Maths: Roots of a polynomial?

Question

Two roots of the polynomial x^3 + ax^2 + 15x -7 = 0 are equal and rational. Find "a"

Please show working - thanks in advance :)

Answer 1

By rational roots theorem, possible roots are:

±1 , ±7

f(7) = 343 + 49a + 105 - 7 = 0

49a = -441

a = -9

x^3 -9x^2 + 15x -7 = 0

(x - 1)^2(x - 7) = 0

x = 1 , 7

notice that x = 1 has a multiplicity of 2.

Answer 2

By the Rational Root Theorem, rational roots must be +/- 1 or +/- 7. Since you have a rational root of multiplicity 2, all three roots must be rational. Start with 7 and perform synthetic division on the polynomial; with 7 in the "box" and the upper coefficient string {1, a, 15, -7}, the bottom string reduces to {1, a+7, 7a + 49, 49a + 441}. This last term must equal 0 by the Remainder and Factor Theorems, so if 49a + 441 = 0, then a = -9. Substitute that in the string and recheck the division (it works). The depressed polynomial coming from the bottom string is {1, -2, 1}, which is the coefficient string for the quadratic x^2 - 2x + 1. This is a perfect square trinomial, factoring into (x-1)^2, so your roots are 7 (once) and 1 (twice). a, therefore, is -9.

Answer 3

if 2 roots are rational then 3rd must be rational.

possible roots are -7 , -1, 1, 7

now product of 2 roots (as same) so roots shall be +1 or - 1



so 3rd root has to be 7 ( as product has to be +ve)



so f(x) = x^3 + ax^2 + 15x -7 = 0 = f(7)



or 343 + 49 a + 105 - 7 = 0



or a = 9



check:

x^3 - 9x^2 + 15x - 7 = (x-1)^2(x-7)