00:14
So because the term cannot be rational decomposition, so we'll have a equals to x plus 1, and we have bx plus c over x square from what we know on the book.
00:35
So now we times the denominator on both sides, we have x square minus 2, just equals to a x squared plus x plus 1 and bx plus c times x plus 1 so that is a x square plus a x plus a plus b x square plus b plus c x plus c so we can see the coefficient of x squared, that is 1 on the left, and on the red we have a square, bx square.
01:28
So we have a plus b, and the coefficient of x, that is a 0 on the left, and on the red we have a and b plus x, that is a plus b, plus c, and the constant minus 2 equals to a plus c.
01:54
So from this one, we have a equals to 1 minus b, and we substitute that in this formula, we have c plus 1 equals to 0...