How do you factor $w^{2} + 14 w - 1632 = 0$ $w^2 + 14w-1632=0$ ?

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How do you factor $w^{2} + 14 w - 1632 = 0$ $w^2 + 14w-1632=0$ ?

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Answer 1 · 2016-07-19T11:22:02

$(w - 34) (w + 48) = 0$ $(w-34)(w+48) = 0$

Explanation:

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Method 1 - Completing the square

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

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

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

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

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

$= (w - 34) (w + 48)$ $= (w-34)(w+48)$

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Method 2 - Find pair of factors

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

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

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

So:

$\sqrt{1632} \approx 41.5$ $sqrt(1632) ~~ 41.5$

$\sqrt{1632} - 7 \approx 34.5$ $sqrt(1632)-7 ~~ 34.5$

$\sqrt{1632} + 7 \approx 48.5$ $sqrt(1632)+7 ~~ 48.5$

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

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

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How do you factor $w^{2} + 14 w - 1632 = 0$ $w^2 + 14w-1632=0$ ?

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How do you factor $w^{2} + 14 w - 1632 = 0$ $w^2 + 14w-1632=0$ ?