Finding Zeros of Polynomial Functions

Finding zeros = finding where the graph crosses the x-axis

The zeros of $f(x)$ are the values of $x$ where $f(x) = 0$. On the graph, these are the x-intercepts.

Step-by-step solution: Find all zeros of $f(x) = x^4 - 5x^2 + 4$

Step 1 — Notice this is a "quadratic in disguise." Let $u = x^2$:

$u^2 - 5u + 4 = 0$

Step 2 — Factor the quadratic in $u$: Find two numbers that multiply to 4 and add to $-5$: $-1$ and $-4$.

$u^2 - 5u + 4 = (u - 1)(u - 4) = 0$

Step 3 — Solve for $u$: $u = 1$ or $u = 4$.

Step 4 — Replace $u$ with $x^2$ and solve for $x$:

$x^2 = 1 \implies x = \pm 1$

$x^2 = 4 \implies x = \pm 2$

Step 5 — List all zeros: $x = -2, -1, 1, 2$ (four zeros).

Step 6 — Check: $f(1) = 1 - 5 + 4 = 0$ ✓. $f(-2) = 16 - 20 + 4 = 0$ ✓.

The complete factorization

$x^4 - 5x^2 + 4 = (x^2 - 1)(x^2 - 4) = (x-1)(x+1)(x-2)(x+2)$

Each linear factor $(x - r)$ gives one zero at $x = r$.

The u-substitution trick

Whenever you see $ax^4 + bx^2 + c$ (only even powers of $x$), substitute $u = x^2$ to turn it into a standard quadratic $au^2 + bu + c$. Solve for $u$, then take square roots.

Try it in the visualization

The graph shows the quartic crossing the x-axis at all four zeros. The factored form and u-substitution steps are shown. Toggle "local extrema" to see the max and min points between the roots.