Here we have to find zeroes of the given function.
Let's start solving!
[tex]f(x) = x ^{3} + 9x ^{2} + 6x - 16[/tex]
- To find x-intercept or zeroes, Substitute f(x) =0
[tex]0 = x ^{3} + 9x ^{2} + 6x - 16[/tex]
- Move the expression to the left hand side and change its sign
[tex]0 - x ^{3} - 9x ^{2} - 6x + 16 = 0[/tex]
[tex] - x ^{3} + x ^{2} - 10x ^{2} + 10x - 16x + 16 = 0[/tex]
[tex] - x ^{2} (x - 1) - 10x(x - 1) - 16(x - 1) = 0[/tex]
[tex] - (x - 1) \times (x ^{2} + 10x + 16) = 0[/tex]
[tex] - (x - 1) \times (x ^{2} + 8x + 2x + 16) = 0[/tex]
[tex] - (x - 1) \times (x \times (x + 8) + 2(x + 8)) = 0[/tex]
[tex] - (x - 1) \times (x + 8) \times (x + 2) = 0[/tex]
[tex](x - 1)(x + 8)(x + 2) = 0[/tex]
[tex]x - 1 = 0 \\ x + 8 = 0 \\ x + 2 = 0[/tex]
[tex]x = 1 \\ x = - 8 \\ x = - 2[/tex]
[tex] x_{1} = 8, x_{2} = - 2, x_{3} = 1[/tex]
