Factor Theorem & Irrational roots

[jacktosh]

Does the factor theorem only work with rational roots since rational roots can be solved for zero? I'm thinking that the factor theorem does not work with irrational roots since an irrational number is non-terminating & never ends so it can never truly equal zero. Is this correct? Thanks.

 

[Steven G]

The rational root theorem only finds rational roots.
If I have x^2 -10, I can use the factor theorem to see that the zeros are +/- sqrt(10) so the factor theorem can yield irrational roots (as well as complex roots)

Just to be sure can you please state the factor theorem? The more I think about your question the more I think that you might not know the correct definition.

 

[pka]

Does the factor theorem only work with rational roots since rational roots can be solved for zero? I'm thinking that the factor theorem does not work with irrational roots since an irrational number is non-terminating & never ends so it can never truly equal zero. Is this correct?

Consider \(\displaystyle x^2-2=0\\(x-\sqrt{2})(x+\sqrt{2})=0\\\text{The roots are }x=\pm\sqrt{2}\\\text{the factors are }(x-\sqrt{2})~\&~(x+\sqrt{2})\)

 

[jacktosh]

Excellent example. I totally get it. I didn't think about complex zeros but now I can see that too--say if I had something like x^2+1. If I ran "i" thru this equation I'd get i^2 +1 ... or -1+1 = 0. Thus by the factor thm the complex number "i" is a root. Thanks Jomo.

 

[Steven G]

Excellent example. I totally get it. I didn't think about complex zeros but now I can see that too--say if I had something like x^2+1. If I ran "i" thru this equation I'd get i^2 +1 ... or -1+1 = 0. Thus by the factor thm the complex number "i" is a root. Thanks Jomo.

...and since complex roots come in pairs we also have -i as a root.