# Thread: So i have this polynomial.....

1. ## So i have this polynomial.....

p(x)= x^3 + ax^2 + bx + 6 i know (x-2) is a factor and when divided by (x-1) the remainder is 4, i have done the first part and solved for a and b
a=-4 and b=1
the next step asks me to find the two other linear factors of p(x), i keep thinking quadratics but cant see it, i also tried dividing p(x) into (x-2) to get the quadratic, i get a remainder of 32..... after that im stuck, please tell me if im going good and what to do next.

2. ## Re: So i have this polynomial.....

Originally Posted by johnreal
p(x)= x^3 + ax^2 + bx + 6 i know (x-2) is a factor and when divided by (x-1) the remainder is 4, i have done the first part and solved for a and b
a=-4 and b=1
the next step asks me to find the two other linear factors of p(x), i keep thinking quadratics but cant see it, i also tried dividing p(x) into (x-2) to get the quadratic, i get a remainder of 32..... after that im stuck, please tell me if im going good and what to do next.
$\displaystyle \frac{x^3-4x^2+x+6}{x-2}=x^2-2x-3=x^2-3x+x-3=$

$\displaystyle =x(x-3)+1\cdot(x-3)=(x+1)(x-3)$

3. ## Re: So i have this polynomial.....

thank you so much i re did my whole equation like yours and i finally got it thanks soooo much, since ive started here princeps you have been so much help, thanks