How do you factor and solve #3x^2+11x-4=0 #?

In this quadratic, the coefficient of the squared term gives us a big hint on how to factor this. Since 3 is a prime number, we know that the variable in one binomial will keep a coefficient of 3 and the other will have a coefficient of 1.

So we know the factored form will look something like this.
#(3x + somethi\ng)(x + somethi\ng)#

4 has 3 factors: 1, 2, and 4. Our 4 is negative. This means one (and only one) of the factors is negative.

We can think about what numbers can subtract (since one of the numbers must be negative, it has to be subtraction) to get 11. Whatever the positive number is, it must be larger than 11. Since none of the factors of 4 are larger than 11, we know that we will be putting this positive number in the last position so that it can be multiplied by 3. In this case, our only possible factorized form is #color(red)((3x - 1)(x + 4))#. It is the only case in which we will be able to multiply a factor by 3 and get a value larger than 11 (2 would give us 6 and 1 would give us 3).

If we check, we can see that this is a correct factorization.
#(3x-1)(x+4)#
#3x^2 + 12x - x - 4#
#3x^2 + 11x - 4#

Now to solve for x.

#(3x-1)(x+4) = 0#

We can divide 0 by either binomial to cancel it out and focus on solving just one binomial. When we have this option, we have to do it both ways and see the different outcomes- we need both outcomes for a complete answer.

#3x-1 = 0 -> 3x = 1 -> x = color(red)(1/3)#
#x+4 = 0 -> x = color(red)(-4)#