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Is it possible to factor any further?

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Henry C. | Seasoned Researcher/TechnicianSeasoned Researcher/Technician
5.0 5.0 (56 lesson ratings) (56)


With what level of Algebra are you working? I mean what grade and text book are you using?

The previous two solutions will provide you with the binomial factors for the equation ax^2 + bx + c = 0, but your question is concerned only with factoring an expression. I recall these sort of exercises were used often, in basic mathematics courses by the Hake Saxon style texts, to provide experience with the associative and distributive properties of some algebraic structures.

If I am not mistaken with the assumption that this is middle school algebra, the answer is no. Because there are no more common factors among any of the terms. Notice that (2x-5) is not equal to (2x+5), also that x is clearly not equal to -1.

Kirill Z. | Physics, math tutor with great knowledge and teaching skillsPhysics, math tutor with great knowledge...
4.9 4.9 (174 lesson ratings) (174)

This expression hasn't been factored yet. Jaison is right: collect similar terms, you will end up with 2x2-7x-5. Solve quadratic equation 2x2-7x-5=0, then 2x2-7x-5 is factored as 2(x-x1)(x-x2), where x1 and x2 are the solutions.

Quadratic formula for ax2+bx+c=0:  x1,2=[-b-±√(b2-4ac)]/(2a).

Jaison N. | A PhD to Teach You Math and PhysicsA PhD to Teach You Math and Physics
4.9 4.9 (150 lesson ratings) (150)

Nope, distribute the x and -1, collect like terms together, and then plug everything into the quadratic formula. 

Remember one thing about factoring quadratics: if you're stumped, the quadratic formula is always true.


You're assuming that the expression is associated with a homogeneous system, i.e. equal to zero. The question doesn't mention that.

Sorry, but he is not. What this expression equals to has nothing to do with its factorizability, if I may call it this way.

It's a safe assumption. Most of the students here on Wyzant are high school students so we can safely assume that what Jesus wrote is the whole problem. I'm 33 so my memories of high school are a bit fuzzy but I don't recall ever being exposed to using the term homogeneous in a math class until Diff Eq in college.

Now I could assume that the quadratic is equal to an arbitrary constant C and find a pair of roots parameterized by C. But that may confuse the student.

Kirill: I saw you profile and noticed that you went to Moscow Institute of Physics and Technology. I'm pretty sure that one of my best friends did his BS and MS there as well. I'm seeing Ivan today so I'll check. Anyway we can email over Wyzant to see if we have a mutual acquaintance?

Sure, Jaison. By the way, if, say, x2+3x+2=c then we can factor left side any time we want to get (x+1)(x+2)=c. This is what I meant when I said to Henry that you need not assume it equals to zero.

I appreciate what you meant Kirill.  My point was merely that one does not need to assume anything about the expressions' relationship with any other function, because the question only asks if the expression may be factored any further.

To demonstrate the symmetry with your example: Can the expression x(x+2)+(x+2) be factored any further....

I also see your point, Henry. It was interesting to look at things that way, truly. Thanks.