Find the zeros of qx x3 10x2 3x 108SolutionAs per the ra

Solution

As per the rational roots theorem, the zeroes of p(x) are of the form a/b where a is a factor of the constant term 108 and b is a factor of the coefficient of the leading term x³ which is 1. The factors of 108 are ± 1,2,3,4,6,9,12,27,36,54 and 108 and the facors of 1 are ± 1. Further q(4) = 4³ -10*4² – 3*4 +108 = 64- 160-12 +108 = 172-172 = 0. Therefore, 4 is a zero of q(x). Now, q(x) = x²(x-4) -6x(x-4)-27(x-4) = (x-4)(x²-6x-27). Further, x²-6x-27 = x²+3x -9x -27 = x(x+3)-9(x+3) = (x+3)(x-9). Therefore, q(x) = (x+3)(x-4)(x-9). Hence the zeroes of q(x) are -3,4, and 9.

Alternatively, we may draw a graph of q(x) using a graphing calculator. We would observe that the graph crosses X-Axis at x = -3, 4 and 9. Hence the zeroes of q(x) are -3,4, and 9.