Find the value of 'K' for given Polynomial.

1. If P(x) = x^3 + 5 x^2 + 5x + k Find the value of k for given polynomial.
if Following are the factor of polynomial.

a). x + 1

b) x – 1

c). x – 1/3

d). x + ¼

e). x

f). 3x + 6

Solution.

a). P(x) = x^3 + 5 x^2 + 5x + 1

Put x + 1 = 0 => x = -1

P(-1) = 0
(-1)^3 + 5(-1)^2 + 5(-1) + k = 0

-1 + 5 - 5 + k = 0

k = 1

b). P(x) = x^3 + 5 x^2 + 5x + k

Put x - 1 = 0 => x = 1

P(1) = 0
(1)^3 + 5(1)^2 + 5(1) + k = 0

1 + 5 + 5 + k = 0

k = -11

c). P(x) = x^3 + 5 x^2 + 5x + k

Put x – 1/3 = 0 => x = 1/3

P(1/3) = 0
(1/3)^3 + 5(1/3)^2 + 5(1/3) + k = 0

1/27 + 5/9 + 5/3 + k = 0
( 1 + 15 + 45 ) / 27 + k = 0

k = -61 / 27

d). P(x) = x^3 + 5 x^2 + kx + 1

Put x + 1/4 = 0 => x = -1/4

P(-1/4) = 0
(-1/4)^3 + 5(-1/4)^2 + k(-1/4) + 1 = 0

-1/64 + 5/16 - k/4 + 1 = 0
( -1 + 20 - 16k + 64 ) / 64 = 0
83 - 16 k = 0 => 16k = 83

k = 83/16

e). P(x) = x^3 + 5 x^2 + 5x + k

Put x = 0

P(0) = 0
(0)^3 + 5(0)^2 + 5(0) + k = 0
k = 0

f). P(x) = x^3 + 5 x^2 + kx + k

Put 3x + 6 = 0 => 3x = -6

X = -6/3 => x = -2

P(-2) = 0
(-2)^3 + 5(-2)^2 + k(-2) + k = 0

-8 + 5 * 4 – k * 2 + k = 0
-8 + 20 – 2k + k = 0

12 - k = 0 => k = 12

maths-ncert-solution

2. Find the value of k, P(x) = 3x^3 – a x^2 + 2 x – 2 k if x – a is factor of polynomial.

Sol. P(x) = 3x^3 – a x^2 + 2 x – 2 k

Put x – a = 0 => x = a

P(a) = 0
3 a^3 – a * a^2 + 2 * a - 2 k = 0

3 a^3 – a^3 + 2 a - 2 k = 0
2 a^3 + 2 a = 2 k

k = a^3 + a => k = a (a a^2 + 1)

3. Find the value of k, P(x) = 27 x^3 – 9 x^2 + 3 x – k if x – 1/3 is a factor of polynomial.

Sol. P(x) = 27 x^3 – 9 x^2 + 3 x – k

Put x – 1/3 = 0 => x = 1/3

P(1/3) = 0
27 * (1/3)^3 – 9 * (1/3)^2 + 3 * (1/3) - k = 0

1 – 1 + 1 – k = 0

1 - k = 0 => k = 1

4. Find the value of k, P(x) = x^3 + k x^2 + 2 k x + 1 if x + 1 is a factor of polynomial.
Sol. P(x) = x^3 + k x^2 + 2 k x + 1
Put x = -1
P(-1) = 0
(-1)^3 + k * (-1)^2 + 2 k * (-1) + 1 = 0
-1 + k - 2 k + 1 = 0
k = 0

5. Find the value of k, P(x) = k x^2 + 3 x – 2 if x - 1/3 is a factor of polynomial.
Sol. P(x) = k x^2 + 3 x - 2
Put x = 1/3
P(1/3) = 0
k * (1/3)^2 + 3 * (1/3) - 2 = 0
k/9 + 1 - 2 = 0 k/9 - 1 = 0
k/9 = 1 => k = 9

6. Find the value of k, P(x) = k^2 + x - 2 if x - 1 is a factor of polynomial.
Sol. P(x) = k^2 + x - 2
Put x = 1
P(1) = 0
(k)^2 + (1) - 2 = 0
k^2 + 1 - 2 = 0
k^2 - 1 = 0 => k^2 = 1
k = 1

7. Find the value of k, P(x) = 4 x^2 + k x – 2 if x - 1/2 is a factor of polynomial.
Sol. P(x) = 4 x^2 + k x - 2
Put x = 1/2
P(1/2) = 0
4 * (1/2)^2 + k * (1/2) - 2 = 0
1 + k/2 - 2 = 0 => k/2 - 1 = 0
k/2 = 1 => k = 2

maths-ncert-solution

8. Find the value of k, P(x) = k^2 + x – 2 if x - 1 is a factor of polynomial.

Sol. P(x) = k^2 + x - 2
Put x = 1
P(1) = 0
(k)^2 + (1) - 2 = 0
k^2 + 1 - 2 = 0 => k^2 - 1 = 0
k^2 = 1 => k = 1

9. Find the value of k, P(x) = 9 x^2 + k x + 2 if x + 1/3 is a factor of polynomial.
Sol. P(x) = 9 x^2 + k x + 2
Put x = -1/3
P(-1/3) = 0
9 * (-1/3)^2 + k * (-1/3) + 2 = 0
1 - k + 2 = 0
-k + 3 = 0
k = 3

10. Find the value of k, P(x) = x^3 + k x^2 + k x + 1 if x - 1 is a factor of polynomial.
Sol. P(x) = x^3 + k x^2 + k x + 1
Put x = 1
P(1) = 0
(1)^3 + k * (1)^2 + k * (1) +1 = 0
1 + k + k + 1 = 0
2 k + 2 = 0 => k + 1 = 0
k = -1

Milan Tomic

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