# How do you factor w^2 + 14w-1632=0?

Jul 19, 2016

$\left(w - 34\right) \left(w + 48\right) = 0$

#### Explanation:

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Method 1 - Completing the square

$0 = {w}^{2} + 14 w - 1632$

$= {\left(w + 7\right)}^{2} - 49 - 1632$

$= {\left(w + 7\right)}^{2} - 1681$

$= {\left(w + 7\right)}^{2} - {41}^{2}$

= ((w+7)-41))((w+7)+41)

$= \left(w - 34\right) \left(w + 48\right)$

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Method 2 - Find pair of factors

We want to find a pair of factors of $1632$ which differ by $14$.

Since $14$ is somewhat smaller than $1632$ they will be approximately $\sqrt{1632} - 7$ and $\sqrt{1632} + 7$.

Note that ${40}^{2} = 1600$ and ${41}^{2} = 1681$

So:

$\sqrt{1632} \approx 41.5$

$\sqrt{1632} - 7 \approx 34.5$

$\sqrt{1632} + 7 \approx 48.5$

Of the nearby numbers, find that $34$ and $48$ are both factors of $1632$ and they differ by $14$, so:

${w}^{2} + 14 w - 1632 = \left(w + 48\right) \left(w - 34\right)$