Geometrical meaning of the zeroes of a polynomial.

Zeroes are found from intersections with the (x)-axis, not the (y)-axis. So if there is no (x)-axis intersection, there are (0) real zeroes.

The governing concept is the graphical meaning of a zero of a polynomial. A number r is called a zero of p(x) when p(r) = 0. On the graph y = p(x), the point corresponding to x = r therefore has coordinates (r, p(r)) = (r, 0). Hence its vertical coordinate, or y-value, is zero. Geometrically, every real zero is represented by a point where the polynomial graph meets or touches the x-axis. The options 1 and -1 are possible y-values at other points, but not at a zero. The value cannot be undetermined because the defining condition of a zero itself fixes p(r) as 0.

A point on the x-axis has \(y=0\). So set \(2x-6=0\), giving \(2x=6\) and \(x=3\). Hence the zero is 3. The option -3 arises from a sign or arithmetic mistake; 6 or -6 are values of y (the y-intercept is -6), not the x-intercept. Exam tip: to find the zero of a linear polynomial from its line, always put \(y=0\) and solve for x.

The x-axis corresponds to y = 0, so set y = 0 in \(y = x + 7\): \(0 = x + 7\) ⇒ \(x = -7\). Thus the zero (x-intercept) is -7. Option A (7) is the common sign-error distractor; options C and D do not satisfy the equation. Exam tip: to find an x-intercept of any line, put y = 0 and solve for x.

If the graph touches the x-axis at (5,0) then the polynomial evaluates to zero at x=5, i.e. \(p(5)=0\). Hence x=5 is a zero. "Touching" typically implies the root has even multiplicity, but the essential fact here is the value is zero. Options B and D are incorrect because the contact point has x-coordinate 5, not 0 or -5; option C is wrong because touching the axis means the polynomial is zero at that point. Exam tip: use the x-coordinate of the contact point and substitute into \(p(x)\) to verify a zero; if it only touches (doesn't cross), expect even multiplicity.

Each distinct (x)-axis intersection represents one real zero. Therefore (4) intersections give (4) real zeroes.

By definition a number a is a zero of a polynomial iff \(p(a)=0\). Here \(p(2)=5\), which is not zero, so 2 is not a zero. The degree of the polynomial (linear, quadratic, etc.) does not affect this: if \(p(a)\neq0\) it cannot be a root. Exam tip: To check whether a value is a root, evaluate \(p(a)\) directly — roots give result 0; remainders on dividing by \((x-a)\) equal \(p(a)\).

For a polynomial function the graph contains points of the form (x,p(x)). If p(a)=0 then the point (a,p(a)) becomes (a,0), so the graph must pass through (a,0). The closest distractor, (-a,0), is also on the x-axis but would be correct only if p(-a)=0. Exam tip: Always express a point as (x,p(x)) and substitute the given x to find the corresponding point on the graph.

Zeroes are the (x)-values where the graph cuts the (x)-axis. So the zeroes are (1), (3), and (6).

Points where a parabola meets the x-axis are its real zeros. Two distinct intersections therefore mean two distinct real zeros; here the zeros are x = -1 and x = 5. Option B (one) would correspond to a tangent (a repeated root), option C (none) is impossible because intersections are given, and option D (three) is impossible for a quadratic since a parabola (degree 2) can have at most two real zeros. Exam tip: Count x-axis intersections for real zeros; for quadratics the maximum is 2 and a single intersection implies a repeated root.

Touching the x-axis means the parabola's vertex lies on the axis and the quadratic has a repeated root — one distinct real zero (a double root). In terms of discriminant, this corresponds to \\(\Delta=0\\). So there is exactly one real zero: x=2. Why other options are wrong: 'Two' would require the parabola to intersect the axis at two distinct points; 'None' applies when the parabola does not meet the axis at all; 'Infinite' is impossible for a polynomial of finite degree. Exam tip: for a double root of a quadratic check that \\(\Delta=0\\).

Real zeroes of a quadratic polynomial are found from points where it meets the (x)-axis. Two distinct intersections give two real zeroes.

