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factoring 256x^8 - 1 completely
- Thread starter jennal
- Start date
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wantedcriminal03
Guest
Don't know if this is right or not, but I randomly get:
\(\displaystyle 256x^8 - 1\)
\(\displaystyle =16^2(x^8)-1\)
\(\displaystyle =(16x^4)^2-1^2\)
\(\displaystyle y^2-x^2=(y-x)(y+x)\)
\(\displaystyle =(16x^4+1)(16x^4-1)\)
\(\displaystyle 256x^8 - 1\)
\(\displaystyle =16^2(x^8)-1\)
\(\displaystyle =(16x^4)^2-1^2\)
\(\displaystyle y^2-x^2=(y-x)(y+x)\)
\(\displaystyle =(16x^4+1)(16x^4-1)\)
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wantedcriminal03
Guest
Oh yeh! Thanks for the finish Denis :wink: