Question

For what value of k, (–4) is a zero of the polynomial x2 – x – (2k + 2)?​

Answer 1

Answer:

In the given polynomial f(x)=x  

2

−x−(2k+2), substitute x=−4 as shown below:

f(−4)=(−4)  

2

−(−4)−(2k+2)=16+4−2k−2=18−2k

Therefore, f(−4)=18−2k.

Since it is given that x=−4 is a zero of the polynomial f(x)=x  

2

−x−(2k+2), therefore f(−4)=0 that is:

18−2k=0

⇒2k=18

⇒k=  

2

18

​

 

⇒k=9

Hence k=9.

Step-by-step explanation:

Answer 2

let p)(X) = x²-x-(2k+2)

(-4) is a zero of p(x)

so p(-4) = 0

=> (-4)² -(-4) - (2k+2) = 0

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

=> -(2k+2) = -20

=> 2k +2 = 20

=> k = 9