Complex and Rational Zeros of Polynomials (3.4)(2) Continuing what you started on the worksheet

First, a little POD Using Descartes’ Great Big Rule o’ Signs, give the possible solution types for x 3 + x x - 24 = 0. Using either theorem on bounds, give possible bounds for the real roots.

First, a little POD Using Descartes’ Great Big Rule o’ Signs, give the possible solution types for x 3 + x x - 24 = 0. There will be one positive root. There will be two or no negative roots. If there are no negative roots, then there will be two imaginary roots. Using our second theorem on bounds, the real roots will fall between -25 and 25. Graph it and see.

Now, on to a method to find actual rational roots of polynomials (woo-hoo!) First the theorem (because you’re worth it): Theorem on Rational Zeros of a Polynomial: If the polynomial f(x) = a n x n + a n-1 x n-1 + a n-2 x n-2 + ……. + a 1 x + a 0 has integer coefficients and if c/d is a rational zero of f(x) such that c and d have no common prime factor, then 1.The numerator c of the zero is a factor of the constant term a 0 and 2.The denominator d of the zero is a factor of the leading coefficient a n.

Now, on to a method to find actual rational roots of polynomials (woo-hoo!) What in tarnation does this mean, you ask. Let’s see. If there is a rational zero for the polynomial: That rational zero, c/d, will be a completely reduced rational, real zero of the polynomial. Factors of the final constant will be among the c values Factors of the leading coefficient will be among the d values.

Now, on to a method to find actual rational roots of polynomials (woo-hoo!) Try it. Start with 3x x x 2 - 8x - 8 = 0. What are the factors of -8? Of 3? So, the possible rational zeros of the polynomial are ±1, ±1/3, ±2, ±2/3, ±4, ±4/3, ±8, ±8/3.

Now, on to a method to find actual rational roots of polynomials (woo-hoo!) Start with 3x x x 2 - 8x - 8 = 0. The possible zeros of the polynomial are ±1, ±1/3, ±2, ±2/3, ±4, ±4/3, ±8, ±8/3. Use synthetic division to test each one. (Yes, I know, a lot of testing. Just do it.) When you find one, see if you can factor the quotient-- it saves time. If not, try another of those possible zeros with synthetic division. In this case you may need to do two rounds of SD. If you like, try all the integers possibilities first.

Now, on to a method to find actual rational roots of polynomials (woo-hoo!) Start with 3x x x 2 - 8x - 8 = 0. The possible zeros of the polynomial are ±1, ±1/3, ±2, ±2/3, ±4, ±4/3, ±8, ±8/3. (What does Descartes Rule tell us about possible zeros?) What have you found? What is the complete factorization? What are all the real zeros? How do these match what Descartes led us to?

Now, on to a method to find actual rational roots of polynomials (woo-hoo!) Start with 3x x x 2 - 8x - 8 = 0. So, the possible zeros of the polynomial are ±1, ±1/3, ±2, ±2/3, ±4, ±4/3, ±8, ±8/3. What is the complete factorization? What are all the real zeros? Zeros: Factors:

Try another Sometimes you get a surprise. Try the process with x 3 - 4x - 2 = 0. The only possible rational roots are ±1 and ±2. Do any of them work? What’s going on?

Try another x 3 - 4x - 2 = 0 The only possible rational roots are ±1 and ±2. Do any of them work? Nope. What does that mean for the roots of the polynomial? They’re not rational!

Another approach x 3 - 4x - 2 = 0 How would you know this is NOT the graph of the cubic function? Hint: Descartes.