**Chapter 3 Polynomials Practice Set 3.5**

Chapter 3 Polynomials Practice Set 3.5

**Question 1.Find the value of the polynomial 2x â€“ 2x ^{3} + 7 using given values for x.i. x = 3ii. x = -1iii. x = 0Solution:**

i. p(x) = 2x â€“ 2x

^{3}+ 7

Put x = 3 in the given polynomial.

âˆ´ p(3) = 2(3) â€“ 2(3)

^{3}+ 7

= 6 â€“ 2 x 27 + 7

= 6 â€“ 54 + 7

âˆ´ P(3) = â€“ 41

ii. p(x) = 2x â€“ 2x^{3} + 7

Put x = -1 in the given polynomial.

âˆ´ p(- 1) = 2(- 1) â€“ 2(-1)^{3} + 7

= â€“ 2 â€“ 2(-1) + 7

= -2 + 2 + 7

âˆ´ p(-1) = 7

iii. p(x) = 2x â€“ 2x^{3} + 7

Put x = 0 in the given polynomial.

âˆ´ p(0) = 2(0) â€“ 2(0)^{3} + 7

= 0 â€“ 0 + 7

âˆ´ 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}â€“ 2y + 5ii. p(y) = x

^{4}â€“ 2x

^{2}+ xSolution:

i. p(x) = x

^{3}

âˆ´ p(1) = 1

^{3}= 1

p(x) = x

^{3}

âˆ´ p(0) = 0

^{3}= 0

p(x) = x

^{3}

âˆ´ p(-2) = (-2)

^{3}= -8

ii. p(y) = y^{2} â€“ 2y + 5

âˆ´ p(1) = 1^{2} â€“ 2(1) + 5

= 1 â€“ 2 + 5

âˆ´ P(1) = 4

p(y) = y^{2} â€“ 2y + 5

âˆ´ p(0) = 0^{2} â€“ 2(0) + 5

= 0 â€“ 0 + 5

âˆ´ p(0) = 5

p(y) = y^{2} â€“ 2y + 5

âˆ´ p(- 2) = (- 2)^{2} â€“ 2(- 2) + 5

= 4 + 4 + 5

âˆ´ p(-2) = 13

iii. p(x) = x^{4} â€“ 2x^{2} â€“ x

âˆ´ p(1) = (1)^{4} â€“ 2(1)^{2} â€“ 1

= 1 â€“ 2 â€“ 1

âˆ´ p(1) = -2

âˆ´ p(x) = x^{4} â€“ 2x^{2} â€“ x

âˆ´ p(0) = (0)^{4} â€“ 2(0)^{2} â€“ 0

= 0 â€“ 0 â€“ 0

âˆ´ p(0) = 0

p(x) = x^{4} â€“ 2x^{2} â€“ x

âˆ´ p(-2) = (-2)^{4} â€“ 2(-2)^{2} â€“ (-2)

= 16 â€“ 2(4) + 2

= 16 â€“ 8 + 2

âˆ´ 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

âˆ´ p(2) = (2)

^{3}+ 2(2) + a

âˆ´ 12 = 8 + 4 + a â€¦ [âˆµ p(2)= 12]

âˆ´ 12 = 12 + a

âˆ´ a = 12 â€“ 12

âˆ´ a = 0

**Question 4.For the polynomial mx ^{2} â€“ 2x + 3 if p(-1) = 7, then find m.Solution:**

p(x) = mx

^{2}â€“ 2x + 3

âˆ´ p(- 1) = m (- 1)

^{2}â€“ 2(- 1) + 3

âˆ´ 7 = m(1) + 2 + 3 â€¦[âˆµ p(-1) = 7]

âˆ´ 7 = m + 5

âˆ´ m = 7 â€“ 5

âˆ´ m = 2

**Question 5.Divide the first polynomial by the second polynomial and find the remainder using remainder theorem.**

i. (x

^{2}â€“ 1x + 9); (x + 1)

ii. (2x

^{3}â€“ 2x

^{2}+ ax â€“ a); (x â€“ a)

iii. (54m

^{3}+ 18m

^{2}â€“ 27m + 5); (m â€“ 3)

Solution:

i. p(x) = x

^{2}â€“ 7x + 9

Divisor = x + 1

âˆ´ take x = â€“ 1

âˆ´ By remainder theorem,

âˆ´ Remainder =p(-1)

p(x) = x

^{2}â€“ 7x + 9

âˆ´ p(-1) = (- 1)

