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Expert-verified Found in: Page 216 ### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280 # (a) Show that a polynomial of degree $$3$$ has at most three real roots.(b) Show that a polynomial of degree $$n$$ has at most three real roots.

(a) The given equation has at most three real roots.

(b) A polynomial of degree $$n$$ has at most $$n$$ real roots.

See the step by step solution

## Step 1: Given data

A polynomial of degree $$3$$ and $$n$$.

## Step 2: Concept of Rolle’s Theorem

“Let a function $$f$$ satisfies the following conditions,

1. A function $$f$$ is continuous on the closed interval $$(a\;,\;b)$$.

2. A function $$f$$ is differentiable on the open interval $$(a\;,\;b)$$.

3. $$f(a) = f(b)$$

Then, there is a number $$c$$ in open interval $$(a\;,\;b)$$ such that $${f^\prime }(c) = 0$$.”

## Step 3: Calculation to show that polynomial of degree $$3$$ has at most three real roots

(a)

Let the polynomial of degree $$3$$ as $$P(x)$$.

Suppose, $$P(x)$$ has four distinct real roots and $${b_1},{b_2},{b_3}$$ and $${b_4}$$ where. $${b_1} < {b_2} < {b_3} < {b_4}$$.

Then, $$f\left( {{b_1}} \right) = f\left( {{b_2}} \right) = f\left( {{b_3}} \right) = f\left( {{b_4}} \right) = 0.$$

Since the polynomial is continuous on $$(a,b)$$ and differentiable on $$(a,b)$$, then Rolle's Theorem implies that there is a number $${c_1},{c_2}$$ and $${c_3}$$ with $${b_1} < {c_1} < {b_2},{b_2} < {c_2} < {b_3}$$ and $${b_3} < {c_3} < {b_4}$$ in $$(a,b)$$ such that $${P^\prime }\left( {{c_1}} \right) = {P^\prime }\left( {{c_2}} \right) = {P^\prime }\left( {{c_3}} \right) = 0$$.

So, $${P^\prime }(x) = 0$$ must have at least three real solutions.

Thus, the derivative of $$P(x)$$ is a polynomial with degree 2.

It must have two roots but this contradicts the fact that $${f^\prime }(x) = 0$$ have at least three real solutions.

Thus, it can be concluded that $$P(x)$$ cannot have four real roots.

Hence, the given equation has at most three real roots.

## Step 4: Calculation to show that polynomial of degree $$n$$ has at most three real roots

(b)

Let the polynomial with degree $$n$$ be $$P(x)$$.

Prove this by induction method.

When $$n = 1$$, it is obvious that a polynomial of degree 1, has at most one root.

When $$n = k$$, assume that a polynomial of degree $$k$$ has at most $$k$$ roots.

When $$n = k + 1$$, to prove a polynomial of degree $$k + 1$$ has at most $$k + 1$$ roots.

Let the polynomial $$P(x)$$ with degree $$k + 1$$.

It is clear that $${P^\prime }(x)$$ is the derivative of $$P(x)$$ has degree $$k$$.

Assume that $$P(x)$$ has $$k + 2$$ distinct real roots and let the roots are $${a_1},{a_2}, \ldots ,{a_{k + 2}}$$ then, $$P\left( {{a_1}} \right) = P\left( {{a_2}} \right) = \cdots = P\left( {{a_{k + 1}}} \right) = P\left( {{a_{k + 2}}} \right) = 0$$.

Since the polynomial is continuous on $$(a,b)$$ and differentiable on $$(a,b)$$, then Rolle's Theorem implies that there is a number $${c_1},{c_2}, \ldots ,{c_{n + 1}}$$ with $${a_1} < {c_1} < {a_2},{a_2} < {c_2} < \cdots < {a_{k + 1}} < {c_{k + 1}} < {a_{k + 2}}$$ in $$(a,b)$$ such that $${P^\prime }\left( {{c_1}} \right) = {P^\prime }\left( {{c_2}} \right) = \cdots = {P^\prime }\left( {{c_{k + 1}}} \right) = {P^\prime }\left( {{c_{k + 2}}} \right) = 0$$.

So, $${P^\prime }(x) = 0$$ must have at least $$k + 2$$ real solutions.

Thus, the derivative of $$P(x)$$ is a polynomial with degree $$k + 1$$.

It must have $$k + 1$$ roots but this contradicts the fact that $${f^\prime }(x) = 0$$ must have at least $$k + 2$$ real solutions.

Thus, it can be concluded that the given equation cannot have at least $$k + 2$$ real roots rather it has at least $$k + 1$$ real roots.

Hence, a polynomial of degree $$n$$ has at most $$n$$ real roots. ### Want to see more solutions like these? 