + y)(x^{2} + y^{2} + xy)

Question 10.

Factorise each of the following:

(i) 27y^{3} + 125z^{3}

(ii) 64m^{3} – 343n^{3}

Answer:

(i) 27y^{3} + 125z^{3}

Using identity, x^{3} + y^{3} = (x + y)(x^{2} + y^{2} + xy)

= (3y + 5z) [(3y)^{2} – 3y × 5z + (5z)^{2}]

= (3y + 5z) [9y^{2} – 15yz + 25z^{2}]

(ii) 64m^{3} – 343n^{3}

Using identity, x^{3} + y^{3} = (x + y)(x^{2} + y^{2} + xy)

= (4m – 7n) [(4m)^{2} + 4m × 7n + (7n)^{2}]

= (4m – 7n) [16m^{2} + 28mn + 49n^{2}]

Question 11.

Factorise: 27x^{3} + y^{3} + z^{3} – 9xyz

Answer:

27x^{3} + y^{3} + z^{3} – 9xyz

= (3x)^{3} + y^{3} + z^{3} – 3 × (3x) yz

Using identity, x^{3} + y^{3} + z^{3} – 3xyz

= (x + y + z)(x^{2} + y^{2} + z^{2} – xy – yz – xz)

27x^{3} + y^{3} + z^{3} – 9xyz

= (3x + y + z) [(3x)^{2} + (y)^{2} + (z)^{2} – 3xy – yz – 3xz]

= (3x + y + z) (9x^{2} + y^{2} + z^{2} – 3xy – yz – 3xz)

Question 12.

Verify that: x^{3} + y^{3} + z^{3} – 3xyz = \(\frac {1}{2}\)(x + y + z) [(x – y)^{2} + (y – z)^{2} + (z – x)^{2}]

Answer:

We know that,

x^{3} + y^{3} + z^{3} – 3xyz

= (x + y + z)(x^{2} + y^{2} + z^{2} – xy – yz – xz)

⇒ x^{3} + y^{3} + z^{3} – 3xyz

= \(\frac {1}{2}\) × (x + y + z) 2(x^{2} + y^{2}+ z^{2} – xy – yz – xz)

= \(\frac {1}{2}\)(x + y + z)(2x^{2} + 2y^{2} + 2z^{2} – 2xy – 2yz – 2xz)

= \(\frac {1}{2}\)(x + y + z) [(x^{2} + y^{2} – 2xy) + (y^{2} + z^{2} – 2yz) + (x^{2} + z^{2} – 2xz)]

= \(\frac {1}{2}\)(x + y + z) [(x – y)^{2}+ (y – z)^{2} + (z – x)^{2}]

Question 13.

If x + y + z = 0, show that x^{3} + y^{3} + z^{3} = 3xyz.

Answer:

We know that,

x^{3} + y^{3} + z^{3}

= (x^{2} + y^{2}+ z^{2} – xy – yz – xz)

Now put (x + y + z) = 0,

(x^{2} + y^{2}+ z^{2} – xy – yz – xz) – 3xyz = (0)(x^{2} + y^{2}+ z^{2} – xy – yz – xz)

⇒ x^{3} + y^{3} + z^{3} = 3xyz

Question 14.

Without actually calculating the cubes, find the value of each of the following:

(i) (-12)^{3} + (7)^{3} + (5)^{3}

(ii) (28)^{3} + (-15)^{3} + (-13)^{3}

Answer:

(i) (-12)^{3} + (7)^{3} + (5)^{3}

Let x = – 12, y = 7 and z = 5

We observed that,

x + y + z = -12 + 7 + 5 = 0

We know that if, x + y + z = 0,

then x^{3} + y^{3} + z^{3} = 3xyz

(-12)^{3} + (7)^{3} + (5)^{3} = 3(-12)(7)(5)

= – 1260

(ii) (28)^{3} + (-15)^{3} + (-13)^{3}

Let x = 28, y = – 15 and z = – 13

We observed that,

x + y + z = 28 – 15 – 13 = 0

We know that if, x + y + z = 0,

then x^{3} + y^{3} + z^{3} = 3xyz

(28)^{3} + (-15)^{3} + (-13)^{3}

= 3(28)(-15)(-13) = 16380

Question 15.

Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:

(i) Area: 25a^{2} – 35a +12

(ii) Area: 35y^{2} + 13y – 12

Answer:

(i) Area: 25a^{2} – 35a +12

Since, area is product of length and breadth therefore by factorising the given area, we can know the length and breadth of rectangle.

25a^{2} – 35a +12 = 25a^{2} – 15a – 20a + 12

= 5a(5a – 3) – 4(5a – 3) = (5a – 4) (5a -3 )

Possible expression for length = 5a – 3

Possible expression for breadth = 5a – 4

(ii) Area: 35y^{2} + 13y – 12

35y^{2} + 13y – 12 = 35y^{2} – 15y + 28y – 12

= 5y(7y – 3) + 4(7y – 3) = (5y + 4)(7y – 3)

Possible expression for length = (5y + 4)

Possible expression for breadth = (7y – 3)

Page – 50

Question 16.

What are the possible expressions for the dimensions of the cuboids whose volumes are given below?

(i) Volume: 3x^{2} – 12x

(ii) Volume : 12ky^{2} + 8ky – 20k

Answer:

(i) Volume: 3x^{2} – 12x

Since, volume is product of length, breadth and height therefore by factorising the given volume, we can know the length, breadth and height of the cuboid.

3x^{2} – 12x = 3x(x – 4)

Possible expression for length = 3

Possible expression for breadth = x

Possible expression for height = (x – 4)

(ii) Volume : 12ky^{2} + 8ky – 20k

Since, volume is product of length, breadth and height therefore by factorising the given volume, we can know the length, breadth and height of the cuboid.

12ky^{2} + 8ky – 20k = 4k(3y^{2} + 2y – 5) = 4k(3y^{2} +5y – 3y – 5)

= 4k[y(3y + 5) – 1(3y + 5)]

= 4k (3y + 5) (y – 1)

Possible expression for length = 4k

Possible expression for breadth = (3y + 5)

Possible expression for height = (y – 1)