Question

Simplify (2x^2+x-1)

Answer 1

It is already simplified.

Answer 2

-1 + x + 2/(3 + x)

Answer 3

sorry I meant 2x^2+x-1 / x^2 + 4x +3

Answer 4

\[\frac{(2x^2+x-1)}{(x^2+4x+3)} = \frac{(2x^2+x-1)}{(x+1)(x+3)}\] by noticing that 1*3 = 3 and 1+3 = 4. Now we factor the top to see if we can get a matching term up there which will cancel out. We need \[(2x+a)(x+b)=(2x^2+x-1)\] for some values of a and b. If we multiply that out\[2x^2+2bx+ax+ab = 2x^2+x-1\] we see that ab=-1 and 2b+a = 1. Solutions for that are b=1, a =-1, so our factoring of the top is \[(2x-1)(x+1) = 2x^2+2x-x-1=2x^2+x-1\] and our fraction becomes
\[\frac{(2x^2+x-1)}{(x+1)(x+3)}=\frac{(2x-1)(x+1)}{(x+1)(x+3)} = \frac{2x-1}{x+3}\]

Answer 5

and we'll have a zero at x = 1/2 and a pole at x = -3 if this is a transfer function.