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| − | | + | #REDIRECT[[Rational root theorem]] |
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| − | Given a [[polynomial]] <math>P(x) = a_n x^n + a_{n - 1}x^{n - 1} + \ldots + a_1 x + a_0</math> with [[integer | integral]] [[coefficient]]s, <math>a_n \neq 0</math>. The '''Rational Root Theorem''' states that if <math>P(x)</math> has a [[rational number| rational]] [[root]] <math>r = \pm\frac pq</math> with <math>p, q</math> [[relatively prime]] [[positive integer]]s, <math>p</math> is a [[divisor]] of <math>a_0</math> and <math>q</math> is a divisor of <math>a_n</math>.
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| − | As a consequence, every rational root of a [[monic polynomial]] with integral coefficients must be integral.
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| − | This gives us a relatively quick process to find all "nice" roots of a given polynomial, since given the coefficients we have only a finite number of rational numbers to check.
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| − | == Proof ==
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| − | Given <math>\frac{p}{q}</math> is a rational root of a polynomial <math>f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_0</math>, where the <math>a_n</math>'s are integers, we wish to show that <math>p|a_0</math> and <math>q|a_n</math>. Since <math>\frac{p}{q}</math> is a root, <cmath>0=a_n\left(\frac{p}{q}\right)^n+\cdots +a_0</cmath> Multiplying by <math>q^n</math>, we have: <cmath>0=a_np^n+a_{n-1}p^{n-1}q+\cdots+a_0q^n</cmath> Examining this in modulo <math>p</math>, we have <math>a_0q^n\equiv 0\pmod p</math>. As <math>q</math> and <math>p</math> are relatively prime, <math>p|a_0</math>. With the same logic, but with modulo <math>q</math>, we have <math>q|a_n</math>, which completes the proof.
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| − | ==Problems==
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| − | ===Easy===
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| − | 1. Factor the polynomial <math>x^3-5x^2+2x+8</math>.
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| − | ===Intermediate===
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| − | 2. Find all rational roots of the polynomial <math>x^4-x^3-x^2+x+57</math>.
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| − | 3. Prove that <math>\sqrt{2}</math> is irrational, using the Rational Root Theorem.
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| − | ===Answers===
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| − | 1. <math>(x-4)(x-2)(x+1)</math>
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| − | 2. <math>\text{There are no rational roots for the polynomial.} </math>
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| − | 3. A polynomial with integer coefficients and has a root as <math>\sqrt{2}</math> must also have <math>-\sqrt{2}</math> as a root. The simplest polynomial is <math>(x+\sqrt{2})(x-\sqrt{2})</math> which is <math>x^2-2=0</math>. We see that the only possible rational roots are <math>\pm 1</math> and <math>\pm 2</math>, and when substituted, none of these roots work.
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| − | {{stub}}
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