The zeroes of f(x)=x^3+3x^2+2x are 0,1,2 1,2 -2,-1,0 -1,-2 1,3,2

To find the zeroes of the polynomial function $f(x) = x^3 + 3x^2 + 2x$, we need to solve the equation $f(x) = 0$.<br /><br />We can start by factoring the polynomial. Notice that $x$ is a common factor in each term:<br /><br />$f(x) = x(x^2 + 3x + 2)$<br /><br />Now, we can set each factor equal to zero and solve for $x$:<br /><br />$x = 0$<br /><br />$x^2 + 3x + 2 = 0$<br /><br />To solve the quadratic equation $x^2 + 3x + 2 = 0$, we can use the quadratic formula:<br /><br />$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$<br /><br />In this case, $a = 1$, $b = 3$, and $c = 2$. Plugging these values into the quadratic formula, we get:<br /><br />$x = \frac{-3 \pm \sqrt{3^2 - 4(1)(2)}}{2(1)}$<br /><br /> \frac{-3 \pm \sqrt{9 - 8}}{2}$<br /><br />$x = \frac{-3 \pm \sqrt{1}}{2}$<br /><br />$x = \frac{-3 \pm 1}{2}$<br /><br />So, the solutions are:<br /><br />$x = \frac{-3 + 1}{2} = -1$<br /><br />$x = \frac{-3 - 1}{2} = -2$<br /><br />Therefore, the zeroes of the polynomial function $f(x) = x^3 + 3x^2 + 2x$ are $0$, $-1$, and $-2$.