D ETERMINING R ATIONAL R OOTS Apply the rational roots theorem!

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Presentation transcript:

D ETERMINING R ATIONAL R OOTS Apply the rational roots theorem!

T HE R ATIONAL R OOTS T EST Suppose P(x) = a n x n + a n-1 x n-1 + … + a 0.where a n, a n-1, …, a 0 are integers. If P(x) has a rational root x = p/q, then p must be a factor of a 0 and q must be a factor of a n. This theorem allows us to find the possible roots of a polynomial with integer coefficients.

E XAMPLE 1 Let’s try to find the roots of x 3 +4x 2 +x-6. The leading coefficient is 1, so q = 1. The constant term is 6, so our possible values of p are ±1, ±2, ±3, and ±6. Since q = 1, these are also our possible fractions. Let’s test these possibilities by plugging them in to find out which of them are actually factors of our polynomial.

E XAMPLE 1 (C ONT.) Let’s start with x = 1. If we plug x = 1 into our polynomial, we get – 6, which equals zero. 1 is a root of the equation (we got lucky!). Now, we can either continue testing possible roots until we find them all or factor out an (x-1) from the equation using long division. Let’s do the latter.

L ONG D IVISION We’re left with (x-1)(x 2 + 5x + 6). The second part is a quadratic equation, which we can factor. This leaves us with (x-1)(x+2)(x+3), so the three roots of the polynomial are 1, - 2, and -3. We also could have found these roots by investigating the other possible rational roots the way we did for x = 1.

E XAMPLE 2 Let’s try to find the rational roots of 2x 3 + 4x 2 – 2x + 1. The constant term is 1 and the leading coefficient is 2, so the possible rational roots are ±1/2. Let’s test 1/2 first. 2(1/2) 3 + 4(1/2) 2 – 2(1/2) + 1 = 1/4 + 1 – = 5/4 1/2 is not a root of our equation. Let’s test -1/2. 2(-1/2) 3 + 4(-1/2) 2 – 2(-1/2) + 1 = -1/ = 11/4. Neither of the possibilities is a root of the polynomial, so this polynomial has no rational roots.

E XAMPLE 3 Let’s find the rational roots of f(x) = x 5 -2x 3 -6x 2 -35x-30. The leading coefficient is 1 and the constant coefficient is -30, so the possible roots are ±1, ±2, ±3, ±5, ±6, ±10, ±15, and ±30. Let’s test these roots by plugging them back into the equation. Try this yourself! You should find that f(x) = 0 when x = -1, -2, and 3. These are the rational roots of our equation.

T RY IT Y OURSELF Find the rational roots of the following polynomials. 1. x 3 -2x 2 -13x x 4 -x 3 -11x 2 -x x 3 -17x 2 -26x-8

S OLUTIONS 1. -1, -2, and and /2, -2/3, and 4.