Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.5 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 3 Polynomials.

## Practice Set 3.5 Algebra 9th Std Maths Part 1 Answers Chapter 3 Polynomials

Question 1.

Find the value of the polynomial 2x – 2x^{3} + 7 using given values for x.

i. x = 3

ii. x = -1

iii. x = 0

Solution:

i. p(x) = 2x – 2x^{3} + 7

Put x = 3 in the given polynomial.

∴ p(3) = 2(3) – 2(3)^{3} + 7

= 6 – 2 x 27 + 7

= 6 – 54 + 7

∴ P(3) = – 41

ii. p(x) = 2x – 2x^{3} + 7

Put x = -1 in the given polynomial.

∴ p(- 1) = 2(- 1) – 2(-1)^{3} + 7

= – 2 – 2(-1) + 7

= -2 + 2 + 7

∴ p(-1) = 7

iii. p(x) = 2x – 2x^{3} + 7

Put x = 0 in the given polynomial.

∴ p(0) = 2(0) – 2(0)^{3} + 7

= 0 – 0 + 7

∴ P(0) = 7

Question 2.

For each of the following polynomial, find p(1), p(0) and p(- 2).

i. p(x) = x^{3}

ii. p(y) = y^{2} – 2y + 5

ii. p(y) = x^{4} – 2x^{2} + x

Solution:

i. p(x) = x^{3}

∴ p(1) = 1^{3} = 1

p(x) = x^{3}

∴ p(0) = 0^{3} = 0

p(x) = x^{3}

∴ p(-2) = (-2)^{3} = -8

ii. p(y) = y^{2} – 2y + 5

∴ p(1) = 1^{2} – 2(1) + 5

= 1 – 2 + 5

∴ P(1) = 4

p(y) = y^{2} – 2y + 5

∴ p(0) = 0^{2} – 2(0) + 5

= 0 – 0 + 5

∴ p(0) = 5

p(y) = y^{2} – 2y + 5

∴ p(- 2) = (- 2)^{2} – 2(- 2) + 5

= 4 + 4 + 5

∴ p(-2) = 13

iii. p(x) = x^{4} – 2x^{2} – x

∴ p(1) = (1)^{4} – 2(1)^{2} – 1

= 1 – 2 – 1

∴ p(1) = -2

∴ p(x) = x^{4} – 2x^{2} – x

∴ p(0) = (0)^{4} – 2(0)^{2} – 0

= 0 – 0 – 0

∴ p(0) = 0

p(x) = x^{4} – 2x^{2} – x

∴ p(-2) = (-2)^{4} – 2(-2)^{2} – (-2)

= 16 – 2(4) + 2

= 16 – 8 + 2

∴ p(-2) = 10

Question 3.

If the value of the polynomial m^{3} + 2m + a is 12 for m = 2, then find the value of a.

Solution:

p(m) = m^{3} + 2m + a

∴ p(2) = (2)^{3} + 2(2) + a

∴ 12 = 8 + 4 + a … [∵ p(2)= 12]

∴ 12 = 12 + a

∴ a = 12 – 12

∴ a = 0

Question 4.

For the polynomial mx^{2} – 2x + 3 if p(-1) = 7, then find m.

Solution:

p(x) = mx^{2} – 2x + 3

∴ p(- 1) = m (- 1)^{2} – 2(- 1) + 3

∴ 7 = m(1) + 2 + 3 …[∵ p(-1) = 7]

∴ 7 = m + 5

∴ m = 7 – 5

∴ m = 2

Question 5.

Divide the first polynomial by the second polynomial and find the remainder using remainder theorem.

i. (x^{2} – 1x + 9); (x + 1)

ii. (2x^{3} – 2x^{2} + ax – a); (x – a)

iii. (54m^{3} + 18m^{2} – 27m + 5); (m – 3)

Solution:

i. p(x) = x^{2} – 7x + 9

Divisor = x + 1

∴ take x = – 1

∴ By remainder theorem,

∴ Remainder =p(-1)

p(x) = x^{2} – 7x + 9

∴ p(-1) = (- 1)^{2} – 7(- 1) + 9

= 1 + 7 + 9

∴ Remainder =17

ii. p(x) = 2x^{3} – 2x^{2} + ax – a

Divisor = x – a

∴ take x = a

By remainder theorem,

Remainder = p(a)

p(x) = 2x^{3} – 2x^{2} + ax – a

∴ p(a) = 2a^{3} – 2a^{2} + a(a) – a

= 2a^{3}– 2a^{2} + a^{2} – a

∴ Remainder = 2a^{3} – a^{2} – a

iii. p(m) = 54m^{3} + 18m^{2} – 27m + 5

Divisor = m – 3

∴ take m = 3

∴ By remainder theorem,

Remainder = p(3)

p(m) = 54m^{3} + 18m^{2} – 27m + 5

∴ p(3) = 54(3)^{3} +18(3)^{2} – 27(3) + 5

= 54(27) + 18(9) – 81 + 5

= 1458 + 162 – 81 + 5

∴ Remainder = 1544

Question 6.

