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

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Answer 1 · 2016-07-19T11:22:02

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

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Method 1 - Completing the square

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

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

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

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

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

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

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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:

$sqrt(1632) ~~ 41.5$

$sqrt(1632)-7 ~~ 34.5$

$sqrt(1632)+7 ~~ 48.5$

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

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

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