Question

Find the value of k, if x-1 is a factor of p(x) in each of the following cases: p(x)=x²+x+k

Answer 1

Answer:

k = -2

Step-by-step explanation:

x-1 is a factor of p(x)

=> p(1) = 0

p(x) = x² + x + k

=> P(1) = 1² + 1 + k

=> P(1) = 1 + 1 + k

=> P(1) = 2 + k

p(1) = 0

=> 2 + k = 0

=> k = -2

verification :

putting k = -2 in p(x)

= x² + x - 2

= x² + 2x - x - 2

= x(x + 2) - 1(x + 2)

= (x - 1)(x + 2)

x -1 is a factor

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO DETERMINE

The value of k, if x-1 is a factor of p(x) in each of the following cases: p(x)=x²+x+k

CALCULATION

$\sf{Let \: \: f(x )= x - 1}$

$\sf{ p(x) = {x}^{2} + x + k\: }$

Now for the zero of polynomial f(x) we have

$\sf{ f(x) = 0\: \: }$

$\implies \sf{x - 1 = 0 \: \: }$

$\implies \sf{x = 1 \: \: }$

Hence by the Remainder Theorem the required Remainder is

$= \sf{p(1) \: \: }$

$= \sf{ {(1)}^{2} + 1 + k \: \: }$

$= \sf{ 2 + k \: \: }$

Since x-1 is a factor of p(x)

$\therefore \sf{ \: \: \: 2 + k = 0\: \: }$

$\implies \sf{ k = - 2\: \: }$

RESULT

$\boxed{ \: \: \sf{ k = - 2\: \: }}$

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