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Factoring Quadratics

Factoring Quadratics 1.17 - Solution

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To factor a trinomial with a leading coefficient of 1,1, we need to find two numbers whose product is the independent term. x2+5x+6\begin{gathered} x^2+5x+\textcolor{#ff8c00}{6} \end{gathered} In this case, we have that the constant term is 6.\textcolor{#ff8c00}{6}. This is a positive number, so for a product to be positive, the factors must have the same sign (both positive or both negative).

Factor Constants Product of Constants
11 and 66 6\textcolor{#ff8c00}{6}
-1\text{-}1 and -6\text{-}6 6\textcolor{#ff8c00}{6}
22 and 33 6\textcolor{#ff8c00}{6}
-2\text{-}2 and -3\text{-}3 6\textcolor{#ff8c00}{6}

Next, let's consider the coefficient of the linear term. x2+5x+6\begin{gathered} x^2+\textcolor{#ff00ff}{5}x+6 \end{gathered} In this case, since the linear coefficient is 5,\textcolor{#ff00ff}{5}, we need the sum of the factors to be 5.\textcolor{#ff00ff}{5}.

Factors Sum of Factors
11 and 66 77
-1\text{-}1 and -6\text{-}6 -7\text{-}7
22 and 33 5\textcolor{#ff00ff}{5}
-2\text{-}2 and -3\text{-}3 -5\text{-}5
We found two numbers whose product is 6\textcolor{#ff8c00}{6} and whose sum is 5.\textcolor{#ff00ff}{5}. These numbers are 22 and 3.3. With this we can factor the given trinomial.
x2+5x+6x^2+5x+6
x2+2x+3x+6x^2+2x+3x+6
x(x+2)+3x+6x(x+2)+3x+6
x(x+2)+3(x+2)x(x+2)+3(x+2)
(x+2)(x+3)(x+2)(x+3)