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To determine whether the Rational Root Theorem provides a complete list of all roots for the given polynomial functions, we need to carefully analyze each polynomial function and identify its roots. The Rational Root Theorem states that any rational solution of a polynomial equation with integer coefficients is of the form [tex]\( \pm \frac{p}{q} \)[/tex], where [tex]\(p\)[/tex] is a factor of the constant term and [tex]\(q\)[/tex] is a factor of the leading coefficient.
Let's examine each polynomial function one by one.
1. [tex]\( f(x) = 4x^2 - 25 \)[/tex]
- The constant term is [tex]\(-25\)[/tex] and the leading coefficient is [tex]\(4\)[/tex].
- Possible rational roots are [tex]\( \pm 1, \pm 5, \pm 25\)[/tex] (factors of [tex]\(-25\)[/tex]) divided by [tex]\( \pm 1, \pm 2, \pm 4\)[/tex] (factors of [tex]\(4\)[/tex]).
- The possible rational roots to test are [tex]\( \pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm 5, \pm \frac{5}{2}, \pm \frac{5}{4}, \pm 25, \pm \frac{25}{2}, \pm \frac{25}{4}\)[/tex].
We find that [tex]\( f(x) = 4x^2 - 25 \)[/tex] has roots [tex]\( x = \pm \frac{5}{2} \)[/tex]. Since these roots are rational numbers that fit the potential rational roots derived from the Rational Root Theorem, the theorem does provide a complete list of roots.
2. [tex]\( g(x) = 4x^2 + 25 \)[/tex]
- The constant term is [tex]\(25\)[/tex] and the leading coefficient is [tex]\(4\)[/tex].
- Possible rational roots are [tex]\( \pm 1, \pm 5, \pm 25\)[/tex] (factors of [tex]\(25\)[/tex]) divided by [tex]\( \pm 1, \pm 2, \pm 4\)[/tex] (factors of [tex]\(4\)[/tex]).
- The possible rational roots to test are [tex]\( \pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm 5, \pm \frac{5}{2}, \pm \frac{5}{4}, \pm 25, \pm \frac{25}{2}, \pm \frac{25}{4}\)[/tex].
In this case, [tex]\( g(x) = 4x^2 + 25 \)[/tex] does not have real roots because the polynomial is always positive (as [tex]\( 4x^2 + 25 > 0 \)[/tex] for all [tex]\( x \in \mathbb{R} \)[/tex]). Solving this equation, we find that the roots are [tex]\( x = \pm \frac{5i}{2} \)[/tex], which are imaginary numbers. The Rational Root Theorem only deals with rational (real) roots, so it does not provide the complete list of roots for this polynomial.
3. [tex]\( h(x) = 3x^2 - 25 \)[/tex]
- The constant term is [tex]\(-25\)[/tex] and the leading coefficient is [tex]\(3\)[/tex].
- Possible rational roots are [tex]\( \pm 1, \pm 5, \pm 25\)[/tex] (factors of [tex]\(-25\)[/tex]) divided by [tex]\( \pm 1, \pm3\)[/tex] (factors of [tex]\(3\)[/tex]).
- The possible rational roots to test are [tex]\( \pm 1, \pm \frac{1}{3}, \pm 5, \pm \frac{5}{3}, \pm 25, \pm \frac{25}{3}\)[/tex].
We find that [tex]\( h(x) = 3x^2 - 25 \)[/tex] has roots [tex]\( x = \pm \frac{5\sqrt{3}}{3} \)[/tex]. These roots are not rational numbers, so the Rational Root Theorem does not provide the complete list of roots for this polynomial either.
Summary:
1. For [tex]\( f(x) = 4x^2 - 25 \)[/tex], the Rational Root Theorem provides a complete list of all roots.
2. For [tex]\( g(x) = 4x^2 + 25 \)[/tex], the Rational Root Theorem does not provide a complete list of all roots (since the roots are imaginary).
3. For [tex]\( h(x) = 3x^2 - 25 \)[/tex], the Rational Root Theorem does not provide a complete list of all roots (since the roots are irrational).
