Question

X2+8x+k,ifx+1 is @ factor of it then find the value of k

Answer 1

$p(x) = {x}^{2} + 8x + k$
x+1 is a factor so...
x+1=0
x= -1 will satisfy the given polynomial
ie. p(-1)=0
${( - 1)}^{2} + 8( - 1) + k = 0 \\ 1 - 8 + k = 0 \\ - 7 + k = 0 \\ k = 7$

Answer 2

Given,

p(x) = x²+8x+k

(x+1) is a factor of the given polynomial.

To find,

The value of k.

Solution,

The value of k will be 7.

According to the question,

p(x) = x²+8x+k

(x+1) is a factor of the given polynomial.

Then,

x+1 = 0

x = -1

Putting this value of x in the given polynomial should make its value to be 0.

So,

p(-1) = 0

p(x) = x²+8x+k

p(-1) = (-1)²+8(-1)+k

p(-1) = 1-8+k

p(-1) = k-7

k-7 = 0

k = 7

Hence, the value of k is 7.