Answer
$3x^{2}+4x-7=(3x+7)(x-1)$
Work Step by Step
$3x^{2}+4x-7$
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:
$...=(3x+7)\dfrac{(3x-3)}{3}=...$
$...=(3x+7)(x-1)$
