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# ​ Factor.

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Factor.

SamJones  Jan 27, 2018
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For factoring, I generally use the "AC-method." The method does not have a formal name, so I will assign it myself! In a quadratic, which is in the form \(\textcolor{red}{a}x^2+\textcolor{blue}{b}x+\textcolor{green}{c}\), you first consider factors of \(\textcolor{red}{a}\textcolor{green}{c}\) that sum to \(\textcolor{blue}{b}\).

In this particular case for \(\textcolor{red}{5}x^2+\textcolor{blue}{22}x+\textcolor{green}{8}\),  \(\textcolor{red}{a}\textcolor{green}{c}=\textcolor{red}{5}*\textcolor{green}{8}=40\) and \(\textcolor{blue}{b}=\textcolor{blue}{22}\). We need to find two numbers that multiply to get 40 and add to get 22. Let's list a few factors.

 Factors of 40 Factor's Sum 1,40 41 2,20 22 4,10 14 5,8 13

Of the factors of 40 listed, only 2 and 20 are factors that multiply to 40 and add to 22. We can use this information to manipulate the given trinomial.

 \(5x^2+22x+8\Rightarrow(5x^2+2x)+(20x+8)\) Notice here that I have not actually changed the value of this expression; I have simply changed the way it looks. The next step is to factor via grouping. I've made the groups easier to identify by adding parentheses to the expression. Factor out the GCF from the individual groups. \(x(5x+2)+4(5x+2)\) Notice how a common factor x and 4 both have a common factor of 5x+2. It is possible to combine them together then! \((x+4)(5x+2)\)

Factoring is one of those computations where there are a multitude of ways to arrive at the same answer. This is just the method I generally default to--even if it is not necessarily most efficient.

TheXSquaredFactor  Jan 27, 2018

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