## Calculus with Applications (10th Edition)

$21m^{2}+13mn+2n^{2}=(3m+n)(7m+2n)$
$21m^{2}+13mn+2n^{2}$ Since the coefficient of $m^{2}$ is not equal to $1$, begin by multiplying the whole expression by $21$, which is the actual coefficient of $m^{2}$. Leave the product between $21$ and the second term expressed: $21(21m^{2}+13mn+2n^{2})=...$ $...=441m^{2}+21(13mn)+42n^{2}$ Open two parentheses containing initially the square root of the first term, which is $21m$, followed by the sign of the second term, in the first parentheses, and the product of the signs of the second and third terms, on the second parentheses: $...=(21m+)(21m+)$ Find two values whose product is equal to the third term, $42n^{2}$ and whose sum is equal to the coefficient of the expression inside parentheses in the second term, $13$. These two numbers are $7n$ and $6n$, because $(7n)(6n)=42n^{2}$ and $7+6=13$. $...=(21m+7n)(21m+6n)$ The expression was affected initially by multiplying it by $21$. Divide it by $21$ to obtain the answer: $...=\dfrac{(21m+7n)(21m+6n)}{21}=...$ Substitute $21$ by $3\cdot7$ and divide the first parentheses by $7$ and the second parentheses by $3$: $...=\dfrac{(21m+7n)(21m+6n)}{3\cdot7}=...$ $...=\dfrac{(21m+7n)}{7}\dfrac{(21m+6n)}{3}=...$ $...=(3m+n)(7m+2n)$