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$$(3x^2 + 6x) / (x^3 - 4x)$$

To find the hole(s) of a rational expression, I must first create a common factor in both the numerator and denominator; however, can't come up with anything for the above expression.

I should add that I am aware of the common factor of $x$, but was told that there is a second common factor; thus, a second hole.

Ben
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2 Answers2

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Factoring polynomials is analogous to factoring integers. For example, when you see $12-4$, this can be written as $4(3-1)$ since $4\cdot 3 = 12$ and $4\cdot -1 = -4$, and you recover the original expression $12-4$ (this is a good technique to check you answer, once you factor check by distributing to see that you recover the original expression). Now what allowed us to "factor-out" the $4$? It was because it was present in the factorization of both summands, the $12$ and the $-4$. Similarly, with an expression like $x^2+x$, we look for common things in each summand, i.e. an $x$ (treating the variable $x$ as we did with the integer $4$) and we have $x^2+x = x(x+1)$. Try this with your numerator and denominator.

Kevin Sheng
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  • Yes, $x$ gives a hole of (0,-1.5); however, I'm wondering if there is a second common factor. – Ben Oct 07 '15 at 02:31
  • Well, after you factor the denominator you get a linear term (the $x$) and a quadratic ($x^2 -4$). You can now use the quadratic formula to find the roots of this quadratic (which is equivalent to factoring it). Then check if the numerator shares any common (linear) factors with the denominator. – Kevin Sheng Oct 07 '15 at 02:33
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$3x^2 + 6x = x(3x+6) = 3x(x+2)$ as I think you've noticed. What of the denominator? Well the denominator is $x^3-4x = x(x^2-4)$. This is called a difference of two squares. In general, $a^2-b^2 = (a+b)(a-b)$ which you can check by multiplying $(a+b)$ and $(a-b)$. Can you see where to go from here?

Cameron Williams
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