Question

If p(x)= X^3-4x^2+x+6, then show that p(3) and hence factorise p(x)

Answer 1

x=3 satisfies p(x) so (x-3) is one of the factor.

Now you can divide p(x)/(x-3)

$(x ^{ 3} - 4 {x}^{2} + 3x + 6) \div (x - 3)$
Next you can factories obtained quotient
$x ^{2} - x - 2 = (x + 1)(x - 2)$
Thus now you have obtained all factors
$(x - 3)(x - 2)(x + 1) = 0$