Solving Quadratic Equation Using Box Factoring

AI Thread Summary
The discussion revolves around solving the quadratic equation f(x)=6x^2+9x-6 using the Box Method for factoring. The user initially attempts to factor the equation but arrives at an incorrect factorization, (6x-3)(3x+6), which leads to an incorrect leading term. It is clarified that the correct approach involves factoring out the greatest common factor first, which resolves the confusion. The user expresses gratitude for the clarification, indicating that understanding the method correctly alleviated their frustration. The conversation highlights the importance of recognizing common factors in quadratic equations for accurate factoring.
nordqvist11
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Homework Statement




f(x)=6x^2+9x-6

Where does this function intersect the x-axis (i.e. what are the roots or zeroes of f(x))?

Solve by factoring using the Box Method. When I use the method that is shown on this website http://www.purplemath.com/modules/factquad2.htm for this particular problem I end up with this as my factored down version of the equation.
(6x-3)(3x+6)
But that is wrong because I would end up with 18x^2 as my leading term instead of the 6x^2 (which is in the desired equation) when I FOIL them out.

My question is that does using the method on that website to factor down a quadratic work for this equation? If so, could someone tell me what I'm doing wrong.

Thanks :)

Homework Equations





The Attempt at a Solution


[PLAIN]http://5img.com/img836/4599/69quadproblem.jpg
 
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Welcome to PF, nordqvist11! :smile:

Your equation looks like the last example on the page you mentioned.
It says there that you should take out the common factor first...
 
You're right it does :). Thanks a lot, I got a real headache because of that.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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