How do you factor the expression $49p^2 + 63pq - 36q^2$ ?

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Answer 1 · 2017-01-23T00:09:19

$49p^2+63pq-36q^2 = (7p+12q)(7p-3q)$

Note that $49p^2 = (7p)^2$ and $36q^2 = (6q)^2$

Further note that $63 = 3^2*7$ is divisible by $7$ and by $3$ but not by $6$ .

So let's look at this quadratic in terms of $7p$ and $3q$ ...

$49p^2+63pq-36q^2 = (7p)^2+3(7p)(3q)-4(3q)^2$

Note that $4-1=3$ and $4*1=4$ , so we find:

$(7p)^2+3(7p)(3q)-4(3q)^2 = ((7p)+4(3q))((7p)-(3q))$

$color(white)((7p)^2+3(7p)(3q)-4(3q)^2) = (7p+12q)(7p-3q)$

Answer 2 · 2017-01-23T00:58:46

$-36(q+7/12p)(q-7/3p)$

Making $q=lambda p$ and substituting

$49p^2 + 63 pq - 36q^2=(49 + 63 lambda - 36 lambda^2)p^2$

but $(49 + 63 lambda - 36 lambda^2)=-36(lambda+7/12)(lambda-7/3)$

so

$-36(lambda+7/12)(lambda-7/3)p^2=-36(lambdap+7/12p)(lambdap-7/3p) = -36(q+7/12p)(q-7/3p)$

How do you factor the expression 49p^2 + 63pq - 36q^249p^2 + 63pq - 36q^2?