solve for k: when x^3+kx+1 is divided by x-2, and the remainder is -3

Answer 1

I do not find any quadratic equation. To solve there is synthetic division and careful algebra

Want x³+kx+1 (call equation) when divided by x-2 =f(x)+3/((x-2)

x-2 into equation first step is x², x²(x-2)=x³-2x², substract x³-2x² from equation,

x-2 must now go into 2x²+kx+1, that is 2x, 2x(x-2)=2x²-4x, subtract from 2x²+kx+1

x-2 must now go into (k+4)x+1, that is k+4, (k+4)(x-2) = (k+4)x-2(k+4), subtract from (k+4)x+1

now have2k+8+1 and that need sto be equal to -3

2k=-12, k= -6

(x³-6x+1)/(x-2)=x²+2x-2+(-3/(x-2)

Answer 2

Use synthetic division and set the remainder equal to -3 and solve to k.