SOLUTION: I have this complicated math expression that I cant seem to solve. It's: 14x^2-x-3/2^2-7x+3=? nothing will factor into either of these number, so I got: 7x^2-x-3/x^2-7x+3 I

Question 720088: I have this complicated math expression that I cant seem to solve.
It's: 14x^2-x-3/2^2-7x+3=?
nothing will factor into either of these number, so I got: 7x^2-x-3/x^2-7x+3
I need help badly. Its hard for me to start these problems.
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
First of all, please put multiple-term numerators and denominators in parentheses. What you posted, without parentheses, meant:

but I'm guessing you intended:

which should be posted as: (14x^2-x-3)/(2x^2-7x+3)
Clear, unambiguous problems are more likely to get quick responses.

The reason we try to factor these fractions is that only factors may be canceled when reducing fractions! So that 2 you canceled out was an error. 2 is not a factor of the whole numerator and denominator. It is only a factor of a term in each.

It is also an error to say that the numerator and denominator do not factor. They do factor:

and looking at this we can see that there is a common factor to cancel:

leaving:

P.S. Here some hints that might help you figure out how to factor trinomials like the original numerator and denominator:

Don't forget that positive numbers can have two positive factors or two negative factors. In the original denominator the +3 at the end has factors of 1*3 and (-1)*(-3) and, as it turned out, it was the negative factors that were needed in order to factor that expression.
Be sure to try all possible combinations. In the original numerator can be factored as: 14x*x, 7x*2x, (-14x)*(-x) or ((-7x)*(-2x) and the -3 at the end can be factored as 1*(-3) or (-1)*3. This makes 8 different combinations. If you overlook some of them you might not see the one that works.
If you have trouble seeing all the different combinations or if there are so many combinations you'd like a faster way then the quadratic formula, , can be used to help factor:
- The expression in the square root, , has several uses and it even has its own name, the "discriminant". A use for the discriminant in factoring is that it can be a quick way to check if the quadratic expression will factor at all. This can save you from having to check a lot of combinations, none of which will end up working. If the value of the discriminant is not a perfect square then the quadratic expression will not factor!
- If the discriminant is a perfect square then the quadratic expression will factor. Not only that, its value and the quadratic formula can be used to help figure out how it will factor.
Here's some examples to show you how:
- With a = 3, b = -4 and c = 64, the discriminant for this is:
  
  This is not a perfect square so will not factor. This saves us a lot of time because there are many factor of 64
- (the original numerator which you couldn't see how to factor)
  With a = 14, b = -1 and c = -3, the discriminant for this is:
  
  Simplifying...
  
  Since it is a perfect square. Now we know that will factor. We can also use this value, 169, to help us figure out how it factors. We do this by finishing the quadratic formula. Since we already know the value of the discriminant and since we already know its square root, this goes pretty quickly:
  
  which is short for:
  or
  Simplifying...
  or
  or
  These tell us how to factor.
  From x = 1/2 we get the factor: (2x+1)
  From x = -3/7 we get the factor: (7x + (-3)) or (7x - 3)
  Note where the numerators and denominators go into the factors!
- .
  This discriminant:
  This is a perfect square so our expression will factor. Continuing with the quadratic formula:
  
  or
  or
  or
  From x = (-1)/2 we get a factor of: (2x + (-1)) or (2x - 1)
  With x = -2 we have a whole number. If you end up with x equal to a whole number like this, then make it a fraction by putting it over a 1. So we -2 rewrite it as a fraction: -2/1 and then find the factor: (1x + (-2)) or (x - 2)

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