How do you factor #2x^2+11x+15#?

To factor this expression we will need to factor each of the terms individually so that when we reverse the factorization, the terms will re-multiply into the given expression.

Given: #2x^2 + 11x + 15#

Because we start out with #x^2# we know we will end up with two brackets: #( ...)( ...)#

We know that each bracket will need an #x# inside: #(x...)(x...)#

In this case we can see that one of the #x# terms will need to be multiplied by 2 since we start with #2x^2#. So #(2x...)(x...)#

We know that each bracket will need a numerical factor. In this case the number to be factored is: #15=(15)(1) =(3)(5)=(5)(3)#

So we can write in:#(2x...15)(x...1) = (2x...3)(x...5) = (2x...5)(x...3)#

But we need to have factors that when multiplied by #x# will add or subtract to result in the central term of the expression - in this case #11x#.

We can see that #2*3 = 6 and 5*1 = 5# will result in #11# when added. This indicates both numeric terms need to be #+#.

The given expression also agrees with the double #+#, because both signs are #+#.

Then we can write: #2x^2 + 11x + 15 = (2x+5)(x+3)#

To check for correctness, simply re-multiply the answer to result in the given expression.