f(r) = a d n) d + + a0 = 0

Transcription

1 Math /Foundations of Algebra/Fall 07 Polynomials at the Foundations: Roots Next, we turn to the notion of a root of a olynomial in Q[x]. Definition 8.. r Q is a rational root of fx) Q[x] if fr) 0. Examles. a) The non-zero) constant olynomials have no roots. b) Every linear olynomial has one rational root. sx t 0; r t/s As a very easy consequence of division with remainders, we have: Proosition 8.. r is a root of fx) if and only if x r divides fx). Proof. If x r divides fx), then: fx) x r)qx) and fr) r r) qr) 0 so r is a root. On the other hand, if x r does not divide fx), then: fx) x r)qx) s and fr) r r)qr) s s for some non-zero constant remainder term) s, so r is not a root. Corollary 8.. If degfx)) d, then fx) has at most d roots. Proof. Let r,..., r n be different roots of fx). Then: fx) x r )f x) and r,..., r n are roots of f x) they are not roots of x r!), so: fx) x r )f x) x r )x r )f x) and r,..., r n are roots of f x). Eventually, fx) x r ) x r n )f n x) and then taking degrees, we get: degf n x)) n d, so n d. We can ut olynomials fx) Q[x] in a convenient form without changing their roots. If all the coefficients of fx) are in lowest terms: fx) a d n d x d a 0 n 0 multily through by the least common multile of the denominators to get a olynomial with integer coefficients and no common factors. Definition 8.4. A olynomial fx) Q[x] with integer coefficients that share no common factor is in lowest terms. If we additionally require that such a olynomial have a ositive leading term, then the lowest terms of a olynomial fx) Q[x] is unique.

2 Examle. The olynomial x x times is: 4 6x 8x 9 which is in lowest terms, desite the fact that several airs of the coefficients share rime factors. Proosition 8.5. Let fx) a d x d a 0 be given in lowest terms. If r a/n is a root of fx), then n divides a d and a divides a 0. Proof. If a/n is a root, then: Multilying through by n d, gives: fr) a d a n) d a0 0 a d a d a d a d n a an d a 0 n d 0 and if we isolate the first term and last term searately, we get: a d a d n a d a d a 0 n d ) 0 and a a d a d a n d ) a 0 n d 0 But since a and n are relatively rime and: n a 0 a d and a a d n d, from the two equations, we get n a 0 and a a d as desired. This Proosition gives us the Rational Roots Test: To detect whether fx) a d x d a 0 Q[x] in lowest terms) has a rational root, it suffices to check all the rational numbers: a n where a divides a 0 and n divides a d. Examles. a) For olynomials of the form: fx) x d b the only rational roots are integers dividing b. In other words, the only way that b has a dth root as a rational number is if it has a dth root as an integer! This gives another roof that has no rational square root or any dth root), since,,, are clearly not dth roots of. b) For quadratic olynomials ax bx c in lowest terms, the quadratic equation seems to contradict the rational roots test! The formula: r b ± b 4ac a suggests that the rational roots have denominator a, which does not divide the leading coefficient! But the contradiction is resolved if we look more closely at the situation. By Examle a), if b 4ac is

3 rational, then it is an integer. If b is even, then b 4ac is divisible by 4, and the overall numerator is even. If b is odd, then b 4ac is odd, and the numerator is once again even. Thus the quadratic formula roduces a fraction that is not in lowest terms, and the contradiction is resolved by dividing numerator and denominator by. Examle. Alying the quadratic formula to x x 4 gives: r ± { 48) and 7 } 6 Suose, however, that the rational roots test fails? Question. What are the roots when there are no rational roots? There are several answers to this. One is to comlete the rational numbers using analysis to the real numbers, and then to the comlex numbers by adjoining i, in which case one can and we will) rove: The Fundamental Theorem. Every olynomial has a comlex root. But this does not tell us what the roots are. The strength of the quadratic formula is that it exlicitly gives us the roots in terms of square roots of rational numbers. There is a similar exlicit formula for cubic olynomials, which we will give below, and for fourth degree olynomials, which we will exlore later. But there is no such formula for roots of olynomials of degree five or more! This requires thinking abstractly about roots, and is the foundation of modern algebra. Recall the roof of the Quadratic Formula. To find roots of: ax bx c 0 i) Divide by a and rename the coefficients b/a and q c/a. x x q 0 ii) Use ) x x x) and substitute for 4 x x: x x) q x ) ) q 0 4 iii) Rename y x and solve: y q and y ± q 4 ) and then substitute back: x y ) q ± b ± b 4ac a

4 We initially aroach a cubic olynomial in the same way: The Cubic Formula: To find the roots of: ax bx cx d 0 i) Divide by a and rename: r b/a, s c/a, t d/a: x rx ) sx t 0 ii) Use x r/) x rx r /)x r /) and substitute: x ) ) )) r r r x sx t ) ) ) x r s r x t r 0 iii) Rename y x r : ) ) y s r y r ) t r ) y s r y t ) ) r s r 0 iv) Rename s r and q t r s r y y q 0 to get: But now how do we roceed? By an extraordinary change of variables! y z z ) 0 z ) z z ) z q ) z z z q z z z z q z ) z z 6 qz ) iv) Solve with the quadratic formula assuming z 0). q ± q 4 z q q ) ± ) and then substitute back for y: y z z q q ) ) ± q ± q ) )

5 When we rationalize the denominator on the right, we get a surrise: q q ) ) ) q q ) q q ) ) q q ) and so, finally, we get: ) y q q ) ) q q ) ) And you can in rincile substitute all the way back to a, b, c, d to get a very ugly but exlicit analogue of the quadratic formula. Examles. a) y 6y. 6 and q ) y 4 b) y 7y 6. This looks retty bad: y but the roots are, and. All three are rational, and yet the cubic formula requires taking the square root of a negative number! Remark. We will look into this more closely when we have the comlex numbers, and then again when we think about the symmetries of roots. In articular, you should be worried in the final formula about which cube roots we are taking for each of the air of terms. Exercises. 8.. Find quartic degree four) olynomials in Q[x] with 0, and rational roots, but show that there is no quartic olynomial with exactly rational roots and one irrational root. 8.. Given x x q, suose that: q ) ) 0 ) Show that q/ has a rational cube root r and that y y q x r) x r) 8.. Try to make sense out of the the cubic formula when: a) 0 b) q 0. Notice that in these two cases you know what the roots are! 8.4. Find a formula for the roots of x 4 x q 0.