Hi, I need help with this question:
Factor the expressions px^3 + (p-q)x^2 + (-2p-q)x + 2q and abx^3 + (a-2b-ab)x^2 + (2b-a-2)x + 2
I've tried a few different ways, but none of them work out. Thank you for any help!
$\displaystyle px^3 + px^2 -2px - qx^2 - qx + 2q $$\displaystyle px^3 + (p-q)x^2 + (-2p-q)x + 2q $
$\displaystyle px(x^2 + x - 2) - q(x^2 - x - 2) $
$\displaystyle px(x^2 + x - 2) - q(x^2 - x - 2) $
$\displaystyle (px - q)(x + 2)(x - 1) $
i was just about to make that suggestion
i noticed almost immediately that x - 1 was a factor by the factor theorem. performing long division twice will get you the answer you have
coincidentally, (x - 1) is a factor of the second one as well
janvdl's method is still easiest by far though. just make sure you factor correctly. factoring out a -q from the second set changes all the signs inside the bracket
EDIT: Good Job Janvdl. You are the best mathematician in this post!
Hello, starswept!
$\displaystyle abx^3 + (a-2b-ab)x^2 + (2b-a-2)x + 2$
Expand: .$\displaystyle abx^3 + ax^2 - 2bx^2 - abx^2 + 2bx - ax - 2x + 2$
Re-arrange: .$\displaystyle \underbrace{abx^3 - abx^2} + \underbrace{ax^2 - ax} - \underbrace{2bx^2 + 2bx} - \underbrace{2x + 2}$
. . Factor: .$\displaystyle abx^2(x-1) + ax(x-1) - 2bx(x-1) - 2(x-1)$
. . Factor: .$\displaystyle (x-1)\,(\underbrace{abx^2 + ax} - \underbrace{2bx - 2})$
. . Factor: .$\displaystyle (x-1)\,[ax(bx +1) - 2(bx+1)]$
. . Factor: .$\displaystyle (x-1)\,(ax-2)\,(bx+1)$