Question

How do you factor sin^2x + sin x -2

Answer 1

let sin(x) be N

N^2 + N - 2
(the next step is called completing the square. if you multiply everything out it is equal.
(N+0.5)^2 - 0.25 - 2
(N+0.5)^2-2.25
(N+0.5)^2-1.5^2

then sub back sin(x):
\[\sin^{2} (x) +\sin(x)-2 = (\sin(x)+\frac{ 1 }{ 2 })^{2} -(\frac{ 1 }{ 2 })^{2}\]

... well it doesnt look very simplified like that but since it reduced the sin^2(x) to a first power i guess it is correct.

Answer 2

completing the square:

say you have \[x^2 +4x + 5 = 0\] and you want it simplified. You will need to reduce the power of x to simplify it. recall how expanding brackets works:\[(x+3)^2 = (x+3)(x+3) = x^2+3x+3x+9:\]\[ (x+3)^2= x^2 + 6x + 9\]

if we are presented with a second order polynomial equation like this expanded equation then we will have to work backwards to put it back into the brackets...
inspect this and see how both sides are equal:
\[x^2 + 2x = (x+1)^2-2\]\[x^2 + 2x = x^2+x + x + 2-2\]\[x^2 + 2x = x^2+2x\]

to sum up:
to simplify \[x^2 +Kx\]
you turn it into this:
\[ (x+K/2)^2 - (K/2)^2\]
comprende?