(2x-1) is a factor of the polynomial 6x^6 + x^5 - 92x^4 + 45x^3 + 184x^2 +4x -48? why?

Answer 1

You could do long division, but that is a pain. What does it mean to be a factor? To have p(x) be a factor of q(x), it means that q(x) = p(x)*f(x) where f(x) is some leftover stuff. But, more importantly, if p(x) = 0, then q(x) = 0.

So taking q(x) = 6x^6 + x^5 - 92x^4 + 45x^3 + 184x^2 +4x -48 and p(x) = 2x-1 then all you need to do is show that q(x) = 0 whenever p(x) = 0. But p(x) = 0 when x = 1/2.

So, to make a long story short, all you need to do is show that q(1/2) = 0 (I'll let you do the calculation).

Answer 2

2x-1

6 1 -92 45 184 4 -48

3 | 4 - 92 45 184 4 -48
3 2 | -90 45 184 4 -48
3 2 -45 | 0 184 4 -48
3 2 -45 0 | 184 4 -48
3 2 -45 0 92 | 96 -48
3 2 -45 0 92 48 | 0

yes, (2x-1) is a factor of 6x^6 + x^5 - 92x^4 + 45x^3 + 184x^2 +4x -48 because it divides into the equation without a remainder