Question

For what value of k,(-4) is a zero of the polynomial x^2-x-(2k+2)?

Answer 1

p(x)=x²-x-(2k+2)

Zero of p(x)= -4

p(-4)=(-4)²-(-4)-2k-2

=>16+4-2k-2=0

=>-2k+18=0

=>-2k= -18

=>k= -18/-2

=>k=9

Thus,the value of k is 9.

Answer 2

Zero of the polynomial → (- 4 )

So,

$\huge\sf\green{x = (- 4 )}$

Put above value of x in the polynomial .

$=>\sf\blue{p(x) = x^2 - x - (2k+2) = 0}$

$=>\sf{p(-4) = (-4)^2 - (-4) - (2k+2) = 0}$

$=>\sf\orange{16 + 4 - 2k - 2 = 0}$

$=>\sf\orange{ 20 - 2 - 2k = 0}$

$=>\sf\orange{ 18 - 2k = 0}$

$=>\sf\orange{ 2k = 18}$

$=>\sf\orange{ k = \dfrac{18}{2}}$

$=> k = 9$

Therefore, For k = 9 zero of the polynomial will be - 4