# Math Help - Is this correct factoring?

1. ## Is this correct factoring?

Find all integers b for which a2 + ba – 50 can be factored.
Factors of – 50 = (- 50, 1), (50 – 1), (- 25, 1), (25, 1), (- 10, 1), (10, -1), (- 5,1), (5,-1), (-2,1),(2,-1)
(a – 50)(a + 1) = a2 – 50a + a – 50 = a2 – 49a – 50
(a + 50)(a – 1 ) = a2 + 50a - a + 50 = a2 + 49a – 50
(a – 25)(a + 1) = a2 + a - 25a – 25 = a2 – 24a – 25
(a + 25)(a - 1) = a2 - a + 25a – 25 = a2 + 24a – 25
(a – 10)(a + 1) = a2 + a - 10a – 10 = a2 – 9a – 10
(a + 10)(a - 1) = a2 - a + 10a – 10 = a2 + 9a – 10
(a – 5)(a + 1) = a2 + a - 5a – 5 = a2 – 4a – 5
(a + 5)(a - 1) = a2 - a + 5a – 5 = a2 + 4a – 5
(a – 2)(a + 1) = a2 + a - 2a – 2 = a2 – a – 2
(a + 2)(a - 1) = a2 - a + 2a – 2 = a2 + a – 2

2. Originally Posted by mcruz65
Find all integers b for which a2 + ba – 50 can be factored.
Factors of – 50 = (- 50, 1), (50 – 1), (- 25, 1), (25, 1), (- 10, 1), (10, -1), (- 5,1), (5,-1), (-2,1),(2,-1)
(a – 50)(a + 1) = a2 – 50a + a – 50 = a2 – 49a – 50
(a + 50)(a – 1 ) = a2 + 50a - a + 50 = a2 + 49a – 50
(a – 25)(a + 1) = a2 + a - 25a – 25 = a2 – 24a – 25
(a + 25)(a - 1) = a2 - a + 25a – 25 = a2 + 24a – 25
(a – 10)(a + 1) = a2 + a - 10a – 10 = a2 – 9a – 10
(a + 10)(a - 1) = a2 - a + 10a – 10 = a2 + 9a – 10
(a – 5)(a + 1) = a2 + a - 5a – 5 = a2 – 4a – 5
(a + 5)(a - 1) = a2 - a + 5a – 5 = a2 + 4a – 5
(a – 2)(a + 1) = a2 + a - 2a – 2 = a2 – a – 2
(a + 2)(a - 1) = a2 - a + 2a – 2 = a2 + a – 2
1. Explain the argument behind this

CB

3. ## Am I approaching this problem correctly?

The problem is the following:

Find all integers b for which a^2 + ba – 50 can be factored.

Am I solving this in the proper fashion?

4. Originally Posted by mcruz65
The problem is the following:

Find all integers b for which a^2 + ba – 50 can be factored.

Am I solving this in the proper fashion?
Factored over what? All quadratics can be factored over the complex numbers, so its unlikely to be that. Factored over the reals? Over the integers or rationals?

The rational roots theorem tells you that if this quadratic has a factorisation with linear factors with only integer coefficients (which is equivalent to asking for factorisation over the rationals) then those factors correspond to roots:

$\pm 1, \ \pm 2,\ \pm 5,\ \pm 10,\ \pm 25,\pm 50$

this is the approach I believe you have adopted, and it is a workable approach for the problem. But you have to say what you are doing not just slap down an undigested bunch of expressions.

CB