x(2x-5)-(2x+5)

Question

Is it possible to factor any further?

Answer 1

Jesus,

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.

Answer 2

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:  x_1,2=[-b-±√(b²-4ac)]/(2a).

Answer 3

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.