If the polynomial y^{3} – 5y^{2} + 7y + m is divided by y + 2 and the remainder is 50, then find the value of m.

Solution:

p(y) = y^{3} – 5y^{2} + 7y + m

Divisor = y + 2

∴ take y = – 2

∴ By remainder theorem,

Remainder = p(- 2) = 50

P(y) = y^{3} – 5y^{2} + 7y + m

∴ P(-2) = (- 2)^{3} – 5(- 2)^{2} + 7(- 2) + m

∴ 50 = -8 – 5(4) – 14 + m

∴ 50 = -8 – 20 – 14 + m

∴ 50 = – 42 + m

∴ m = 50 + 42

∴ m = 92

Question 7.

Use factor theorem to determine whether x + 3 is a factor of x^{2} + 2x – 3 or not.

Solution:

p(x) = x^{2} + 2x – 3

Divisor = x + 3

∴ take x = – 3

∴ Remainder = p(-3)

p(x) = x^{2} + 2x – 3

∴ p(-3) = (-3)^{2} + 2(- 3) – 3

= 9 – 6 – 3

∴ p(-3) = 0

∴ By factor theorem, x + 3 is a factor of x^{2} + 2x – 3.

Question 8.

If (x – 2) is a factor of x^{3} – mx^{2} + 10x – 20, then find the value of m.

Solution:

p(x) = x^{3} – mx^{2} + 10x – 20 x – 2 is a factor of x3 – mx2 + lOx – 20.

∴By factor theorem,

Remainder = p(2) = 0

p(x) = x^{3} – mx^{2} + 10x – 20

∴ p(2) = (2)^{3} – m(2)^{2} + 10(2) – 20

∴ 0 = 8 – 4m + 20 – 20

∴ 0 = 8 – 4m

∴ 4m = 8

∴ m = 2

Question 9.

By using factor theorem in the following examples, determine whether q(x) is a factor of p(x) or not.

i. p(x) = x^{3} – x^{2} – x -1 ; q(x) = x – 1

ii. p(x) = 2x^{3} – x^{2} – 45 ; q(x) = x – 3

Solution:

i. p(x) = x^{3} – x^{2} – x – 1

Divisor = q(x) = x – 1

∴ take x = 1

Remainder = p(1)

p(x) = x^{3} – x^{2} – x – 1

∴ P(1) = (1)^{3} – (1)^{2} – 1 – 1

= 1 – 1 – 1 – 1

= -2 ≠ 0

∴ By factor theorem, x – 1 is not a factor of x^{3} – x^{2} – x – 1.

ii. p(x) = 2x^{3} – x – 45

Divisor = q(x) = x – 3

take x = 3

Remainder = p(3)

p(x) = 2x^{3} – x^{2} – 45

P(3) = 2(3)^{3} – (3)^{2} – 45

= 2(27) – 9 – 45

= 54 – 9 – 45

= 0

∴ By factor theorem, x – 3 is a factor of 2x^{3} – x^{2} – 45.

Question 10.

If (x^{31} + 31) is divided by (x + 1), then find the remainder.

Solution:

p(x) = x^{31} + 31

Divisor = x + 1

∴ take x = – 1

∴ By remainder theorem,

Remainder = p(-1)

p(x) =x^{31} + 31 …

∴ p(-1) = (-1)^{31} + 31

= -1 + 31 = 30

∴ Remainder = 30

Question 11.

Show that m – 1 is a factor of m^{21} – 1 and m^{22} – 1. [3 Marks]

Solution:

i. p(m) = m^{21} – 1

Divisor = m – 1

∴ take m = 1

Remainder = p(1)

p(m) = m^{21} – 1

∴ P(1) = 1^{21} – 1 = 1 – 1 = 0

∴ By factor theorem, m -1 is a factor of m^{21} -1.

ii. p(m) = m^{22} – 1

Divisor = m – 1

∴ take m = 1

Remainder = p(1)

p(m) = m^{22} – 1

∴ P(1) = 1^{22} – 1 = 1 – 1 = 0

∴ By factor theorem, m -1 is a factor of m^{22} – 1.

