Algebra Help Please?

Question

Factor completly
2x^2 - x - 10
Can you show me how to work out the problem. Thanks.

Answer 1

2x^2 - x - 10
=(x+2)(2x-5)

Hint:
use trial and error method and distributive property.

Answer 2

2x^2 - x - 10
You want to find two factors that multiply to be -20 and add up to be -1. These numbers are -5 and 4.
2x^2 + 4x - 5x - 10 (which says the same thing as the original equation)
You can group the equation in two different parts: (2x^2 + 4x) -(5x+10)
With the first group, factor out 2x -> 2x(x+2)
With the second group, factor out 5 -> 5(x+2)
The new equation is 2x(x+2)-5(x+2)
You can now write the equation as (2x-5)(x+2)

Answer 3

Find the possible factors first, of the the bit with x square, and the bit that's got no x.
2x square has two factors:2x and x.
-10 has -5 and 2
cross multiply: 2x times 2=4x
x times -5= -5x
Make sure the two end-products combined become the bit with only x, that is -x, or -1(x).
4x+(-5)x = -x
Therefore, you can take the cross multiplied bits, switch them, and you have 2x-5 and x+2.
And you get: (2x-5)(x+2)

Answer 4

we hav
2x^2-x-10
so by splitting the middle term we hav,
2x^2+4x-5x-10
(2x^2+4x)(-5x-10)
2x(x+2)-5(x+2)
(x+2)(2x-5)

hope its clear!

Answer 5

2x^2 - x - 10
Factorising :
(2x - 5)(x + 2)