Practice Set 3.5 Algebra 9th Std Maths Part 1 Answers Chapter 3 Polynomials
Solution & Step-by-Step Answer:
i. p(x) = 2x – 2x3 + 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 – 2x3+ 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 – 2x3+ 7
Put x = 0 in the given polynomial.
∴ p(0) = 2(0) – 2(0)3+ 7
= 0 – 0 + 7
∴ P(0) = 7
Solution & Step-by-Step Answer:
i. p(x) = x3 ∴ p(1) = 13 = 1 p(x) = x3 ∴ p(0) = 03 = 0 p(x) = x3 ∴ p(-2) = (-2)3 = -8
ii. p(y) = y2– 2y + 5
∴ p(1) = 12– 2(1) + 5
= 1 – 2 + 5
∴ P(1) = 4
p(y) = y2– 2y + 5
∴ p(0) = 02– 2(0) + 5
= 0 – 0 + 5
∴ p(0) = 5
p(y) = y2– 2y + 5
∴ p(- 2) = (- 2)2– 2(- 2) + 5
= 4 + 4 + 5
∴ p(-2) = 13
iii. p(x) = x4– 2x2– x
∴ p(1) = (1)4– 2(1)2– 1
= 1 – 2 – 1
∴ p(1) = -2
∴ p(x) = x4– 2x2– x
∴ p(0) = (0)4– 2(0)2– 0
= 0 – 0 – 0
∴ p(0) = 0
p(x) = x4– 2x2– x
∴ p(-2) = (-2)4– 2(-2)2– (-2)
= 16 – 2(4) + 2
= 16 – 8 + 2
∴ p(-2) = 10
Solution & Step-by-Step Answer:
p(m) = m3 + 2m + a ∴ p(2) = (2)3 + 2(2) + a ∴ 12 = 8 + 4 + a … [∵ p(2)= 12] ∴ 12 = 12 + a ∴ a = 12 – 12 ∴ a = 0
Solution & Step-by-Step Answer:
p(x) = mx2 – 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
Solution & Step-by-Step Answer:
i. p(x) = x2 – 7x + 9 Divisor = x + 1 ∴ take x = – 1 ∴ By remainder theorem, ∴ Remainder =p(-1) p(x) = x2 – 7x + 9 ∴ p(-1) = (- 1)2 – 7(- 1) + 9 = 1 + 7 + 9 ∴ Remainder =17
ii. p(x) = 2x3– 2x2+ ax – a
Divisor = x – a
∴ take x = a
By remainder theorem,
Remainder = p(a)
p(x) = 2x3– 2x2+ ax – a
∴ p(a) = 2a3– 2a2+ a(a) – a
= 2a3– 2a2+ a2– a
∴ Remainder = 2a3– a2– a
iii. p(m) = 54m3+ 18m2– 27m + 5
Divisor = m – 3
∴ take m = 3
∴ By remainder theorem,
Remainder = p(3)
p(m) = 54m3+ 18m2– 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
Solution & Step-by-Step Answer:
p(y) = y3 – 5y2 + 7y + m Divisor = y + 2 ∴ take y = – 2 ∴ By remainder theorem, Remainder = p(- 2) = 50 P(y) = y3 – 5y2 + 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
Solution & Step-by-Step Answer:
p(x) = x2 + 2x – 3 Divisor = x + 3 ∴ take x = – 3 ∴ Remainder = p(-3) p(x) = x2 + 2x – 3 ∴ p(-3) = (-3)2 + 2(- 3) – 3 = 9 – 6 – 3 ∴ p(-3) = 0 ∴ By factor theorem, x + 3 is a factor of x2 + 2x – 3.
Solution & Step-by-Step Answer:
p(x) = x3 – mx2 + 10x – 20 x – 2 is a factor of x3 – mx2 + lOx – 20. ∴By factor theorem, Remainder = p(2) = 0 p(x) = x3 – mx2 + 10x – 20 ∴ p(2) = (2)3 – m(2)2 + 10(2) – 20 ∴ 0 = 8 – 4m + 20 – 20 ∴ 0 = 8 – 4m ∴ 4m = 8 ∴ m = 2
Solution & Step-by-Step Answer:
i. p(x) = x3 – x2 – x – 1 Divisor = q(x) = x – 1 ∴ take x = 1 Remainder = p(1) p(x) = x3 – x2 – 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 x3 – x2 – x – 1.
ii. p(x) = 2x3– x – 45
Divisor = q(x) = x – 3
take x = 3
Remainder = p(3)
p(x) = 2x3– x2– 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 2x3– x2– 45.
Solution & Step-by-Step Answer:
p(x) = x31 + 31 Divisor = x + 1 ∴ take x = – 1 ∴ By remainder theorem, Remainder = p(-1) p(x) =x31 + 31 … ∴ p(-1) = (-1)31 + 31 = -1 + 31 = 30 ∴ Remainder = 30
Solution & Step-by-Step Answer:
i. p(m) = m21 – 1 Divisor = m – 1 ∴ take m = 1 Remainder = p(1) p(m) = m21 – 1 ∴ P(1) = 121 – 1 = 1 – 1 = 0 ∴ By factor theorem, m -1 is a factor of m21 -1.
ii. p(m) = m22– 1
Divisor = m – 1
∴ take m = 1
Remainder = p(1)
p(m) = m22– 1
∴ P(1) = 122– 1 = 1 – 1 = 0
∴ By factor theorem, m -1 is a factor of m22– 1.
Solution & Step-by-Step Answer:
p(x) = nx2 – 5x + m (x – 2) is a factor of nx2 – 5x + m. ∴ By factor theorem, P(2) = 0 ∴ p(x) = nx2 – 5x + m ∴ p(2) = n(2)2 – 5(2) + m ∴ 0 = n(4) – 10 + m ∴ 4n – 10 + m = 0 …(i) Also, ( x = ) is a factor of nx2 – 5x + m. ∴ By factor theorem, p() = 0 p(x) = nx2 – 5x + m ∴ p() = n()2 – 5 + m 0 = – + 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
Solution & Step-by-Step Answer:
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) = 2x2– 5√3 x + 5, then find the value of p(5√3 ).
Solution:
p(x) = 2x2– 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
Solution & Step-by-Step Answer:
∴ Remainder = 17
2. p(x) = 3x2+ 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 t3– 3t2+ 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) = t3– 3t3+ kt + 50
By remainder theorem,
Remainder = p(3) = 33 – 32+ 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
Solution & Step-by-Step Answer:
Here, p(x) = x3 + 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 x3 + 4x – 5.