Let $P(x) = x^4 + 4x^3 − 8x^2 − 1$. Which of the following is false?
(A) $P(x)$ has a real root in $(−4, 1)$
(B) $P(x)$ has a real root $< −4$
(C) $P(x)$ has a real root $> 1$
(D) $P(x)$ has at least two real roots
What I had learned so far was to find the type of roots for three degree polynomials. Will the method of finding the type of roots for four degree polynomials be same?
On
By Descarte's rule $f(x)=0$ has atmost one real positive root as their is only one sign change in $f(x).$ The number of sign changes in $f(-x)$ is one so, this equation has atmost one negative root. Thus this equation has at most two real roots.
$f'(x)=4x(x^2+3x-4)=0$ has roots as $x=0,1,-4$. Next, $f''(x)=12x^2+24x$ implies that $f''(0)=0, f''(1)>0, f''(-4)>0$ because of change in the sign of $f'(x)$ around zero from positive to negative, there is a max at $x=0$. $f_{max}=f(0)=-1$ and $f_{min}=f(-4)<0$ , $f_{min}=f(1)<0.$ There are only two real roots. Further, as $f(0)=-1<0, f(2)>0$ there is one real root in $(0,2)$, Next, f(-4)<0 and f(-6)>0, there is another real root in $(-5,-6).$
Since $f(x)$ has negative min at $x=-4$, so one real root can be asserted to be $<-4$. Similarly, $f(x)$ has negative min at $x=1$, $f(x)$ can be asserted to have a real rot for $x>1$.
So options (B) and (C) are correct and the equation has exactly two real roots, so the statement (D) is also true.
Since the coefficient of $x^4$ is positive, we now that $$\lim_{x\to\pm\infty} P(x) = +\infty > 0. $$ Furthermore, \begin{align*} P(-4) &= -129 < 0,\\ P(1) &= -4 < 0. \end{align*} Hence, $P$ must change sign and hence have zeroes both in $(-\infty, -4)$ and $(1,+\infty)$, so B), C) and D) are true.
Since the question implies one of the statements is false, it has to A).
If you want to verify that A) is false, see if $P$ has a positive local maximum in $(-4,1)$ using the zeroes of the degree three polynomial $P'$.