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Question

Factorize each of the following algebraic expression: x2 + 14x + 45

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Solution

$\mathrm{To}\mathrm{factorise}{\mathrm{x}}^{2}+14\mathrm{x}+45,\mathrm{we}\mathrm{will}\mathrm{find}\mathrm{two}\mathrm{numbers}\mathrm{p}\mathrm{and}\mathrm{q}\mathrm{such}\mathrm{that}\mathrm{p}+\mathrm{q}=14\mathrm{and}\mathrm{pq}=45.\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}9+5=14\phantom{\rule{0ex}{0ex}}\mathrm{and}\phantom{\rule{0ex}{0ex}}9×5=45\phantom{\rule{0ex}{0ex}}\mathrm{Splitting}\mathrm{the}\mathrm{middle}\mathrm{term}14\mathrm{x}\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{quadratic}\mathrm{as}9\mathrm{x}+5\mathrm{x},\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}{\mathrm{x}}^{2}+14\mathrm{x}+45={\mathrm{x}}^{2}+9\mathrm{x}+5\mathrm{x}+45\phantom{\rule{0ex}{0ex}}=\left({\mathrm{x}}^{2}+9\mathrm{x}\right)+\left(5\mathrm{x}+45\right)\phantom{\rule{0ex}{0ex}}=\mathrm{x}\left(\mathrm{x}+9\right)+5\left(\mathrm{x}+9\right)\phantom{\rule{0ex}{0ex}}=\left(\mathrm{x}+5\right)\left(\mathrm{x}+9\right)$

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