^{2}â€“ 7(- 1) + 9

= 1 + 7 + 9

âˆ´ Remainder =17

ii. p(x) = 2x^{3} â€“ 2x^{2} + ax â€“ a

Divisor = x â€“ a

âˆ´ take x = a

By remainder theorem,

Remainder = p(a)

p(x) = 2x^{3} â€“ 2x^{2} + ax â€“ a

âˆ´ p(a) = 2a^{3} â€“ 2a^{2} + a(a) â€“ a

= 2a^{3}â€“ 2a^{2} + a^{2} â€“ a

âˆ´ Remainder = 2a^{3} â€“ a^{2} â€“ a

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

Divisor = m â€“ 3

âˆ´ take m = 3

âˆ´ By remainder theorem,

Remainder = p(3)

p(m) = 54m^{3} + 18m^{2} â€“ 27m + 5

âˆ´ p(3) = 54(3)^{3} +18(3)^{2} â€“ 27(3) + 5

= 54(27) + 18(9) â€“ 81 + 5

= 1458 + 162 â€“ 81 + 5

âˆ´ Remainder = 1544

**Question 6.If the polynomial y ^{3} â€“ 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}â€“ 5y

^{2}+ 7y + m

Divisor = y + 2

âˆ´ take y = â€“ 2

âˆ´ By remainder theorem,

Remainder = p(- 2) = 50

P(y) = y

^{3}â€“ 5y

^{2}+ 7y + m

âˆ´ P(-2) = (- 2)

^{3}â€“ 5(- 2)

^{2}+ 7(- 2) + m

âˆ´ 50 = -8 â€“ 5(4) â€“ 14 + m

âˆ´ 50 = -8 â€“ 20 â€“ 14 + m

âˆ´ 50 = â€“ 42 + m

âˆ´ m = 50 + 42

âˆ´ m = 92

**Question 7.Use factor theorem to determine whether x + 3 is a factor of x ^{2} + 2x â€“ 3 or not.Solution:**

p(x) = x

^{2}+ 2x â€“ 3

Divisor = x + 3

âˆ´ take x = â€“ 3

âˆ´ Remainder = p(-3)

p(x) = x

^{2}+ 2x â€“ 3

âˆ´ p(-3) = (-3)

^{2}+ 2(- 3) â€“ 3

= 9 â€“ 6 â€“ 3

âˆ´ p(-3) = 0

âˆ´ By factor theorem, x + 3 is a factor of x

^{2}+ 2x â€“ 3.

**Question 8.If (x â€“ 2) is a factor of x ^{3} â€“ mx^{2} + 10x â€“ 20, then find the value of m.Solution:**

p(x) = x

^{3}â€“ mx

^{2}+ 10x â€“ 20 x â€“ 2 is a factor of x3 â€“ mx2 + lOx â€“ 20.

âˆ´By factor theorem,

Remainder = p(2) = 0

p(x) = x

^{3}â€“ mx

^{2}+ 10x â€“ 20

âˆ´ p(2) = (2)

^{3}â€“ m(2)

^{2}+ 10(2) â€“ 20

âˆ´ 0 = 8 â€“ 4m + 20 â€“ 20

âˆ´ 0 = 8 â€“ 4m

âˆ´ 4m = 8

âˆ´ 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}â€“ x

^{2}â€“ x -1 ; q(x) = x â€“ 1ii. p(x) = 2x

^{3}â€“ x

^{2}â€“ 45 ; q(x) = x â€“ 3Solution:

i. p(x) = x

^{3}â€“ x

^{2}â€“ x â€“ 1

Divisor = q(x) = x â€“ 1

âˆ´ take x = 1

Remainder = p(1)

p(x) = x

^{3}â€“ x

^{2}â€“ x â€“ 1

âˆ´ P(1) = (1)

^{3}â€“ (1)

^{2}â€“ 1 â€“ 1

= 1 â€“ 1 â€“ 1 â€“ 1

= -2 â‰ 0

âˆ´ By factor theorem, x â€“ 1 is not a factor of x

^{3}â€“ x

^{2}â€“ x â€“ 1.

ii. p(x) = 2x^{3} â€“ x â€“ 45

Divisor = q(x) = x â€“ 3

take x = 3

Remainder = p(3)

p(x) = 2x^{3} â€“ x^{2} â€“ 45

P(3) = 2(3)^{3} â€“ (3)^{2} â€“ 45

= 2(27) â€“ 9 â€“ 45

= 54 â€“ 9 â€“ 45

= 0

âˆ´ By factor theorem, x â€“ 3 is a factor of 2x^{3} â€“ x^{2} â€“ 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

