# Chapter R - Algebra Reference - R.2 Factoring - R.2 Exercises: 13

$3a^{2}+10a+7=(3a+7)(a+1)$

#### Work Step by Step

$3a^{2}+10a+7$ Since the coefficient of $a^{2}$ is not equal to $1$, begin by multiplying the whole expression by $3$, which is the actual coefficient of $a^{2}$. Leave the product between $3$ and the second term expressed: $3(3a^{2}+10a+7)=...$ $...=9a^{2}+3(10a)+21$ Open two parentheses containing initially the square root of the second term, which is $3a$, 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: $...=(3a+)(3a+)$ Find two numbers whose product is equal to the third term, $21$ and whose sum is equal to the coefficient of the expression inside parentheses in the second term, $10$. These two numbers are $7$ and $3$, because $(7)(3)=21$ and $7+3=10$. $...=(3a+7)(3a+3)$ The expression was affected initially by multiplying it by $3$. Divide it by $3$ to obtain the answer: $...=(3a+7)\dfrac{(3a+3)}{3}=...$ $...=(3a+7)(a+1)$

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