Geometrical meaning of the zeroes of a polynomial.

\(p(x)=4\) is a constant polynomial so its graph is the horizontal line \(y=4\). Since \(y\) is never zero, the graph does not cross the x-axis and there are no zeros of the polynomial. Option B is incorrect because intersecting the x-axis requires solutions of \(p(x)=0\); a constant nonzero polynomial has none. Options C and D are also incorrect: passing through the origin needs \(p(0)=0\), and being a parabola requires degree 2, neither of which holds here. Exam tip: To find x-intercepts set \(p(x)=0\); a nonzero constant yields no solution and thus no x-intercepts.

A root (zero) is a value a for which \(p(a)=0\). With \(p(x)=x\), substituting \(x=0\) gives \(p(0)=0\), so \(x=0\) is the root. Option B (1) is incorrect because \(p(1)=1\), not zero. Exam tip: to find zeros set \(p(x)=0\) and solve; graphically zeros are x-values where the curve meets the x-axis (here the origin).

Use the zeroes as (x)-coordinates and take (y=0). So the intersection points are ((-1,0)), ((3,0)), and ((8,0)).

Two real zeroes show two distinct (x)-axis intersections. Hence the parabola cuts the (x)-axis at two points.

One real zero means the graph meets the (x)-axis at only one point. For a parabola, this is usually the touching case.

Both cutting and touching mean meeting the (x)-axis. There are two distinct points, so there are two distinct real zeroes.

If the graph touches the x-axis at \,\((r,0)\\) then the function value there is zero, i.e. \,\(f(r)=0\\). Hence \,\(r\\) is a zero (root) of the polynomial. Option B is wrong because the constant term means the value of the polynomial at \(x=0\\), not the x-coordinate of a general intercept. Option D is incorrect since \(y=0\\) at that point implies it is a root. (Additional note: when the graph only touches and does not cross, the root typically has even multiplicity.) Exam tip: whenever a graph meets the x-axis at a point, that x-coordinate is a root; use whether it crosses or only touches to infer multiplicity.

\(p(12)=0\) means the polynomial's value at \(x=12\) is zero, so the point \((12,0)\) lies on the graph. Hence the graph meets the x-axis at \(x=12\). Note: whether it cuts or just touches depends on the root's multiplicity (odd → cuts, even → touches). Exam tip: for any root \(a\) with \(p(a)=0\), check the point \((a,0)\) on the graph to locate the intersection with the x-axis.

A zero of a polynomial at x = 11 means the graph has y = 0 when x = 11, i.e. the point (11,0) lies on the graph (an x-intercept at x = 11). Option (0,11) has x = 0, so it does not show x = 11 is a zero; (11,1) has x = 11 but y ≠ 0 so it is not a zero; (-11,0) has y = 0 but at x = -11, not x = 11. Exam tip: to verify a zero visually, check for the point (a,0) — the x-intercept at that x-value.

A zero (root) corresponds to an x-value where the polynomial evaluates to 0, i.e., a point on the graph \/(x,0\/). For x = -2 the point must be \/(-2,0\/), so option A is correct. Option D has x = -2 but y = 2 (not zero), so it is not a root. Option B is an x-intercept at x = 2, and option C has y = -2, not y = 0. Exam tip: to find roots from a graph, locate points with y = 0; their x-coordinates are the zeros.

The origin is \((0,0)\), so at that point \(y=p(x)=0\) when \(x=0\). Hence \(x=0\) is a root of the polynomial. Options B and C give other specific x-values which are not implied by the graph passing through the origin. Option D (Cannot be determined) is incorrect because passing through the origin directly gives \(p(0)=0\). Exam tip: always substitute the coordinates of the given point into \(p(x)\) to check which x-value makes the polynomial zero.

Three distinct (x)-axis points show three distinct real zeroes. Their (x)-coordinates are the zeroes.

A zero of a polynomial is an x‑value where the graph meets the x‑axis, i.e. where \(y=0\). The point (4, 2) has \(y=2\), not 0, so \(x=4\) is not a zero. Options C and D are incorrect because degree or whether the graph is a straight line does not change the definition of a zero. Exam tip: always check the y‑coordinate — zeros occur only when y = 0 (x‑intercepts).

The governing idea is that the real zeroes of a polynomial are precisely the x-coordinates of the points where its graph meets the x-axis. At (m, 0), the polynomial value is zero, so m is a real zero. At (n, 0), the value is also zero, so n is another real zero. Since m ≠ n, these are two distinct values and must be counted separately. Therefore the graph has two distinct real zeroes, namely m and n. One would be correct only if the graph met the x-axis at a single distinct x-coordinate. None would mean no intersection, while infinitely many would require infinitely many distinct x-axis intersections, neither of which is stated here.

Zeroes are obtained from the points where the graph meets the (x)-axis. So focus on the (x)-axis while reading the graph.

The point \((7,0)\) means that when \(x=7\), the function value \(y=0\). For a polynomial written as \(y=p(x)\), this gives \(p(7)=0\). Option B (7) is incorrect because 7 is the x-coordinate, not the value of the polynomial at that x. Short exam tip: any point \((a,0)\) on the x-axis is a root of the polynomial, so \(p(a)=0\).

If a polynomial satisfies \(p(a)=0\), then \(x=a\) is a root and the graph crosses the x-axis at \((a,0)\). Here \(p(-4)=0\) means \(y=0\) when \(x=-4\), so the intercept is \((-4,0)\). Option B has the wrong sign for x; option C has x=0 so it's on the y-axis; option D has y≠0 so it is not on the x-axis. Exam tip: For quick MCQs remember a root \(a\) gives the x-intercept \((a,0)\).

Each distinct meeting with the (x)-axis gives one real zero. If it meets twice, it has two real zeroes.

A real zero appears when the graph meets the (x)-axis. With no real zero, the graph will not meet the (x)-axis.

For the graph of \(p(x)\), a zero is an x-value for which \(p(x)=0\). Geometrically, this is exactly the x-coordinate of a point where the graph lies on the x-axis. The y-coordinate in each listed point is 0, so it tells us that the point is on the x-axis; the required zero is obtained from the first coordinate, not from the second one.

The three intersections are \((u,0)\), \((v,0)\), and \((w,0)\). Reading their x-coordinates gives \(u\), \(v\), and \(w\). Thus \(p(u)=0\), \(p(v)=0\), and \(p(w)=0\), so the zeroes are u, v, and w. The choices involving three zeroes, their negatives, or their sum confuse the coordinates or combine them unnecessarily. Therefore option A is correct.