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Factoring
- Thread starter beth56374
- Start date
D
Deleted member 4993
Guest
beth56374 said:I have been trying to figure out how to factor this equation: x^(2n-1)-2x^(n+1)+x
I have been lost on where to even begin... any help is much appreciated!!!
x^(2n-1)-2x^(n+1)+x
\(\displaystyle x^{2n-1} - 2x^{n+1} + x\)
Are you sure that the first power is (2n -1) and not (2n+1)
D
Deleted member 4993
Guest
No wonder!!
x^(2n+1)-2x^(n+1)+x[/quote]
\(\displaystyle x^{2n+1} - 2x^{n+1} + x\)
\(\displaystyle = x \cdot (x^{2n} - 2x^{n} + 1)\)
substitute
u = x[sup:fk8w0pk0]2n[/sup:fk8w0pk0]
\(\displaystyle x \cdot (x^{2n} - 2x^{n} + 1)\)
\(\displaystyle = x \cdot (u^2 - 2u + 1)\)
Now continue...
x^(2n+1)-2x^(n+1)+x[/quote]
\(\displaystyle x^{2n+1} - 2x^{n+1} + x\)
\(\displaystyle = x \cdot (x^{2n} - 2x^{n} + 1)\)
substitute
u = x[sup:fk8w0pk0]2n[/sup:fk8w0pk0]
\(\displaystyle x \cdot (x^{2n} - 2x^{n} + 1)\)
\(\displaystyle = x \cdot (u^2 - 2u + 1)\)
Now continue...