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Chapter 2: Problem 27

Find a polynomial \(p\) of degree 3 such that \(-1,2,\) and 3 are zeros of \(p\) and \(p(0)=1\).

Short Answer

Expert verified
The polynomial of degree 3 with zeros at \(-1, 2,\) and \(3\), and satisfying the condition \(p(0) = 1\) is given by \(p(x) = \frac{1}{6}(x^3 - 4x^2 + x + 6)\).

Step by step solution

01

Write the general form of the polynomial

Since the polynomial has -1, 2 and 3 as its zeros, we can write it in its general factored form: \(p(x) = A(x + 1)(x - 2)(x - 3)\), where A is a constant.
02

Find the constant A using the given condition

We are given that \(p(0)=1\). We will substitute \(x=0\) in the equation and solve for A: \(p(0) = A(0 + 1)(0 - 2)(0 - 3) = 1\) By calculating it we find \((1)(-2)(-3) = 6\) Therefore, A should be \(\frac{1}{6}\) to satisfy the condition \(p(0)=1\): \(A = \frac{1}{6}\)
03

Write the polynomial using the value of A

Now, substitute the value of A back into the polynomial equation: \(p(x) = \frac{1}{6}(x + 1)(x - 2)(x - 3)\) So the polynomial we were looking for is: \(p(x) = \frac{1}{6}(x^3 - 4x^2 + x + 6)\)

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Most popular questions from this chapter

Write each expression as the sum of a polynomial and a rational function whose numerator has smaller degree than its denominator. $$ \frac{2 x+1}{x-3} $$

Suppose \(s(x)=4 x^{3}-2\) (a) Show that the point (1,2) is on the graph of \(s\). (b) Give an estimate for the slope of a line containing (1,2) and a point on the graph of \(s\) very close to (1,2) [Hint: Use the result of Exercise \(18 .]\)

Give an example of polynomials \(p\) and \(q\) of degree 3 such that \(p(1)=q(1), p(2)=q(2),\) and \(p(3)=q(3),\) but \(p(4) \neq q(4)\).

Suppose \(p(x)=2 x^{5}+5 x^{4}+2 x^{3}-1 .\) Show that -1 is the only integer zero of \(p\).

Write the indicated expression as a ratio of polynomials, assuming that $$ r(x)=\frac{3 x+4}{x^{2}+1}, \quad s(x)=\frac{x^{2}+2}{2 x-1}, \quad t(x)=\frac{5}{4 x^{3}+3} $$. $$ (s(x))^{2} t(x) $$
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