In my workbook, the problem wants me to find all solutions for x, if P(x) / Q(x) is zero. P(x) = x^3 - 5x^2 + 2x + 8; Q(x) = x + 1.
So when you divide the two polynomials ( by long division ) you get x^2 - 6x + 8. Since this quadratic expression we got has to be equal to 0 ( by the formulation of the problem ), we apply the quadratic formula and get 2 solutions: 4 and 2.
HOWEVER, the official solution says there is another solution, that we get when we multiply the P(x) / Q(x) = x^2 - 6x + 8 by Q(x), then make Q(x) zero, so that ( x - 1 ) = 0 and thus x_3 = -1. How come for polynomials the rule ''you can't divide by zero'' seems to be ignored ? Expressions like:
[math]\frac{x^2 + 1}{x^2 - 1} = 1[/math]
cannot be defined for x^2 = 1 or x = +- 1 no matter what ( even if we multiply the equation by ( x^2 - 1 ) to get rid of the discomfort of having to divide by zero ). So why does it appear here ? How can it suddenly become plausible to divide by zero ?
I would appreciate any answers as this is a seemingly simple yet essential question for understanding the core of polynomials.
So when you divide the two polynomials ( by long division ) you get x^2 - 6x + 8. Since this quadratic expression we got has to be equal to 0 ( by the formulation of the problem ), we apply the quadratic formula and get 2 solutions: 4 and 2.
HOWEVER, the official solution says there is another solution, that we get when we multiply the P(x) / Q(x) = x^2 - 6x + 8 by Q(x), then make Q(x) zero, so that ( x - 1 ) = 0 and thus x_3 = -1. How come for polynomials the rule ''you can't divide by zero'' seems to be ignored ? Expressions like:
[math]\frac{x^2 + 1}{x^2 - 1} = 1[/math]
cannot be defined for x^2 = 1 or x = +- 1 no matter what ( even if we multiply the equation by ( x^2 - 1 ) to get rid of the discomfort of having to divide by zero ). So why does it appear here ? How can it suddenly become plausible to divide by zero ?
I would appreciate any answers as this is a seemingly simple yet essential question for understanding the core of polynomials.