# Factor 2x²+13x+40

I thought with this sort of problem you are supposed to multipy 40 by 2 and get 80 and then come up with two factors that equal 13. If this doesn't work, you are supposed to factor out a number/variable.

Since neither of these methods work, is this problem unable to be factored, or am I mistaken?

Thanks in advance.

i thought it would be 15x3+40. I am not sure though

I don't think it can be factored. The factors of 40 are

40*1

20*2

10*4

8*5

and none of those number will work.

For

a x^2 + bx + c

calculate the discriminant:

b^2 - 4 a c.

If it is negative it cannot be factored.

In this case it is -151. The factorization involves the square root of the discriminant. If it isn't rational you won't be able to find the factorization by inspection easily either.

## To factor the expression 2x²+13x+40, you can try the method you mentioned, which involves finding two factors of the constant term (40) that add up to the coefficient of the middle term (13). However, in this case, you correctly realized that there are no such factors.

Another approach is to use the quadratic formula:

For the quadratic equation ax²+bx+c=0, the solutions can be found using the quadratic formula:

x = (-b ± √(b²-4ac))/(2a)

In the case of 2x²+13x+40, a = 2, b = 13, and c = 40. Substituting these values into the quadratic formula, we get:

x = (-13 ± √(13²-4(2)(40)))/(2(2))

Simplifying further, we have:

x = (-13 ± √(169-320))/(4)

x = (-13 ± √(-151))/(4)

Since the discriminant, which is the expression under the square root, is negative (-151), there are no real solutions to this quadratic equation, and therefore, the expression cannot be factored using real numbers.

It is also worth noting that even if the discriminant had been positive, it does not necessarily mean that the expression can be factored nicely. Some quadratic expressions might require more advanced methods or could be prime/non-factorable.