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secondary 3 | A Maths
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Date Posted: 4 years ago
Views: 199
Yes
Definitely a factor. We do this using the remainder theorem in the same way we normally do. Writing this up in a moment.
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Date Posted:
4 years ago
I’m sorry but I don’t rly understand the part where m=n so could u explain to me in further details? Thank you in advance!
Usually the variable concerned is x. Here it is m.
The usual question goes like this.
Show that x - 1 is a factor of f(x) = x2 - 1.
What you would do is this. Let x = 1 (because “x - 1 = 0”). Then the remainder is f(x) = f(1) = 1^2 - 1 = 0 (we replaced x with 1) and because there is no remainder, the expression x - 1 must be a factor of f(x).
Here we do the same thing, except that the function is not defined in x, but instead in m, such that f(m) = m5 - n5 where m is a variable and n is a constant.
When divided by m - n, we pretend m - n = 0 so that m = n, and the remainder is f(m) = f(n) = n^5 - n^5 = 0 (we replaced m with n). It follows that m - n is a factor of m5 - n5.
The usual question goes like this.
Show that x - 1 is a factor of f(x) = x2 - 1.
What you would do is this. Let x = 1 (because “x - 1 = 0”). Then the remainder is f(x) = f(1) = 1^2 - 1 = 0 (we replaced x with 1) and because there is no remainder, the expression x - 1 must be a factor of f(x).
Here we do the same thing, except that the function is not defined in x, but instead in m, such that f(m) = m5 - n5 where m is a variable and n is a constant.
When divided by m - n, we pretend m - n = 0 so that m = n, and the remainder is f(m) = f(n) = n^5 - n^5 = 0 (we replaced m with n). It follows that m - n is a factor of m5 - n5.
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