Factoring quadratics with the quadratic formula: Stef's sure-fire method
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Check that $\Delta = b^2 - 4ac$ is a perfect square. (If the numbers are big, use a calculator to take the square root.)
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Use the quadratic formula to get both roots in lowest terms $\dfrac{p}{q}$ and $\dfrac{r}{s}.$
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The desired factorization is $k(qx-p)(sx-r),$ where $k = \dfrac{a}{qs}.$
($k = \pm 1$ when the original quadratic has no common integer factor. )
Example
Let's use this method to factor $36x^2+93x+60$.
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$a = 36, b = 93, c = 60 \ \ \Rightarrow \ \ \Delta = b^2 - 4ac = (93)^2-4(36)(60) = 9,$ which is a perfect square.
✓
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Roots:
$\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} = \dfrac{93 \pm \sqrt{9}}{2(36)} = \dfrac{-93 \pm 3}{72},$
giving
$\dfrac{-90}{72} = \dfrac{-5}{4} = \dfrac{p}{q} \qquad$ #[and][y]# $\qquad \dfrac{-96}{72} = \dfrac{-4}{3} = \dfrac{r}{s}$
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$k = \dfrac{a}{qs} = \dfrac{36}{(4)(3)} = \dfrac{36}{12} = 3,$ so the desired factorization is
$36x^2+93x+60$ \t ${}=k(qx-p)(sx-r)$
\\ \t ${}= 3(4x-(-5))(3x - (-4))$
\\ \t ${}= 3(4x+5)(3x+4)$
Done!
#[Some for you][Algunas para ti]#
Use the quadratic formula to factor the following quadratics.
Calculator needed with ability to show fractions! (Calculators that show fractions automatically show them in lowest terms.)
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RANDOMIZE