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

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

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Answer 1 · 2017-01-23T00:09:19

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

Explanation:

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

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

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

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

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

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

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

Answer 2 · 2017-01-23T00:58:46

$- 36 (q + \frac{7}{12} p) (q - \frac{7}{3} p)$ $-36(q+7/12p)(q-7/3p)$

Explanation:

Making $q = λ p$ $q=lambda p$ and substituting

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

but $(49 + 63 λ - 36 λ^{2}) = - 36 (λ + \frac{7}{12}) (λ - \frac{7}{3})$ $(49 + 63 lambda - 36 lambda^2)=-36(lambda+7/12)(lambda-7/3)$

so

$- 36 (λ + \frac{7}{12}) (λ - \frac{7}{3}) p^{2} = - 36 (λ p + \frac{7}{12} p) (λ p - \frac{7}{3} p) = - 36 (q + \frac{7}{12} p) (q - \frac{7}{3} p)$ $-36(lambda+7/12)(lambda-7/3)p^2=-36(lambdap+7/12p)(lambdap-7/3p) = -36(q+7/12p)(q-7/3p)$

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

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