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Q21E

Expert-verifiedFound in: Page 216

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.

A polynomial of degree \(3\) and \(n\).

**“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 **\({f^\prime }(c) = 0\)**.”**

(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.

(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.

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