Question

find the zeros of cubic polynomial p(x) = 2x³-9x²+13x-6​

Answer 1

Answer:

$x = 1, \: x = \frac{3}{2} , \: x = 2 \: are \: the \: zeroes \: of \: given \: cubic \: polynomial.$

Step-by-step explanation:

Given,

$p(x) = 2x {}^{3} - 9x {}^{2} + 13x - 6 \\ let \: p(x) = 0 \\ 2x {}^{3} - 9x {}^{2} + 13x - 6 = 0 \\ = 2x {}^{3} - 2x {}^{2} - 7x {}^{2} + 7x {}^{2} + 6x - 6 = 0 \\ = 2x {}^{2} (x - 1) - 7x(x - 1) + 6(x - 1) = 0 \\ = (x - 1)(2x {}^{2} - 7x + 6) = 0 \\ = (x - 1)(2x {}^{2} - 3x - 4x + 6) = 0 \\ = (x - 1)(x(2x - 3 ) - 2(2x - 3)) = 0 \\ = (x - 1)(2x - 3)(x - 2) = 0 \\ now \: \\ x - 1 = 0 \\ x = 1 \\ and \: 2x - 3 = 0 \\ x = \frac{3}{2} \\ and \: x - 2 = 0 \\ x = 2$

Hope this will help you!!