Geometrical meaning of the zeroes of a polynomial.

The governing concept is the relationship between polynomial zeroes and x-axis intersections. Whenever p(a) = 0, the point (a, 0) belongs to the graph, so x = a is an x-axis intersection. Here p(5) = 0 gives the point (5, 0), and p(−2) = 0 gives the point (−2, 0). Thus the graph cuts the x-axis at x-values 5 and −2, making option A correct. Option B introduces x = 0 without any information about p(0). Option C changes 5 to −5 and −2 to 2, reversing the signs incorrectly. Option D again replaces −2 with 2. The two given function values directly determine the two requested x-values.

A zero (root) corresponds to an x-coordinate where the graph meets the x-axis, so the y-coordinate must be 0. The point (0, 7) lies on the y-axis with y = 7, not on the x-axis, therefore 7 is not a zero. The closest distractor (C) is wrong because a y-intercept has x = 0 but its y need not be 0. Exam tip: always check that the second coordinate (y) equals 0 before calling an x-value a zero.

For a repeated zero the only distinct (x)-value is (2). Tip: do not count the same point twice for distinct zeroes.

Zeroes are the x-values where the graph meets the x-axis, i.e. where y = 0. Among the given points \\((1,0)\\) and \\((4,0)\\) have y = 0, so the zeroes are 1 and 4. Option B is incorrect because \\((0,6)\\) has y = 6 (an y-intercept), not y = 0. Exam tip: always check the y-coordinate — only points with y = 0 give zeroes.

The zeroes are (-7) and (-3) and (-3) is greater. Tip: the number to the right on the number line is greater.

X-intercepts occur where p(x)=0. Solve \\(x^2-9=0\\). Factor as \\(x^2-9=(x-3)(x+3)\\), giving \\x=\pm3\\. Therefore the x-intercepts are \\(-3,0\\) and \\(3,0\\). Option B is incorrect: (0,-9) is the y-intercept since \\(p(0)=-9\\), and \\(p(9)=81-9=72\\) so (9,0) is not a root. Exam tip: set p(x)=0 and use difference of squares to factor quickly.

Solve \(x^2=0\) to find zeros: the only root is \(x=0\), with multiplicity 2. A repeated root of even multiplicity means the parabola touches (is tangent to) the x-axis at that point and does not cross it. Option B is incorrect because two distinct real roots would be required to cut the axis at two points (discriminant > 0), which is not the case here. Option D is wrong since \(x=1\) is not a root. Exam tip: check factorization or discriminant and note root multiplicities to determine whether the graph crosses or merely touches the x-axis.

The zeros come from setting each linear factor to zero: x-2=0 gives x=2 and x+5=0 gives x=-5. x-axis intercepts have coordinates (x,0), so the graph crosses at (2,0) and (-5,0). Option B simply reverses the signs; options C and D are not x-axis intercepts. Exam tip: set each factor equal to 0 and write the intercepts as (x,0).

The x-axis is the set of points whose y-coordinate is zero. Therefore, when a graph y = p(x) intersects the x-axis, the corresponding x-coordinate satisfies p(x) = 0 and is a real zero. Counting distinct x-axis intersections consequently gives the number of distinct real zeroes visible from the graph. It does not always determine the degree: a polynomial may have a higher degree than the number of real intersections. It also does not immediately give the constant term, which is related to the y-intercept p(0), not the x-intercepts. Colour has no mathematical role here. Thus Option A states the correct use.

A real zero is read from the x-coordinate of an x-axis intersection. The graph has exactly one such intersection, so it has exactly one distinct real zero. Because the intersection occurs at x = −8, the zero is −8, not 8. The minus sign is part of the coordinate and cannot be discarded. Option B confuses the numerical magnitude 8 with the number of zeroes; one intersection does not mean eight zeroes. Option D contradicts the stated intersection. Therefore Option A is correct. This conclusion concerns distinct real zeroes visible on the graph and does not claim anything about possible complex zeroes.

The geometrical rule for polynomial zeroes is that every point (r, 0) on the graph corresponds to p(r) = 0, so r is a real zero. The three listed points have different x-coordinates: −4, 1, and 6. Thus they represent three distinct real zeroes. The y-coordinate is 0 in each case because all three points lie on the x-axis. Option A counts only one point, Option B misses one of the three intersections, and Option D adds an unsupported fourth zero. The degree of the polynomial is not needed to answer the question; simply counting the distinct x-axis intersections gives three. Hence Option C is correct.

The touching point gives only one distinct zero (3). Tip: count a repeated zero once when distinct zeroes are asked.

When the graph does not meet the (x)-axis, (p(x)=0) never occurs. Tip: a real zero is seen on the (x)-axis.

If \(p(a)=0\), the graph has y-coordinate zero at x=a, so the point is \((a,0)\); similarly \(p(b)=0\) gives \((b,0)\). Closest distractor: \((0,a)\) and \((0,b)\) would be points on the y-axis (x=0), not x-intercepts. Exam tip: a zero (root) of a polynomial is the x-value where the graph crosses the x-axis, so represent it as \((\text{root},0)\).

Zeroes are the (x)-values of points where (y=0). Tip: ((0,-5)) is a (y)-intercept, not a zero.

A root (zero) of a polynomial is an x-value for which the polynomial evaluates to 0, i.e. the graph meets the x-axis. At (0,8) the value is 8, not 0, so this point does not indicate a zero. Option B is misleading because x=0 alone doesn't guarantee a root—only if f(0)=0. Exam tip: to identify roots check that the y-coordinate (or f(x)) equals 0, not merely the x- or y-intercept label.

An x-axis intersection occurs when y = p(x) = 0. For p(x) = (x − 4)(x + 2), the product is zero when either factor is zero. Thus x − 4 = 0 gives x = 4, and x + 2 = 0 gives x = −2. Each zero x-coordinate is paired with y = 0, so the intersections are (4, 0) and (−2, 0). Option B reverses the signs and therefore gives the wrong zeroes. Option C lists y-axis points because their first coordinate is 0, and Option D uses the two numbers as y-coordinates rather than forming points on the x-axis. Hence Option A is correct.

Every point on the x-axis has y-coordinate 0. Therefore, to find the intersection of y = 2x − 10 with the x-axis, set y = 0: 0 = 2x − 10. Adding 10 to both sides gives 2x = 10, and dividing by 2 gives x = 5. The intersection point is consequently (5, 0), so option B is correct. Option A comes from an incorrect sign while solving the equation. Option C is the y-intercept, found by setting x = 0, not the x-intercept. Option D does not work because x = 10 gives y = 20 − 10 = 10 rather than 0. The value 5 is also the zero of the corresponding linear polynomial 2x − 10.

The zeroes of the parabola are the x-coordinates of its x-intercepts. Here the zeroes are \\(-1\\) and \\(4\\), so their sum is \\(-1 + 4 = 3\\). A common trap (option C: 4) is to pick one root instead of the sum. Exam tip: read the x-values from intercept points to get the zeroes directly.

The zeroes are (-6) and (2), so the product is (-12). Tip: do not include the (y)-coordinate in the product.