I have $3x^{-1/2} (x^2-3x+2)$ However, I just tried to expand, and the answer is not the same as the original question.
With fractional exponents I take out the smallest exponent, then I add the smallest exponents to those left in the non-factored equation. However, it does not seem to be working. Any ideas?
My answer is $3x^{-\frac{1}{2}}(x-1)(x-2)$:
$3x^{\frac{3}{2}}-9x^{1/2}+6x^{-\frac{1}{2}} = 3x^{\frac{1}{2}}(x-3+\frac{2}{x}) =3x^{\frac{1}{2}}(\frac{x^2-3x+2}{x})= 3x^{-\frac{1}{2}}(x^2-3x+2) = 3x^{-\frac{1}{2}}(x^2 -2x +1 - x +1) = 3x^{-\frac{1}{2}}((x-1)^2-(x-1)) = 3x^{-\frac{1}{2}}(x-1)(x-1-1) = 3x^{-\frac{1}{2}}(x-1)(x-2)$
There are at least three alternative but relatively general-purpose ways to factorize a polynomial such as $x^2 - 3x + 2:$
To factorize $x^2 + bx + c,$ look for $p$ and $q$ such that $pq = c$ and $p + q = b,$ so that you can write $x^2 + bx + c = (x + p)(x + q).$ If $c = 2,$ then $p$ and $q$ are either both positive or both negative; $b = -3$ then implies $p$ and $q$ are both negative. It's a relatively easy guess that $p$ and $q$ are $-1$ and $-2$ (in either order; it does not matter), so $x^2 - 3x + 2 = (x - 1)(x - 2).$
Complete the square. for $x^2 + bx + c,$ evaluate $\left(x + \frac b2\right)^2.$ If this is greater than $x^2 + bx + c,$ interpret the difference as the square of something, and now you can write $x^2 + bx + c$ as the difference of two squares (which is something we know how to factorize). In this example, $b = -3,$ so we compute $\left(x - \frac 32\right)^2 = x^2 - 3x + \frac 94 = (x^2 - 3x + 2) + \frac 14,$ so $$\begin{eqnarray} x^2 - 3x + 2 &=& \left(x - \frac 32\right)^2 - \left(\frac 12\right)^2 \\ &=& \left(\left(x - \frac 32\right) + \frac 12\right) \left(\left(x - \frac 32\right) - \frac 12\right) \\ &=& (x - 1)(x - 2) \\ \end{eqnarray}$$
Use the formula to find the roots of the quadratic equation $x^2 - 3x + 2 = 0.$ The roots are $1$ and $2,$ which implies that $x^2 - 3x + 2 = (x - 1)(x - 2).$
Method 1 is something you might try first if you have a hunch it might be easy. One develops a sense for this after doing enough factorizations. If it looks too hard, or it doesn't work, you try one of the other methods. Method 2 will always find a solution if there is one (so if it finds no solution, you know you have factorized as far as you can go with real coefficients). Method 3 is really just a formulaic way of applying method 2 (since the quadratic formula itself can be derived by completing the square); you can use either one, depending on which one you find easier to remember.
It is worth noting that questions about factoring quadratic polynomials have arisen here multiple times, usually receiving answers that would be applicable to this problem as well. For example, Factoring Quadratics. The Kimmel Method can be used like Method 1 above but also works when the coefficient of $x^2$ is not $1$.
factorise; it's synonymous with factor (in this context). – beep-boop Sep 07 '14 at 19:16