Question

If p(x) = x cube + kx square + hx + 6 and x+1 and x-2 are factors of p(x) then find the value of H and K.

Answer 1

Answer:

$p(x) = {x}^{3} + k {x}^{2} + hx + 6$

Since (x+1) and (x-2) are factors of p(x).

p(-1) =p(2)=0 =>

${( - 1)}^{3} + k {( - 1)}^{2} + h( - 1) + 6 = 0$

And

${2}^{3} + k {(2)}^{2} + h(2) + 6 = 0$

-1+k-h+6 =0 => h-k =5...(1)and 4k+2h+14 =0=> h+2k = -7...(2). Multiply (1) by 2 and add (2) => 2h-2k+h+2k =10-7 => 3h=3 => h=1 and substitute the value of h as 1 in equation (1) => 1-k =5 => k =1-5 = -4. The values of h and k are -1 and 4. Hope it helps you.