# Trinomials that are set equal to zero

ko so i have the problem y^2+13y+40=0
what would i do here? i have checked most math sites and they dont tell me what do do if its equal to zero

IF its equal to zero does that change anything or what?

This is a quadratic equation. That means it can be written in the form y2+13y+40 = (y-a)(y-b) = 0. Only if the right hand side is zero do you know that either y-a=0 or y-b=0, so y=a or y=b. In order to find a and b, you must know how to solve quadratic equations.

Think of it as ax^2 + bx + c = 0.

Now, since a = 1, what adds up to b, but when multiplied gives c?
You might also look up the Foil Method.
http://www.algebrahelp.com/lessons/factoring/trinomial/

While it is true that quadratic with a discriminant greater or equal to zero may be written as the product of two binomials (well, I'm not certain, but sometimes it seems to require a coefficient), factoring can sometimes be very difficult.

This case is easy to factor, but if you come across one that you can't get right, try the following.

Suppose you have a trinomial, ax^2+bx+c=0

To solve use x =( -b + sqrt(b^2 -4ac)) / 2a to find one answer.
Sorry, let me try that with LaTex.

$$\ x = (-b\pm \sqrt{b^2 -4ac})/2a$$

Hmm ... the x isn't showing up, but oh well.

Last edited:
HallsofIvy
Homework Helper
ko so i have the problem y^2+13y+40=0
No, you don't- and that's the problem!

what would i do here? i have checked most math sites and they dont tell me what do do if its equal to zero

IF its equal to zero does that change anything or what?
What you do with an equation depends upon the question asked or what you are told to do with it. It's the question or the instruction that is the "problem", not the equation.

If, for example, you are asked to solve the equation, then you might try to factor the polynomial on the left side. In this particular case, y2+ 13y+ 40= (y+ 8)(y+ 5). If the product of two numbers (here y+ 8 and y+ 5) is equal to 0, one of them must be equal to 0. From that you know that either y+ 8= 0 (in which case y= -8) or y+ 5= 0 (in which case y= -5).

But it might be that the "problem" is just to factor the polynomial. It might be that the problem is to find all values of x for which the equation is NOT true! (All numbers except -8 and -5.)

Don't focus on the equation so much you don't determine what the problem is!

(If my English teacher were to tell me "Jimmy is going to the swimming pool", and then look at me expectantly, I might look back quizzically and then say "Okaaay, what do you want me to do about it!")