# Thread: Factor Fully the following expression...

1. ## Factor Fully the following expression...

Hi,
im really not sure how to approach this problem.

Factor Fully the following expression...
abx^3 + (a-2b-ab)x^2 + (2b-a-2)x + 2

the answer i know is (x-1)(ax-2)(bx+1)
really not sure how to get the answer

2. Hello, ferken!

Factor completely: . $abx^3 + (a-2b-ab)x^2 + (2b-a-2)x + 2$

Multiply out: . $abx^3 + ax^2 - 2bx^2 - abx^2 + 2bx - ax - 2x + 2$

Rearrange terms: . $abx^3 - abx^2 + ax^2 - ax - 2bx^2 + 2bx - 2x + 2$

Factor: . $abx^2{\color{blue}(x-1)} + ax{\color{blue}(x-1)} - 2bx{\color{blue}(x-1)} - 2{\color{blue}(x-1)}$

Factor: . $(x-1)(abx^2 + ax - 2bx - 2)$

Rearrange: . $(x-1)(abx^2 - 2bx + ax - 2)$

Factor: . $(x-1)\bigg[bx{\color{blue}(ax-2)} + 1{\color{blue}(ax-2)}\bigg]$

Factor: . $(x-1)(ax-2)(bx+1)$

3. A slightly different way:

If x= 1, $abx^3+ (a- 2b- ab)x^2+ (2b- a- 2)x+ 2$ $= ab+ a- 2b- ab+ 2b- a-2+ 2$ $= (ab-ab)+ (a- a)+ (-2b+ 2b)+ (-2+2)= 0$ so x- 1 is a factor. Dividing $abx^3+ (a- 2b- ab)x^2+ (2b- a- 2)x+ 2$ by x- 1 gives a quotient of $abx^2+ (a- 2b)x- 2$. Solving $abx^2+ (a- 2b)x- 2= 0$ with the quadratic formula gives $x= \frac{2b-a\pm\sqrt{(a- 2b)^2- 4(ab)(-2)}}{2(ab)}$ $= \frac{2b-a\pm\sqrt{a^2- 4ab+ 4b^2+ 8ab}}{2ab}$ $= \frac{2b-a\pm\sqrt{a^2+ 4ab+ 4b^2}}{2ab}$ $=\frac{2b- a\pm(a+ 2b)}{2ab}$. Taking the "+", that is $\frac{2b-a+a+2b}{2ab}= \frac{2}{b}$ so $x- \frac{2}{a}$ is a factor. Taking the "-", that is $\frac{2b-a-(a+ 2b)}{2ab}= -\frac{1}{b}$ so $x-\frac{1}{b}$ is the third factor. Putting those together, we have $(x- \frac{2}{a})(x- \frac{1}{b})$ which does not have the "ab" times a. It is really $ab(x- \frac{2}{a})(x- \frac{1}{b})= (ax- 2)(bx- 1)$

$abx^3+ (a- 2b- ab)x^2+ (2b- a- 2)x+ 2= (x-1)(ax- 2)(bx- 1)$.