[tex]\[ \begin{array}{c|c} \text{Polynomial} & \text{Complete List by Rational Root Theorem?} \\ \hline f(x)=4 x^2-25 & \text{Yes} \\ g(x)=4 x^2+25 & \text{No} \\ h(x)=3 x^2-25 & \text{No} \\ \end{array} \][/tex]
Let's examine each polynomial function one by one.
1. [tex]\( f(x) = 4x^2 - 25 \)[/tex]
- The constant term is [tex]\(-25\)[/tex] and the leading coefficient is [tex]\(4\)[/tex].
- Possible rational roots are [tex]\( \pm 1, \pm 5, \pm 25\)[/tex] (factors of [tex]\(-25\)[/tex]) divided by [tex]\( \pm 1, \pm 2, \pm 4\)[/tex] (factors of [tex]\(4\)[/tex]).
- The possible rational roots to test are [tex]\( \pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm 5, \pm \frac{5}{2}, \pm \frac{5}{4}, \pm 25, \pm \frac{25}{2}, \pm \frac{25}{4}\)[/tex].
We find that [tex]\( f(x) = 4x^2 - 25 \)[/tex] has roots [tex]\( x = \pm \frac{5}{2} \)[/tex]. Since these roots are rational numbers that fit the potential rational roots derived from the Rational Root Theorem, the theorem does provide a complete list of roots.
2. [tex]\( g(x) = 4x^2 + 25 \)[/tex]
- The constant term is [tex]\(25\)[/tex] and the leading coefficient is [tex]\(4\)[/tex].
- Possible rational roots are [tex]\( \pm 1, \pm 5, \pm 25\)[/tex] (factors of [tex]\(25\)[/tex]) divided by [tex]\( \pm 1, \pm 2, \pm 4\)[/tex] (factors of [tex]\(4\)[/tex]).
- The possible rational roots to test are [tex]\( \pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm 5, \pm \frac{5}{2}, \pm \frac{5}{4}, \pm 25, \pm \frac{25}{2}, \pm \frac{25}{4}\)[/tex].
In this case, [tex]\( g(x) = 4x^2 + 25 \)[/tex] does not have real roots because the polynomial is always positive (as [tex]\( 4x^2 + 25 > 0 \)[/tex] for all [tex]\( x \in \mathbb{R} \)[/tex]). Solving this equation, we find that the roots are [tex]\( x = \pm \frac{5i}{2} \)[/tex], which are imaginary numbers. The Rational Root Theorem only deals with rational (real) roots, so it does not provide the complete list of roots for this polynomial.
3. [tex]\( h(x) = 3x^2 - 25 \)[/tex]
- The constant term is [tex]\(-25\)[/tex] and the leading coefficient is [tex]\(3\)[/tex].
- Possible rational roots are [tex]\( \pm 1, \pm 5, \pm 25\)[/tex] (factors of [tex]\(-25\)[/tex]) divided by [tex]\( \pm 1, \pm3\)[/tex] (factors of [tex]\(3\)[/tex]).
- The possible rational roots to test are [tex]\( \pm 1, \pm \frac{1}{3}, \pm 5, \pm \frac{5}{3}, \pm 25, \pm \frac{25}{3}\)[/tex].
We find that [tex]\( h(x) = 3x^2 - 25 \)[/tex] has roots [tex]\( x = \pm \frac{5\sqrt{3}}{3} \)[/tex]. These roots are not rational numbers, so the Rational Root Theorem does not provide the complete list of roots for this polynomial either.
Summary:
1. For [tex]\( f(x) = 4x^2 - 25 \)[/tex], the Rational Root Theorem provides a complete list of all roots.
2. For [tex]\( g(x) = 4x^2 + 25 \)[/tex], the Rational Root Theorem does not provide a complete list of all roots (since the roots are imaginary).
3. For [tex]\( h(x) = 3x^2 - 25 \)[/tex], the Rational Root Theorem does not provide a complete list of all roots (since the roots are irrational).
[tex]\[ \begin{array}{c|c} \text{Polynomial} & \text{Complete List by Rational Root Theorem?} \\ \hline f(x)=4 x^2-25 & \text{Yes} \\ g(x)=4 x^2+25 & \text{No} \\ h(x)=3 x^2-25 & \text{No} \\ \end{array} \][/tex]
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