00:01
Let's use a partial fraction decomposition for this one.
00:05
So first, before we do that, we should check if we can factor the denominator.
00:12
And in this case, you can pull out of x left over with x squared plus 3.
00:20
Now i would also have to check if this quadratic here in parentheses, if that factors.
00:28
So you can try to factor this, but this one doesn't factor.
00:32
And we get this by looking at the discriminant b squared minus 4 a c and in this problem there's no x term in the quadratic so b is 0 and then minus 4 times the coefficient in front of x squared is 1 and then times 3 which equals c and this is a negative number so that means that this quadratic will not factor so using case 1 for the linear factor this is what the author calls case one non -repeated linear factor and then here this is what the author calls case 3 because it's irreducible quadratic so in the numerator we need bx plus c this time and then that denominator x squared plus 3 so now let's multiply both sides of this new equation by that denominator on the left so on the left side we have 2x minus 3 but on the right we we have that and then we can go ahead and simplify this.
01:58
So let's pull out of x squared.
02:04
And we could pull out a plus b.
02:06
And then we have cx and then we have a 3a.
02:14
So i just combine these terms by factoring out some power of x.
02:19
So here x squared, then x, then the constant term...