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Let $f(x)=x^5+a_1x^4+a_2x^3+a_3x^2$ be a polynomial function. If $f(1)<0$ and $f(-1)>0$. Then

  1. $f$ has at least $3$ real zeroes
  2. $f$ has at most $3$ real zeroes
  3. $f$ has at most $1$ real zero
  4. All zeroes are real

My attempt:-

From the given condition. we get

  1. $f(-1)>0 \implies a_1-a_2+a_3>1$

  2. $f(1)<0 \implies a_1+a_2+a_3<-1$

  3. By intermediate theorem, $f$ has at least a zero in $[-1,1]$ $f(0)=0\implies 0$ is a zero of $f(x).$

How do I draw conclusion from this?

Unknown x
  • 703

1 Answers1

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We know that $$\lim_{x\to\infty}f(x)=\infty$$

and

$$\lim_{x\to-\infty}f(x)=-\infty$$

Therefore there are constants $C_-,C_+$ with $C_-<-1$, $C_+>1$ and so that $F(C_-)<0$, $F(C_+)>0$.

Apply intermediate value theorem on $[-C_-,-1]$, $[-1,1]$, $[1,C_+]$ to get at least 3 real zeroes.