Quartic real roots - factor part into quadratic

[binbagsss]

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

 

[member 587159]

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

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.

 

[binbagsss]

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

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]

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