When a polynomial is ÷ by another polynomial and it gives a remainder but when we put the value of the x in the polynomials it gives 0 remainder.

Question

Let we have an equation $x^2+4x+2$ and we want to divide it with $x-4$. By remainder theorem it gives the value of $34$. But when we put the value of $x=6$ in equations it gives zero remainder. And quotient is $31$. How's that? Where I am mistaking?

Answer 1

You are getting remainder 34. This is because you haven't placed value of x yet. If you place value of x, you will see that divisor is (6-4) = 2. Now obviously you can't get remainder of 34 when divisor is 2.(remainder can't be greater than divisor) You get remainder 0.

The equation you solved earlier was general case. So if x had been any other number greater than 38(so that divisor is greater than 34), then remainder would have been 34. Otherwise you would have to divide remaining term with divisor

Answer 2

$x^2 + 4x +2=(x+8)(x-4)+34$. I think you're mistaking what dividing a polynomial means