## Calculus with Applications (10th Edition)

$3x^{2}+4x-7=(3x+7)(x-1)$
$3x^{2}+4x-7$ Since the coefficient of $x^{2}$ is not equal to $1$, begin by multiplying the whole expression by $3$, which is the actual coefficient of $3$. Leave the product between $3$ and the second term expressed: $3(3x^{2}+4x-7)=...$ $...=9x^{2}+3(4x)-21$ Open two parentheses containing initially the square root of the second term, which is $3x$, 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: $...=(3x+)(3x-)$ 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, $4$. These two numbers are $7$ and $-3$, because $(7)(-3)=-21$ and $7-3=4$. $...=(3x+7)(3x-3)$ The expression was affected initially by multiplying it by $3$. Divide it by $3$ to obtain the answer: $...=(3x+7)\dfrac{(3x-3)}{3}=...$ $...=(3x+7)(x-1)$