<p>For instance, how would you go about factorizing this:</p>
<p>3(x^3/2) - 9(x^1/2) + 6(x^-1/2)</p>
<p>Thanks. :)</p>
<p>What you want to do is eliminate the denominator (x^1/2) on the last term. To do this, we multiply the entire polynomial by x^1/2. This gives us 3x^2-9x+2. From here it is simple to factor, and you will end up with 3(x-1)(x-2).</p>
<p>You divide out 3(x^-1/2) because 3 is the smallest and x^-1/2 is the smallest since the exponent on x is negative. What GreyWolf said is wrong… you can’t just multiply everything by x^1/2, that’s creating a whole new polynomial.</p>
<p>Hey there. Actually, unless my brain isn’t working (very possible after an 11 hour workday and only 4 hours of sleep), I stand by my original answer. You can multiply everything by x^1/2 or any constant factor, as long as you are multiplying the entire polynomial and not just a single term. The best way to show this is by plugging in the solutions, namely x=2 and x=1. You will see they work. This is assuming that the polynomial is set to zero which is generally implied when solving for x.</p>
<p>For example, say we have the polynomial (1/3)x^2 + (2/3)x + (2/3) = 0. Obviously you could use the quadratic formula, but a simpler and more elegant way would be to multiply the entire polynomial by 3, eliminating the denominator in all terms. Note that this should only be done when the polynomial is set to zero as you are multiplying both sides of the equation by three. Doing this we get x^2 + 2x +2 = 0. Which obviously factors to (x+1)(x+1).</p>
<p>I have a pretty strong mathematical background and have done quite well on standardized tests, so I know where I am coming from. Let me know if you have questions, corrections, etc.?</p>
<p>Greywolf, what you are doing is right… to an extent. Multiplying by x^1/2 would not cause any problems in this situation. However, a much better practice is to factor out x^-1/2, giving you the same equation with 1/sqrt(x) in front of the polynomial. When you do this, you are left with a third possible solution. However, since x is in the denominator in this factor, there is no solution for x that would give an answer of 0.</p>
<p>So, like I said, in this specific context, what you did would work. But, multiplying like you are saying could form a bad habit in other polynomials. Factoring out gives another possible solution that isn’t present when just multiplying the polynomial by the inverse of that term.</p>
<p>Tyler309, I see where you are coming from, but the (1/sqrtx) term is obviously trivial. In my post, I specified that one should only use this method when the RHS is zero. Can you give an example of when my method would not work given those parameters (too tired to figure one out).</p>
<p>I agree completely. Just a product of my specific math classes, I suppose. </p>
<p>Ex. x^3+6x^2+5x=0</p>
<p>You could divide by x because the RHS is 0, however factoring would give you an additional solution, as it would be… x(x^2+6x+5)=0
x(x+1)(x+5)=0
x=0,-1,-5<br>
The solution x=0 would not be a result if you divided by x to begin with. I know it might seem overly-simplistic, and I’m sure you already knew all of this, but that was what I was trying to emphasize. I was taught never to get rid of a variable in a polynomial like that, regardless of how it would affect the outcome. 1/sqrt(x) can be easily disregarded if there are no solutions in which x would work. In my opinion, it would be better to have two solutions with the possibility of a third that might or might not work, than to have only two solutions. I understand where you are coming from, though.</p>
<p>And if my post makes little to no sense, maybe I misunderstood you, because I, as well, am exhausted. :P</p>
<p>It’s pretty obvious that you and I are both proficient in math, but I guess I’ll continue this “argument” for lack of a better word. I do not view your example as pertinent. My method was to multiply both sides to eliminate that pesky denominator (Though multiplication and division are inexorably linked). I would never divide both sides as the risk of dividing by zero (Which occurs in your example) and getting false/incomplete results is too large.</p>
<p>For full disclosure, I do usually just factor out simply as your examples show and I think my “method” comes from doing too much competition math where eliminating denominators and multiplying by that clever form of 1 is essential. I absolutely agree that it is better to have extraneous solutions. Ultimately, my question to you is whether there is ever a time when multiplyingg to eliminate the denominator on a polynomial set to zero will ever result in the loss of valid, defined (Over all real numbers) solutions?</p>
<p>My argument was a matter of principle. Although your way may work in competitions, because most of the time you are searching for only zeros, you actually are changing the equation, something that doesn’t occur when you approach the problem the way I did. You might get the same zeros if you just divide; however, he never explicitly states that the equation is equal to 0. Furthermore, if you plug in x=5 into both your and my equations- x=5 is a random point- you will see that my solution is equal to the solution of the beginning equation, while yours is not. There might not be any defined zeros when you factor out 1/sqrt(x), but it does have an impact on the outputs you will receive. Disregarding the undefined portion of the equation may make it simpler, which holds a great importance in competition math where time and simplicity is everything, but it actually does change every single output value in the graph except for the two zeros, and is a bad habit to form.</p>