Question

Algebraically, find all remaining zeros of f(x) = 8x^3 - 6x^2-36x +27 in simplest form.

Answer 1

Answer:

$x=-\frac{3\sqrt{2}}{2},\frac{3}{4},\frac{3\sqrt{2} }{2}$

Step-by-step explanation:

Given expression is,

f(x) = 8x³ - 6x² - 36x + 27

For zeros of the function,

f(x) = 0

8x³ - 6x² - 36x + 27 = 0

2x²(4x - 3) - 9(4x - 3) = 0

(2x² - 9)(4x - 3) = 0

(2x² - 9) = 0

x² = $\frac{9}{2}$

x = $\pm\frac{3}{\sqrt{2}}$

x = $\pm \frac{3\sqrt{2}}{2}$

Or (4x - 3) = 0

x = $\frac{3}{4}$

Therefore, all zeros of the given functions are  $x=-\frac{3\sqrt{2}}{2},\frac{3}{4},\frac{3\sqrt{2} }{2}$.