# How do you find a polynomial of degree three that when divided by x - 2 has a remainder of 3?

Multiply $\left(x - 2\right)$ by any quadratic and add $3$
e.g. $\left(x - 2\right) {x}^{2} + 3 = {x}^{3} - 2 {x}^{2} + 3$
$\left(x - 2\right)$ multiplied by any quadratic will give a cubic.
If you then add $3$, that will be the remainder if you divide the resulting polynomial by $\left(x - 2\right)$