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MATH 57091 - Algebra for High School Teachers Factoring Polynomials with Rational Coefficients I: Tools Professor Donald L. White Department of Mathematical Sciences Kent State University D.L. White (Kent State University) 1/8 Review Given a polynomial f (x) ∈ Q[x], we would like to factor f (x), as completely as possible, over Q, R, or C. Recall some of the tools we introduced previously: If f (x) ∈ Q[x] and d is a common multiple of the denominators of the coefficients of f (x), then p(x) = d · f (x) is in Z[x]. Thus if p(x) = a(x)b(x), then f (x) = d1 a(x)b(x). It therefore suffices to consider factoring polynomials with integer coefficients over Q. Gauss’s Lemma: If f (x) ∈ Z[x] and f (x) = a1 (x)b1 (x), with a1 (x), b1 (x) ∈ Q[x], then f (x) = a(x)b(x), with a(x), b(x) ∈ Z[x] and a1 (x) = αa(x), b1 (x) = βb(x) for some α, β ∈ Q. In order to factor polynomials with integer coefficients over Q, it therefore suffices to consider factoring polynomials with integer coefficients over Z. D.L. White (Kent State University) 2/8 Review [Previously introduced tools, continued] Factor Theorem: Let f (x) ∈ F [x], where F is a field. An element α of F is a root of f (x) if and only if x − α is a factor of f (x). Let F be a field and f (x) ∈ F [x] with deg f (x) = n. If the leading coefficient of f (x) is c and f (x) has n roots r1 , r2 , . . . , rn in F , then f (x) = c(x − r1 )(x − r2 ) · · · (x − rn ). We will now concentrate on finding roots (in Z, Q, R, or C) of polynomials in Z[x]. D.L. White (Kent State University) 3/8 Tools for Finding Roots Our first tool for finding roots of polynomials is the Quadratic Formula. Quadratic Formula: If f (x) = ax 2 + bx + c ∈ C[x] with a 6= 0, then the roots of f (x) are √ √ −b − b 2 − 4ac −b + b 2 − 4ac and . 2a 2a The number D = b 2 − 4ac is the discriminant of f (x). The roots of f (x) are always in C. If b 2 − 4ac 6= 0, then there are two distinct roots. If b 2 − 4ac = 0, then there is one root of multiplicity 2. Let f (x) ∈ Z[x]. The roots of f (x) are in R ⇐⇒ b 2 − 4ac > 0. The roots of f (x) are in Q ⇐⇒ b 2 − 4ac is a perfect square. There are also (much more complicated) formulas for the roots of cubic and quartic polynomials. (See §3.6 of the text.) D.L. White (Kent State University) 4/8 Tools for Finding Roots The graph of a polynomial can also be useful in locating roots. Let f (x) ∈ R[x]. The graph of f (x) is the set of points (a, b) in the plane such that f (a) = b. For a real number α, the following are equivalent: i ii iii x = α is a root of f (x) in R. x − α is a factor of f (x) in R[x]. The graph of f (x) intersects the x-axis at x = α. Moreover, if x = α is a root of odd multiplicity, then the graph of f (x) crosses the x-axis at x = α, and if x = α is a root of even multiplicity, then the graph of f (x) touches the x-axis at x = α. D.L. White (Kent State University) 5/8 Rational Root Test Theorem [Rational Root Test] Let f (x) = an x n + an−1 x n−1 + · · · + a1 x + a0 be a polynomial with integer coefficients. If r s is a rational root of f (x), with r , s ∈ Z and (r , s) = 1, then r | a0 and s | an . If an = 1, then s = ±1 in the theorem, and we have the following corollary. Corollary If f (x) = x n + an−1 x n−1 + · · · + a1 x + a0 is a monic polynomial with integer coefficients, then any rational root of f (x) must be an integer r such that r | a0 . D.L. White (Kent State University) 6/8 Rational Root Test Proof of Theorem: Let sr , with (r , s) = 1, be a rational root of f (x). We then have r n−1 r r r n + an−1 + · · · + a1 f = an + a0 s s s s r n−1 r rn = an · n + an−1 · n−1 + · · · + a1 · + a0 = 0. s s s Multiplying both sides of the last equation by s n , we obtain an r n + an−1 r n−1 s + · · · + a1 rs n−1 + a0 s n = 0. Hence an r n = − an−1 r n−1 s + · · · + a1 rs n−1 + a0 s n , so s | an r n . But (s, r ) = 1, so (s, r n ) = 1, and s | an by Euclid’s Lemma. Similarly, a0 s n = − an r n + an−1 r n−1 s + · · · + a1 rs n−1 , hence r | a0 s n and (r , s n ) = 1, which implies r | a0 . D.L. White (Kent State University) 7/8 Example EXAMPLE: Find all possible rational roots of f (x) = 2x 8 + 3x 3 + 6. Solution: The possible rational roots are the numbers ± sr , where r | 6, s | 2, (r , s) = 1, and r , s are positive. Hence r = 1, 2, 3, 6; s = 1, 2, and the possible rational roots are ±1, ±2, ±3, ±6, ± 12 , ± 23 . (Of course, at most 8 of these can be roots, and possibly no “possible” root is actually a root.) D.L. White (Kent State University) 8/8