Question

(a) Show that a polynomial of degree 3 has at most three real zeroes.

(b) Show that a polynomial of degree \(n\) has at most \(n\) real zeroes.

Short answer

1. It is proved that thepolynomial of degree 3has at most three real zeroes.
2. It is proved that the polynomial of degree \(n\) has at most \(n\) real zeroes.

Step-by-step solution

Rolle’s Theorem

Rolle’s theorem states that for any function \(f\left( x \right)\), there exists a number \(r\) in a domain \(\left( {a,b} \right)\) such that:

\({\rm{for r}} \in \left( {a,b} \right) \to f'\left( r \right) = 0\).

(a) Step 2: Proving by contradiction method

The givenpolynomial is of degree 3.

Let, \(P\left( x \right)\)be a cubic polynomial having zeroes \(a,\,\,b,\,\,c{\rm{ and d}}\,\,\left( {a < b < c < d} \right)\).

Then,

\(P\left( a \right) = P\left( b \right) = P\left( c \right) = P\left( d \right) = 0\)

Now, according to Rolle’s Theorem, there must be three numbers \({r_1}{\rm{,}}\,\,{r_2}{\rm{ and }}{r_3}\) such that:

\(P'\left( {{r_1}} \right) = P'\left( {{r_2}} \right) = P'\left( {{r_3}} \right) = 0\)

But, the derivative of cubic will be a 2-degree polynomial which is giving three real zeroes. This can never be possible. This contradicts the function is having four distinct real solutions.

Hence proved, the polynomial of degree 3 has at most three real zeroes.

(b) Step 3: Proving by contradiction method

The givenpolynomial is of degree n.

Let,\(P\left( x \right)\)be a polynomial of degree\(n\)having\(n + 1\)zeroes such that:\({a_1} < {a_2} < .......... < {a_{n + 1}} < {a_{n + 2}}\).

Then,

\(P\left( {{a_1}} \right) = P\left( {{a_2}} \right) = ........ = P\left( {{a_{n + 2}}} \right) = 0\)

Now, according to Rolle’s Theorem, there must be \(n + 1\) numbers \({r_1}{\rm{,}}\,\,{r_2}{\rm{, }}...........{\rm{, }}{r_{n + 1}}\) such that:

\(P'\left( {{r_1}} \right) = P'\left( {{r_2}} \right) = ........... = P'\left( {{r_{n + 1}}} \right) = 0\)

Here, the derivative of \(n\)-degree polynomial which is giving \(n + 1\) real zeroes. This can never be possible.

As this contradictsthe polynomial is having\(n + 1\)distinct real solutions.

Hence proved, the polynomial of degree \(n\) has at most \(n\) real zeroes.