I Quartic real roots - factor part into quadratic

  • Thread starter binbagsss
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If I have ##f(x)=x^4+(x+2)(x+1)##

basically a quartic without a cubic term for which it can be written as above : ##x^3## + some quadratic which has discrimant ##\geq 0 ##, so that the quadratic has real roots, can one ocnclude that ##f(x)## has real roots too?

thanks
 

Math_QED

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##f(x) = x^4 + (x+1)(x+1)##

has no real roots, since ##f > 0## everywhere (because ##f(x) = x^4 + (x+1)^2 \geq 0)##).

So, no, in general you can't conclude that.
 
1,155
6
##f(x) = x^4 + (x+1)(x+1)##

has no real roots, since ##f > 0## everywhere (because ##f(x) = x^4 + (x+1)^2 \geq 0)##).

So, no, in general you can't conclude that.
how about ##x^4-(x+1)(x+1)## ?
changing the sign to minus a quadratic?
 

mathman

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Did you mean ##(x+1)(x+2)## or ##(x+1)^2## for the last part?
 

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