For real zeroes, the graph must meet the (x)-axis. If it does not meet the (x)-axis, there is no real zero.

The number of distinct times the graph meets the (x)-axis gives the number of real zeroes. Check this first while reading a graph.

An x-intercept of (-4, 0) means the polynomial evaluates to 0 at x = -4, so x = -4 is a root. Option B (4) is incorrect because it has the opposite sign; options C (0) and D (1) would only be correct if those x-values were the intercepts, which they are not. Exam tip: the x-coordinate of an x-intercept gives the root directly; if unsure, substitute the value into the polynomial to verify.

For a function or polynomial the graph contains points of the form \\((a,p(a))\\). If \(p(0)=0\), then at \(x=0\) we have \(y=p(0)=0\), so the graph passes through \\((0,0)\\) — the origin. Option (1,0) is a plausible distractor but is wrong because \(p(1)\) is not given; you cannot assume \(p(1)=0\). Quick exam tip: evaluate the polynomial at the given x; whenever \(p(a)=0\), the graph crosses the x-axis at \\((a,0)\\).

A polynomial intersects the x-axis at a point only if its value (y) at that x is zero. To intersect at x = 0 we must have p(0) = 0. Option B gives p(0)=5 so the value is not zero and the graph does not meet the x-axis at x=0. Option C has a zero at x=1, so the intersection is at x=1, not x=0. Option D yields p(-1)=1, which is not zero. Exam tip: substitute the specified x into the polynomial — if the result is 0, the graph crosses the x-axis at that x-coordinate.

The zero of a polynomial is the x-value where the polynomial equals zero, i.e., where the graph meets the x-axis (y=0). At the point \((a,0)\) the x-coordinate is \(a\), so the zero is \(a\). Option \(0\) is incorrect because that is the y-coordinate at the intersection; \(-a\) and \(a^2\) are not the x-coordinate of the given point. Exam tip: when a graph intersects the x-axis, read off the x-coordinate — that is the zero (root).

Zeroes are the (x)-values where the graph cuts or touches the (x)-axis. Hence they are linked with (x)-intercepts.

Both cutting and touching count as meeting the (x)-axis. If the two points are distinct, there are two distinct real zeroes.

The point \\( (8,0) \\) means that at x=8 the function value y is 0. Hence \\(p(8)=0\\). Option \\(8\\) is a common trap — it is the x-coordinate, not the value of the polynomial. Exam tip: any intersection with the x-axis gives a root, so substitute the x-value to get zero.

\(p(-6)=0\) means the polynomial equals zero at \(x=-6\), so the graph has \(y=0\) when \(x=-6\). Hence the point \((-6,0)\) lies on the graph. Option B is wrong because it uses \(x=6\) instead of \(x=-6\). Options C and D are y‑intercepts (x=0) and do not represent a zero of the polynomial. Exam tip: a root/zero \(p(a)=0\) corresponds to the x‑intercept \((a,0)\).

Real zeroes are obtained from common points of the graph and the (x)-axis. If there is no common point, there is no real zero.

A point where the graph touches the x-axis is a real root because at that point p(x)=0. "Touches" usually indicates an even multiplicity root, but regardless of multiplicity it is still a zero of the polynomial. Thus the distinct real zero is x = -2. Option B (2) is wrong because the given x-coordinate is -2, not +2; option C (0) is wrong because the origin is not the given intercept; option D (None) is wrong because a touching point is indeed a root. Exam tip: remember that "touches" implies an even multiplicity root, while "cuts" implies an odd multiplicity.

The governing concept is the geometrical meaning of a zero of a polynomial. A number x = a is a zero when the value of the polynomial is zero, that is, f(a) = 0. On a graph, f(x) is represented by the y-coordinate, so f(x) becomes zero exactly where the graph lies on the x-axis. The points given are (r, 0) and (s, 0). Their x-coordinates are r and s, and the common y-coordinate 0 shows that the polynomial value is zero at both inputs. Therefore the zeroes are r and s, so option A is correct. The coordinates 0, 0 are not the zeroes; the negative values are unjustified, and r + s and rs are relationships involving the zeroes, not the zeroes themselves.