how do u factor something when there is a number in front of the x squared term?

Question

for example the problem.....

 

5a²+43a+56

Answer 1

It isn't that much different, though the factor possibilities to check out will happen to grow the more that a number can be factor. For example, 5 is relatively easy. As a prime, it only has 1 and 5, so it will factor as (x ± something)(5x ± something else).

If we have 4x², we not only have 1 and 4, but also split into 2 and 2. The more that a number can be factored, the more you should see if you can factor out part or all of the coefficient to simplify the problem.

 

Thankfully, this particular problem is rather easy. 5 is prime, and both signs are adding, so we already have:

(x + M)(5x + N)

M and N multiplied together has to equal 56.

5M and N added together has to equal 43.

We can have a 7 and 8, a 4 and 14, or possibly 2 and 28, and either of those flipped.

 

I'm liking the 7 and 8, because 5 * 7 will give us 35. That brings us close to 43. Let's test to see how much would be left:

43 - 35 = 8

A-ha! and 8 * 1 = 8, so our problem factors out as:

(x + 7)(5x + 8)

 

You can check our answer by FOILing back out again.

Answer 2

Another method is to use factoring by grouping.  

 

First, we multiply 5 by 56 to get 280

 

Then, find two numbers with product 280 and sum 43

 

The numbers are 35 and 8

 

Rewrite 43x as 35x + 8x and factor by grouping:

 

5x² + 43x + 56 = 5x² + 35x + 8x + 56

 

                           = 5x(x + 7) + 8(x + 7)

 

                           = (x + 7)(5x + 8)