Question

If x cube - 8x square + 8x + k is completely divisible by ( x- 2), then find the value of k

Answer 1

Given that

p(x) = x³ - 8x² + 8x + k

g(x) = (x - 2)


Given (x - 2) can completely divide

=> (x - 2) is a factor of p(x)


By remainder theorem,

if x - 2 is a factor

=> p(x) = 0 at x = 2


So,

p(x) = x³ - 8x² + 8x + k

x = 2

=> p(2) = 2³ - 8(2)² + 8(2) + k = 0

=> 8 - 8(4) + 16 + k = 0

=> 8 - 32 + 16 + k = 0

=> -8 + k =0

=> k = 8


Hope it helps dear friend ☺️

Answer 2

Answer :


$given \: that \:p(x) = {x}^{3} - 8 {x}^{2} + 8x + k \: is \: divisible \: by \: (x - 2) \\ \\ = > \: zero \: of \: (x - 2) = \\ \\ = > x - 2 = 0 \\ \\ = > x = 2 \\ \\ so \\ \\ = > we \: can \: say \: that \: \\ \\ = > p(2) = 0 \\ \\ = > put \: 2 \: at \: the \: place \: of \: x \\ \\ = > p(2) = {2}^{3} - 8 {(2)}^{2} + 8(2) + k = 0 \\ \\ = > \: 8 - 32 + 16 + k = 0 \\ \\ = > \: 24 - 32 + k = 0 \\ \\ = > - 8 + k = 0 \\ \\ = > k = 0 + 8 \\ \\ = > \: k = 8$



Hope it would help you