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

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Step-by-step explanation:

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.

Answer 2

Given Polynomial :- p(x) = 2x(-1) + k(-1)+ 2.

Factor Of Polynomial :- x – 1 = 0. ➝ x = 1.

Solution :-

Put the value of x in the given polynomial.

➝ 0 = 2(1)(-1) + k(-1) + 2.

➝ 0 = -2 - k + 2.

➝ 0 = - k.

➝ -k = 0.

➝ k = 0.

Therefore, Value Of k = 0.

Verification :-

➝ p(x) = 2x(-1) + k(-1)+ 2.

➝ p(1) = 2(1)(-1) + 0(-1) + 2.

➝ p(1) = -2 + 0 + 2.

➝ p(1) = 0.

➝ 0 = 0.

LHS = RHS.

Hence, Verified.