Question

Show that (x-1) is a factor of p(x)=3x^3-x^2-3x+1 and hence factorise p(x)?

plz tell the full explanation

Answer 1

x-1=0

x=1

p(1)=3(1)^3-(1)^2-3(1)+1

=3-1-3+1

=0

Remainder is 0

Therefore,(x-1) is a factor of p(x)=3x^3-x^2-3x+1

$3{x}^{3} - {x}^{2} - 3x + 1 \\ = {x}^{2} (3x - 1) - 1(3x - 1) \\ = ({x}^{2} - 1)(3x - 1) \\ =( x + 1)(x - 1)(3x - 1)$