Question

find the value of k if the division of kx³+9x²+4x-10 by x-3 leaves a remainder -22

Answer 1

X-3=0
X=3
Put X in divident
K(3)^3 +9(3)^2+4(3)-10=-22
27k+81+12-10+22=0
27k+105=0
27k=-105
K=-105/27
K=-35/9
Hope this helps

Answer 2

x-3=0
x = 3

Put x=3 in given polynomial.

the division of kx³+9x²+4x-10 by x-3 leaves a remainder -22

kx³+9x²+4x-10 = -22

k(3)³+9(3)²+4(3)-10 = -22

27k+9(9)+12-10 = -22

27k+81+2 = -22

27k = -22-83

27k = -105

k = -105/27

k = -35/9

Hope it helps