Easy Problem

Algebra Level 2

How many solutions (Real or complex) does the equation:- $x^{3} = 1$ have:-

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The answer is 3.

2 solutions

Paola Ramírez
Jan 22, 2015

$x^3=1$

$x^3-1=0$

$(x-1)(x^2+x+1)$

Have $\boxed{3}$ solution: one real by $x-1$ and two complex by $x^2+x+1$

Mardokay Mosazghi
Aug 24, 2014

By the fundamental theorem of algebra Any polynomial of degree n ... has n roots so degree 3 has 3 roots including complex.

What about f(x)=0?

John M. - 6 years, 9 months ago

i don't understand your comment

Mardokay Mosazghi - 6 years, 9 months ago

You said "...any polynomial of degree n ... has n roots..."

Polynomial f(x)=0 has a degree of 0 ( $f(x)=0x^{0}$ ), and yet it has infinitely many solutions. And so, What about f(x)=0?

John M. - 6 years, 9 months ago

@John M. – The fundamental Theorem of Algebra requires the polynomial to be non-constant.

Daniel Liu - 6 years, 9 months ago

@John M. – In case of f(x)=0 you can't find its.. degree because it can also be written as $f(x)=0*x^{n}$ where n can take any value, Since f(x) have infinite many degrees so its has infinite many solutions

Sorry for my bad english

Aman Sharma - 6 years, 9 months ago

Easy Problem

The answer is 3.

2 solutions

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