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
thanks in advance
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
thanks in advance
Hello, ferken!
Factor completely: .$\displaystyle abx^3 + (a-2b-ab)x^2 + (2b-a-2)x + 2$
Multiply out: . $\displaystyle abx^3 + ax^2 - 2bx^2 - abx^2 + 2bx - ax - 2x + 2$
Rearrange terms: .$\displaystyle abx^3 - abx^2 + ax^2 - ax - 2bx^2 + 2bx - 2x + 2$
Factor: .$\displaystyle abx^2{\color{blue}(x-1)} + ax{\color{blue}(x-1)} - 2bx{\color{blue}(x-1)} - 2{\color{blue}(x-1)}$
Factor: .$\displaystyle (x-1)(abx^2 + ax - 2bx - 2)$
Rearrange: .$\displaystyle (x-1)(abx^2 - 2bx + ax - 2)$
Factor: .$\displaystyle (x-1)\bigg[bx{\color{blue}(ax-2)} + 1{\color{blue}(ax-2)}\bigg]$
Factor: .$\displaystyle (x-1)(ax-2)(bx+1)$
A slightly different way:
If x= 1, $\displaystyle abx^3+ (a- 2b- ab)x^2+ (2b- a- 2)x+ 2$$\displaystyle = ab+ a- 2b- ab+ 2b- a-2+ 2$$\displaystyle = (ab-ab)+ (a- a)+ (-2b+ 2b)+ (-2+2)= 0$ so x- 1 is a factor. Dividing $\displaystyle abx^3+ (a- 2b- ab)x^2+ (2b- a- 2)x+ 2$ by x- 1 gives a quotient of $\displaystyle abx^2+ (a- 2b)x- 2$. Solving $\displaystyle abx^2+ (a- 2b)x- 2= 0$ with the quadratic formula gives $\displaystyle x= \frac{2b-a\pm\sqrt{(a- 2b)^2- 4(ab)(-2)}}{2(ab)}$$\displaystyle = \frac{2b-a\pm\sqrt{a^2- 4ab+ 4b^2+ 8ab}}{2ab}$$\displaystyle = \frac{2b-a\pm\sqrt{a^2+ 4ab+ 4b^2}}{2ab}$$\displaystyle =\frac{2b- a\pm(a+ 2b)}{2ab}$. Taking the "+", that is $\displaystyle \frac{2b-a+a+2b}{2ab}= \frac{2}{b}$ so $\displaystyle x- \frac{2}{a}$ is a factor. Taking the "-", that is $\displaystyle \frac{2b-a-(a+ 2b)}{2ab}= -\frac{1}{b}$ so $\displaystyle x-\frac{1}{b}$ is the third factor. Putting those together, we have $\displaystyle (x- \frac{2}{a})(x- \frac{1}{b})$ which does not have the "ab" times a. It is really $\displaystyle ab(x- \frac{2}{a})(x- \frac{1}{b})= (ax- 2)(bx- 1)$
$\displaystyle abx^3+ (a- 2b- ab)x^2+ (2b- a- 2)x+ 2= (x-1)(ax- 2)(bx- 1)$.