Question

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

Answer 1

Answer:

$\huge\boxed{\fcolorbox{blue}{orange}{HELLO\:MATE}}$

Step-by-step explanation:

GIVEN:

→$A \:polynomial x^{2}-x(2k+2)$ = p(x)

[Let]

TO FIND:

→$The\: value \:of \:k\: for \:which\: (-4)\: is \:a \:zero\: of \:p(x)$

SOLUTION:

Let p(x) = 0.

=>$x^{2}-x(2k+2)$ = 0

On substituting x =(-4), we have;

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

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

=>$16=4(2k+2)$

=>$16=8k +8$

=>$8k = 16-8$

=>$8k =8$

=>$k =\dfrac{8}{8}$

=>$k = 1$

Therefore the value of k should be 1 , so that (-4) is a zero of the given polynomial.

$\huge\red{\boxed{ANSWER:k =1}}$