Polynomials - Find zeros of polynomials graphically and verify them algebraically

Look at a graph of $y = p(x)$ that crosses the x-axis at $(-2, 0)$ and $(3, 0)$, and crosses the y-axis at $(0, -6)$. Determine the number of zeroes and specify what they are.

1. Identify the points where the graph intersects the x-axis: $(-2, 0)$ and $(3, 0)$.
2. The zeroes are the x-coordinates of these intersection points: $x = -2$ and $x = 3$.
3. Ignore the y-intercept $(0, -6)$ as it does not indicate a zero.
4. Total number of zeroes = $2$.

The geometric meaning of a zero is the x-coordinate of the point where the graph meets the x-axis. Since there are two such points, the polynomial has two zeroes.

Given the quadratic polynomial $p(x) = x^2 - 4x + 4$, describe the nature of its graph and identify the number of zeroes based on its algebraic form $(x-2)^2$.

1. The polynomial can be rewritten as $p(x) = (x-2)^2$.
2. To find the zeroes, set $p(x) = 0 \implies (x-2)^2 = 0$, which gives $x = 2, 2$.
3. Since there is only one unique value, the graph (a parabola opening upwards because $a=1 > 0$) touches the x-axis at exactly one point: $(2, 0)$.
4. Number of zeroes = $1$ (or two coincident zeroes).

When a quadratic is a perfect square, its parabola just touches the x-axis at a single point, representing one unique real zero.

1. Graphical Observation: From the graph, the parabola intersects the $x$-axis at $x = -1$ and $x = 3$. Therefore, the zeroes are $-1$ and $3$.
2. Algebraic Verification: Substitute $x = -1$: $p(-1) = (-1)^2 - 2(-1) - 3 = 1 + 2 - 3 = 0$. Substitute $x = 3$: $p(3) = (3)^2 - 2(3) - 3 = 9 - 6 - 3 = 0$. Both values are verified as zeroes.

The points $(-1, 0)$ and $(3, 0)$ are the x-intercepts. Since $p(x) = 0$ at these values, they are the zeroes.

1. Graphical Observation: From the graph, we observe that the curve $y = x^3 - 4x$ intersects the $x$-axis at three distinct points: $(-2, 0)$, $(0, 0)$, and $(2, 0)$. Therefore, the zeroes of the polynomial are $-2$, $0$, and $2$.

2. Algebraic Verification: To find the zeroes of $p(x) = x^3 - 4x$, set $p(x) = 0$: $x^3 - 4x = 0$ Factor out the common term $x$: $x(x^2 - 4) = 0$ Factor the difference of squares $(x^2 - 4)$ as $(x - 2)(x + 2)$: $x(x - 2)(x + 2) = 0$ Setting each factor to zero gives: $x = 0, \quad x - 2 = 0 \implies x = 2, \quad x + 2 = 0 \implies x = -2$ The algebraic zeroes are $\{-2, 0, 2\}$, which matches the graphical observation.

The zeroes of a polynomial $p(x)$ are the $x$-coordinates of the points where the graph of $y = p(x)$ intersects the $x$-axis. For the cubic polynomial $x^3 - 4x$, the graph crosses the axis at three points, indicating three real zeroes. Factorization confirms these specific values.