Maths qns (algebra)
-
By Euclidean Algorithm, for a map f: R -> R[x], x --> 2x^2 - 3x - 5, there exist g such that f = (x - a)g + r where a is a root, deg r < deg g.Originally posted by launtpc:Nooo! Not solve, because behind the algebraic equation there is no "=0" so we cannot assume that.
We can only factorise.
x(2x-3)-5
Or can further factorise?
It can be further proved that in this case that r = 0 since we know that f(x) = 2x^2 - 3x - 5 = (x + 1)(2x - 5) where a = 1, g = (2x - 5) and deg r = - infinity which is less than deg g = 1. Therefore the condition is satisfied and thus the factorization holds.
Though the "let f(x) = 0 and then solve" is not fundamentally correct but what's most important at this particular level is just to get 2x^2 - 3x - 5 = (x + 1)(2x - 5). It is definitely acceptable at this level.
Hope I didn't baffle anyone. Thanks for letting me have a small algebra revision.
-
Fuih... so many maths expert here... potential actuarists
-
Originally posted by CoolMyth:By Euclidean Algorithm, for a map f: R -> R[x], x --> 2x^2 - 3x - 5, there exist g such that f = (x - a)g + r where a is a root, deg r < deg g.
It can be further proved that in this case that r = 0 since we know that f(x) = 2x^2 - 3x - 5 = (x + 1)(2x - 5) where a = 1, g = (2x - 5) and deg r = - infinity which is less than deg g = 1. Therefore the condition is satisfied and thus the factorization holds.
Though the "let f(x) = 0 and then solve" is not fundamentally correct but what's most important at this particular level is just to get 2x^2 - 3x - 5 = (x + 1)(2x - 5). It is definitely acceptable at this level.
Hope I didn't baffle anyone. Thanks for letting me have a small algebra revision. 
gong simi? lolz.. =p -
Maths lor.Originally posted by papercut87: gong simi? lolz.. =p -
as i said earlier, mentally can liao.
-
actually F(x)=0 is a boundary condition. Those studying Engineering should know this.
If we say F(x)=2x^2+3x-2, at most we can factorise only BUT cannot solve. Because we dont have the boundary condition. If FACT, F(x)=2x^2+3x-2 can have an INFINITE number of solutions, where F(x)=0 or F(x)=5, or F(x)=-1, etc... and the roots of the equation is normally DEFINED as when the equation of the curve CUTs the X-AXIS, i.e. y=f(x)=0, therefore, that is the boundary condition and therefore WHEN THE QUESTION STATES find the ROOTS of the equation, you can equate it equal to ZERO.
hope that clears everything up.