General

Why do only high school freshman ask questions here.

@LowWillpower: Thank you very much!

x^3 + 5x^2 + 2x + 4 = x^3 + bx^2 + cx + d
0x^3 + (5-b)x^2 + (2-c)x + (4-d) = 0

Clearly, the equation is only satisfied (ie. equals 0) when b=5, c=2, and d=4.

[quote=MyTiramisu]But just curious when we solve an equation like x + 5 = 2x + 4 , the coefficients aren't equal to each other, but the equation holds if x = 1
Then I thought to myself, the coefficients are equal, because in my equation x does not hold a specific value. Just trying to get my head around the logic here[/quote]
Those are quite different scenarios.

In the case of setting 2 equations equal and solving for x, you are assuming both functions are of the form f(x) = x + 5 and f(x) = 2x + 4, if you decide to solve for when they are equal, you are really saying for some x, f(x) is equal.

Essentially to solve a system of equations your number of variables must match unknowns. In the case of putting two functions equal to each other, you equate the f(x) to get only the variable x, solve for x, then substitute x to solve for f(x).

In the case of the two functions you have equal, you were not asked to solve for x, so we are assuming this must be true for all values of x. For that to be the case all you really need to notice is that x^3 terms, x^2 terms, x terms and constants must have the same coefficients in both equations, unless you made your a b c and d depend on x, which is just annoying.

[quote=NaturalTEARS]yes! you're correct.
it's because LHS must always equal to RHS[/quote]

But just curious when we solve an equation like x + 5 = 2x + 4 , the coefficients aren't equal to each other, but the equation holds if x = 1
Then I thought to myself, the coefficients are equal, because in my equation x does not hold a specific value. Just trying to get my head around the logic here

yes! you're correct.
it's because LHS must always equal to RHS