How do you factor #y= 3x^2+29x-44 # ?

1 Answer
Jun 7, 2018

Answer:

#(x+11)*(3x-4)#
The explanation below shows one method on how to factor polynomials when the leading coefficient is not equal to 1.

Explanation:

Here's one way to factor a polynomial when the leading coefficient not equal to 1:

For: #y=ax^2+bx+c#
Start by finding 2 numbers #x_1# and #x_2# where:
#x_1 * x_2 = a*c#
and
#x_1 + x_2 = b#

In this case #a*c = 3*(-44) = -132#

It usually helps to think of the prime factors of the number.
In this case, they are: # 132 = 2*2*3*11 #
You can try a few combinations, keeping in mind that, in this case, the 2 numbers must add to #29#.
The winning combination turns out to be #33# and #-4#
#33*(-4) = -132#
#33+(-4) = 29#

Now write down the current answers as if they were the factors:
#(x+33)*(x-4)#

However, that's not quite the answer yet. If we left it like this, the leading coefficient would be 1 and the last coefficient would be too big. So we need to adjust for that.

Divide the 2 answers by the leading coefficient, in this case, #3#.

#(x+33/3)*(x-4/3)#
If they can be divided evenly, leave it at that. If not, move the denominator as a factor next to the x, like this:

#(x+11)*(3x-4)#

And those are the factors!