Question

Find the value of k, if x-1 is a factor of p(x) in each of the following case:

p(x) = 2x(-1) + k(-1)+ 2​

Answer 1

✬ Value of k = 0 ✬

Step-by-step explanation:

Given:x – 1 is a factor of p(x).

• p(x) is 2x(-1) + k(-1)+ 2

To Find:

• What is the value of k ?

Solution: Since x – 1 is factor of p(x). Let's find the value of x. Put divisor = 0.

➟ x – 1 = 0

➟ x = 1

Now put the value of x in p(x).

$\implies{\rm }$ p(x) = 2x(–1) + k(–1) + 2

$\implies{\rm }$ p(x) = 2(1) (–1) + k (–1) + 2

$\implies{\rm }$ p(x) = 2(–1) + (–k) + 2

$\implies{\rm }$ p(x) = –2 – k + 2

$\implies{\rm }$ p(x) = 0 – k

Since, x – 1 is factor of 2x(-1) + k(-1)+ 2 therefore remainder must be 0. So

$\implies{\rm }$ 0 – k = 0

$\implies{\rm }$ 0 = k

Hence, the value of k is 0.

Answer 2

To Find :-

• The value of k.

Solution :-

• x - 1 is a factor of p(x) = 2x(-1) + k(-1)+ 2 (Given)

Put divisor is equal to zero,

→ x - 1 = 0

→ x = 1

Let, p(x) = 2x(-1) + k(-1) + 2

[ Putting x = 1 ]

→ p(1) = 2(1)(-1) + k(-1) + 2

→ p(1) = 2(-1) + (-k) + 2

→ p(1) = -2 - k + 2

→ p(1) = 0 - k

Thus, remainder = p(1) = 0 - k

Since, x - 1 is a factor of p(x) = 2x(-1) + k(-1) + 2,

.°. Remainder is zero.

→ 0 - k = 0

→ k = 0

Therefore,

The value of k is 0.