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Question

# Factorize each of the following expression: (2x + 1)2 − 9x4

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Solution

## $\left(2\mathrm{x}+1{\right)}^{2}-9{\mathrm{x}}^{4}\phantom{\rule{0ex}{0ex}}=\left(2\mathrm{x}+1{\right)}^{2}-\left(3{\mathrm{x}}^{2}{\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left[\left(2\mathrm{x}+1\right)-3{\mathrm{x}}^{2}\right]\left[\left(2\mathrm{x}+1\right)+3{\mathrm{x}}^{2}\right]\phantom{\rule{0ex}{0ex}}=\left(-3{\mathrm{x}}^{2}+2\mathrm{x}+1\right)\left(3{\mathrm{x}}^{2}+2\mathrm{x}+1\right)\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{can}\mathrm{factorise}\mathrm{the}\mathrm{quadratic}\mathrm{expressions}\mathrm{in}\mathrm{the}\mathrm{curved}\mathrm{brackets}\mathrm{as}:\phantom{\rule{0ex}{0ex}}\left(-3{\mathrm{x}}^{2}+3\mathrm{x}-\mathrm{x}+1\right)\left(3{\mathrm{x}}^{2}+2\mathrm{x}+1\right)\phantom{\rule{0ex}{0ex}}=\left\{3\mathrm{x}\left(-\mathrm{x}+1\right)+1\left(-\mathrm{x}+1\right)\right\}\left(3{\mathrm{x}}^{2}+2\mathrm{x}+1\right)\phantom{\rule{0ex}{0ex}}=\left(-\mathrm{x}+1\right)\left(3\mathrm{x}+1\right)\left(3{\mathrm{x}}^{2}+2\mathrm{x}+1\right)\phantom{\rule{0ex}{0ex}}=-\left(\mathrm{x}-1\right)\left(3\mathrm{x}+1\right)\left(3{\mathrm{x}}^{2}+2\mathrm{x}+1\right)$

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