Answer
$(3x - 1)(x + 4)$
Work Step by Step
To factor a quadratic polynomial in the form $ax^2 + bx + c$, we look at factors of $(a)(c)$ such that, when added together, equal the $b$ term.
For the expression $3x^2 + 11x - 4$, we look for the factors that, when multiplied together, will equal $(a)(c)$, which is $(3)(-4)$ or $-12$, but when adding the factors together will equal $b$ or $11$. We need for one factor to be negative and the other factor to be positive, but the positive number must have the greater absolute value. A negative number multiplied with a positive number equal a negative number, but when added together, can make a positive or negative number, depending on the sign of the factor with the greater absolute value. Let's look at some possible factors:
$(a)(c)$ = $(12)(-1)$
$b = 11$
$(a)(c)$ = $(4)(-3)$
$b = 1$
$(a)(c)$ = $(6)(-2)$
$b = 4$
The first pair works. We will use that pair to split the middle term:
$3x^2 + 12x - x - 4$
Now, we can factor by grouping. We group the first two terms together and the second two terms together:
$(3x^2 + 12x) + (-x - 4)$
We see that $3x$ is a common factor for the first group, and $-1$ is a common factor for the second group, so let's factor those out:
$3x(x + 4) - (x + 4)$
We see that $x + 4$ is common to both groups, so we put that binomial in parentheses. The other binomial will be $3x - 1$, which is composed of the coefficients in front of the binomials. We now have the two factors:
$(3x - 1)(x + 4)$