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Question

# Factorize each of the following quadratic polynomials by using the method of completing the square: 4y2 + 12y + 5

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Solution

## $4{y}^{2}+12y+5\phantom{\rule{0ex}{0ex}}=4\left({y}^{2}+3y+\frac{5}{4}\right)\left[Makingthecoefficientof{y}^{2}=1\right]\phantom{\rule{0ex}{0ex}}=4\left[{y}^{2}+3y+{\left(\frac{3}{2}\right)}^{2}-{\left(\frac{3}{2}\right)}^{2}+\frac{5}{4}\right]\left[Addingandsubtracting{\left(\frac{3}{2}\right)}^{2}\right]\phantom{\rule{0ex}{0ex}}=4\left[\left(y+\frac{3}{2}{\right)}^{2}-\frac{9}{4}+\frac{5}{4}\right]\phantom{\rule{0ex}{0ex}}=4\left[\left(y+\frac{3}{2}{\right)}^{2}-{1}^{2}\right]\left[Completingthesquare\right]\phantom{\rule{0ex}{0ex}}=4\left[\left(y+\frac{3}{2}\right)-1\right]\left[\left(y+\frac{3}{2}\right)+1\right]\phantom{\rule{0ex}{0ex}}=4\left(y+\frac{3}{2}-1\right)\left(y+\frac{3}{2}+1\right)\phantom{\rule{0ex}{0ex}}=4\left(y+\frac{1}{2}\right)\left(y+\frac{5}{2}\right)\phantom{\rule{0ex}{0ex}}=\left(2y+1\right)\left(2y+5\right)$

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