The answer is 3. k = 3
Letf(x)=x3+x2+kx-15
By the remainder theorem, the remainder when f(x) is divided by (x-2)
will be
f(2)
So we have: 3=f(2)=8+4+2k-15=2k-3
Add 3 to both ends to get: 6 = 2k
Divide both sides by 2 and transpose to get:
k=3Left(x)=x3+x2+kx-15
By the remainder theorem, the remainder when
f ( x ) is divided by ( x − 2 ) will be f ( 2 )
So we have:
3 = f (2) = 8 + 4 + 2
k − 15 = 2
k − 3
Add 3
to both ends to get: 6 = 2 k
Divide both sides by 2
and transpose to get:
k = 3
