We can attempt to factor this polynomial to find solutions for [latex]f(x) = 0[/latex].

[latex]\begin{array}{rcl} x^6 - 3x^4 + 2x^2 & = & 0 \quad \text{Factor out the greatest common factor.} \\ x^2 (x^4 - 3x^2 + 2) & = & 0 \quad \text{Factor the trinomial.} \\ x^2 (x^2 - 1)(x^2 - 2) & = & 0 \quad \text{Set each factor equal to zero.} \end{array}[/latex]

[latex]\begin{array}{rcl} x^2 & = & 0 \quad \text{or} \quad (x^2 - 1) = 0 \quad \text{or} \quad (x^2 - 2) = 0 \\ x & = & 0 \quad \quad \quad \quad \quad \quad x^2 = 1 \quad \quad \quad x^2 = 2 \\ x & = & 0 \quad \quad \quad \quad \quad \quad x = \pm 1 \quad \quad x = \pm \sqrt{2} \end{array}[/latex]

This gives us five [latex]x[/latex]-intercepts: [latex](0, 0)[/latex], [latex](1, 0)[/latex], [latex](-1, 0)[/latex], [latex](\sqrt{2}, 0)[/latex], and [latex](-\sqrt{2}, 0)[/latex]. We can see that this is an even function because it is symmetric about the y-axis.

We can always check that our answers are reasonable by using a graphing calculator to graph the polynomial as shown below.

This polynomial is not in factored form, has no common factors, and does not appear to be factorable using techniques previously discussed. Fortunately, we can use technology to find the intercepts. Keep in mind that some values make graphing difficult by hand. In these cases, we can take advantage of graphing utilities.

Looking at the graph of this function, as shown below, it appears that there are [latex]x[/latex]-intercepts at [latex]x= -3, -2, \text{ and } 1[/latex].

We can check whether these are correct by substituting these values for [latex]x[/latex] and verifying that

[latex]h(-3) = h(-2) = h(1) = 0[/latex]

Since [latex]h(x) = x^3 + 4x^2 + x - 6[/latex], we have:

[latex]\begin{array}{rcl} h(-3) & = & (-3)^3 + 4(-3)^2 + (-3) - 6 = -27 + 36 - 3 - 6 = 0 \\ h(-2) & = & (-2)^3 + 4(-2)^2 + (-2) - 6 = -8 + 16 - 2 - 6 = 0 \\ h(1) & = & (1)^3 + 4(1)^2 + (1) - 6 = 1 + 4 + 1 - 6 = 0 \end{array}[/latex]

Each [latex]x[/latex]-intercept corresponds to a zero of the polynomial function and each zero yields a factor, so we can now write the polynomial in factored form.

[latex]h(x) = x^3 + 4x^2 + x - 6 = (x + 3)(x + 2)(x - 1)[/latex]