âˆ´ take x = â€“ 1

âˆ´ By remainder theorem,

Remainder = p(-1)

p(x) =x

^{31}+ 31 â€¦

âˆ´ p(-1) = (-1)

^{31}+ 31

= -1 + 31 = 30

âˆ´ Remainder = 30

**Question 11.Show that m â€“ 1 is a factor of m ^{21} â€“ 1 and m^{22} â€“ 1. [3 Marks]Solution:**

i. p(m) = m

^{21}â€“ 1

Divisor = m â€“ 1

âˆ´ take m = 1

Remainder = p(1)

p(m) = m

^{21}â€“ 1

âˆ´ P(1) = 1

^{21}â€“ 1 = 1 â€“ 1 = 0

âˆ´ By factor theorem, m -1 is a factor of m

^{21}-1.

ii. p(m) = m^{22} â€“ 1

Divisor = m â€“ 1

âˆ´ take m = 1

Remainder = p(1)

p(m) = m^{22} â€“ 1

âˆ´ P(1) = 1^{22} â€“ 1 = 1 â€“ 1 = 0

âˆ´ By factor theorem, m -1 is a factor of m^{22} â€“ 1.

**Question 12.**

âˆ´ 0 = n- 10 +4m â€¦ [Multiplying both sides by 4]

âˆ´ n = 10 â€“ 4m â€¦â€¦(ii)

Substituting n = 10 â€“ 4m in equation (i),

4(10 â€“ 4m) â€“ 10 + m = 0

âˆ´ 40 â€“ 16m â€“ 10 + m = 0

âˆ´ -15m+ 30 = 0

âˆ´ -15m = -30

âˆ´ m = 2

Substituting m = 2 in equation (ii),

n = 10 â€“ 4(2)

= 10 â€“ 8

âˆ´ n = 2

âˆ´ m = n = 2

**Question 13.i. If p(x) = 2 + 5x, then find the value of p(2) + p(- 2) â€“ p(1).Solution:**

p(x) = 2 + 5x

âˆ´ P(2) = 2 + 5(2)

= 2 + 10

= 12

p(x) = 2 + 5x

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

= 2 â€“ 10 = â€“ 8

p(x) = 2 + 5x

P(1) = 2 + 5(1)

= 2 + 5 = 7

âˆ´ P(2) + P(- 2) â€“ p(1) = 12 + (- 8) â€“ 7

âˆ´ P(2) + p(- 2) â€“ p(1) = â€“ 3

ii. If p(x) = 2x^{2} â€“ 5âˆš3 x + 5, then find the value of p(5âˆš3 ).

Solution:

p(x) = 2x^{2} â€“ 5âˆš3 x + 5

âˆ´ p(5âˆš3) = 2(5âˆš3)^{2} â€“ 5âˆš3 (5âˆš3 ) + 5

= 2 (25 x 3) â€“ 25 x 3 + 5

= 150-75 + 5

âˆ´ p( 5âˆš3 ) = 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 = â€“ 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:**

âˆ´ Remainder = 17

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

Substituting x = â€“ 2, we get

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

= 12 â€“ 2 + 7

âˆ´ p(-2) = 17

3. Yes, remainder = p(-2)

**Another Example:If the polynomial t ^{3} â€“ 3t^{2} + kt + 50 is divided by (t â€“ 3), the remainder is 62. Find the value of k.Solution:**

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

p(t) = t

^{3}â€“ 3t

^{3}+ kt + 50

By remainder theorem,

Remainder = p(3) = 33 â€“ 3

^{2}+ k x 3 + 50

= 27 â€“ 3 x 9 + 3k + 50

= 27 â€“ 27 + 3k + 50

= 3k + 50

But remainder is 62.

âˆ´ 3k + 50 = 62

âˆ´ 3k = 62 â€“ 50

âˆ´ 3k = 12

âˆ´ k = 4

**Question 2.Verify that (x â€“ 1) is a factor of the polynomial x ^{3} + 4x â€“ 5. (Textbook pg. no. 51)Solution:**

Here, p(x) = x

^{3}+ 4x â€“ 5

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

p(1) = (1)

^{3}+ 4(1) â€“ 5

= 1 + 4 â€“ 5

P(1) = 0

âˆ´ By remainder theorem,

Remainder = 0

âˆ´ (x -1) is the factor of x

^{3}+ 4x â€“ 5.