Question 12.

If x – 2 and x – \(\frac { 1 }{ 2 }\) both are the factors of the polynomial nx^{2} – 5x + m, then show that m = n = 2.

Solution:

p(x) = nx^{2} – 5x + m

(x – 2) is a factor of nx^{2} – 5x + m.

∴ By factor theorem,

P(2) = 0

∴ p(x) = nx^{2} – 5x + m

∴ p(2) = n(2)^{2} – 5(2) + m

∴ 0 = n(4) – 10 + m

∴ 4n – 10 + m = 0 …(i)

Also, ( x = \(\frac { 1 }{ 2 }\) ) is a factor of nx^{2} – 5x + m.

∴ By factor theorem,

p(\(\frac { 1 }{ 2 }\)) = 0

p(x) = nx^{2} – 5x + m

∴ p(\(\frac { 1 }{ 2 }\)) = n(\(\frac { 1 }{ 2 }\))^{2} – 5\(\frac { 1 }{ 2 }\) + m

0 = \(\frac { n }{ 4 }\) – \(\frac { 5 }{ 2 }\) + m

∴ 0 = n- 10 +4m … [Multiplying both sides by 4]

∴ n = 10 – 4m ……(ii)

Substituting n = 10 – 4m in equation (i),

4(10 – 4m) – 10 + m = 0

∴ 40 – 16m – 10 + m = 0

∴ -15m+ 30 = 0

∴ -15m = -30

∴ m = 2

Substituting m = 2 in equation (ii),

n = 10 – 4(2)

= 10 – 8

∴ n = 2

∴ m = n = 2

Question 13.

i. If p(x) = 2 + 5x, then find the value of p(2) + p(- 2) – p(1).

Solution:

p(x) = 2 + 5x

∴ P(2) = 2 + 5(2)

= 2 + 10

= 12

p(x) = 2 + 5x

P(- 2) = 2 + 5(- 2)

= 2 – 10 = – 8

p(x) = 2 + 5x

P(1) = 2 + 5(1)

= 2 + 5 = 7

∴ P(2) + P(- 2) – p(1) = 12 + (- 8) – 7

∴ P(2) + p(- 2) – p(1) = – 3

ii. If p(x) = 2x^{2} – 5√3 x + 5, then find the value of p(5√3 ).

Solution:

p(x) = 2x^{2} – 5√3 x + 5

∴ p(5√3) = 2(5√3)^{2} – 5√3 (5√3 ) + 5

= 2 (25 x 3) – 25 x 3 + 5

= 150-75 + 5

∴ p( 5√3 ) = 80

Question 1.

1. Divide p(x) = 3x^{2} + x + 7 by x + 2. Find the remainder.

2. Find the value of p(x) = 3x^{2} + x + 7 when x = – 2.

3. See whether remainder obtained by division is same as the value of p(-2). Take one more example and verify. (Textbook pg. no. 50)

Solution:

∴ Remainder = 17

2. p(x) = 3x^{2} + x + 7

Substituting x = – 2, we get

p(-2) = 3(2)^{2} + (-2) + 7

= 12 – 2 + 7

∴ p(-2) = 17

3. Yes, remainder = p(-2)

Another Example:

If the polynomial t^{3} – 3t^{2} + kt + 50 is divided by (t – 3), the remainder is 62. Find the value of k.

Solution:

When given polynomial is divided by (t – 3) the remainder is 62. It means the value of the polynomial when t = 3 is 62.

p(t) = t^{3} – 3t^{3} + kt + 50

By remainder theorem,

Remainder = p(3) = 33 – 3^{2} + k x 3 + 50

= 27 – 3 x 9 + 3k + 50

= 27 – 27 + 3k + 50

= 3k + 50

But remainder is 62.

∴ 3k + 50 = 62

∴ 3k = 62 – 50

∴ 3k = 12

∴ k = 4

Question 2.

Verify that (x – 1) is a factor of the polynomial x^{3} + 4x – 5. (Textbook pg. no. 51)

Solution:

Here, p(x) = x^{3} + 4x – 5

Substituting x = 1 in p(x), we get

p(1) = (1)^{3} + 4(1) – 5

= 1 + 4 – 5

P(1) = 0

∴ By remainder theorem,

Remainder = 0

∴ (x -1) is the factor of x^{3} + 4x – 5.