1. ## Factoring

Basically what I need is an explenation and the steps to solve two different types of factoring problems.

I could care less about the answer, I have it in the back of the sheet anyway.

1. x^3 - 6x^2 + 11x - 6 = 0

2. 4(x^3/2)(y^2/3) + 18(x^1/2)(y^5/3)

Once again I need the steps for the first one, to help factor that kind, and the explenation with the fraction factoring.

2. Originally Posted by Freaky-Person
1. x^3 - 6x^2 + 11x - 6 = 0
\displaystyle \begin{aligned}x^3-6x^2+11x-6&=x^3-(x^2+5x^2)+(5x+6x)-6\\ &=(x^3-x^2)-(5x^2-5x)+(6x-6)\\ &=x^2(x-1)-5x(x-1)+6(x-1)\\ &=(x-1)(x^2-5x+6)\\ &={\color{blue}(x-1)(x-2)(x-3)}\,\blacksquare \end{aligned}

3. Originally Posted by Freaky-Person
2. 4(x^3/2)(y^2/3) + 18(x^1/2)(y^5/3)
$\displaystyle 4x^{3/2}y^{2/3}+18x^{1/2}y^{5/3}=4x^{1/2}\cdot x\cdot y^{2/3}+18x^{1/2}\cdot y^{2/3}\cdot y$

Does that make sense?

4. Originally Posted by Krizalid
$\displaystyle 4x^{3/2}y^{2/3}+18x^{1/2}y^{5/3}=4x^{1/2}\cdot x\cdot y^{2/3}+18x^{1/2}\cdot y^{2/3}\cdot y$

Does that make sense?
Alrighty.. No I didn't get that.

All you really needed to do was factor out the correct amount... I just didn't get what amount that would be with the fractions..

If it helps, the answer is -> [2(x^1/2)(y^2/3)][2x+9y]

5. Originally Posted by Krizalid
\displaystyle \begin{aligned}x^3-6x^2+11x-6&=x^3-(x^2+5x^2)+(5x+6x)-6\\ &=(x^3-x^2)-(5x^2-5x)+(6x-6)\\ &=x^2(x-1)-5x(x-1)+6(x-1)\\ &=(x-1)(x^2-5x+6)\\ &={\color{blue}(x-1)(x-2)(x-3)}\,\blacksquare \end{aligned}
That great and all, but I don't get WHY you seperated the x^2 and x values, and why by that amount.

6. Originally Posted by Freaky-Person
Basically what I need is an explenation and the steps to solve two different types of factoring problems.

I could care less about the answer, I have it in the back of the sheet anyway.

1. x^3 - 6x^2 + 11x - 6 = 0
The rational root theorem tells us that any rational root of a polynomial is a
+/- a factor of the constant term divided by +/- a factor of the coeficient of the highest power.

So in this case the rational root theorem tells us that any rational roots
of this cubic are amoung 1, -1, 2, -2, 3, -3.

Try $\displaystyle x=3$ in $\displaystyle x^3 - 6x^2 + 11x - 6$ gives $\displaystyle 0$, so $\displaystyle x-3$ is a factor of $\displaystyle x^3 - 6x^2 + 11x - 6$.

Now dividing $\displaystyle x^3 - 6x^2 + 11x - 6$ by $\displaystyle x-3$ gives:

$\displaystyle x^3 - 6x^2 + 11x - 6=(x-3)(x^2-3x+2)$

and the last term can be factored by inspection, or finding the roots using
the quadratic formula or the same procedure as before to get:

$\displaystyle x^3 - 6x^2 + 11x - 6=(x-3)(x^2-3x+2)=(x-3)(x-1)(x-2)$

RonL

7. Originally Posted by Freaky-Person
2. 4(x^3/2)(y^2/3) + 18(x^1/2)(y^5/3)
Take out the common factor $\displaystyle 2 x^{1/2}y^{2/3}$ to get:

$\displaystyle 4x^{3/2}y^{2/3} + 18x^{1/2}y^{5/3}= 2 x^{1/2}y^{2/3}(2 x +9y)$

RonL