Question

For what value of k, (-4) is a zero of p(x) = x2 - x - (2k - 2)?​

Answer 1

Answer:

(-4) is the zeros of given polynomial f(x)

f(x)= x²-x-(2k-2)

f(x)= (-4)²-(-4) -(2k-2)

= 16 +4-(2k-2)

= 20 -(2k -2)

f(-4) =0

20-(2k-2) =0

-(2k-2) = -20

2k -2=20

2k = 20+2

k= 22/2

k= 11

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Answer 2

$p(x) = - 4 \\ find \: the \: volue \: of \: k\\ p( - 4) = x^{2} - x - (2k - 2) = 0 \\ \: \: ( - 4 {)}^{2} - ( - 4) - 2k + 2 = 0\\ 16 + 4 - 2k + 2 = 0 \\ 22 - 2k = 0 \\ - 2k = - 22 \\ k = \frac{ - 22}{ - 2} \\ k = 11 \: \: answer$