Geometrical meaning of the zeroes of a polynomial.

The point \((r,0)\) has x‑coordinate \(r\) and y‑coordinate 0. For the polynomial \(p(x)\), the y‑value at x=\(r\) is \(p(r)\); therefore \(p(r)=0\). Option B (\(p(0)=r\)) incorrectly relates the value at x=0 to r and need not hold. Option C (\(r=0\)) is only true if the intercept happens at the origin, not in general. Option D (\(p(r)=r\)) would mean the function value equals the x‑coordinate at that point, which contradicts y=0 here. Exam tip: Whenever you see an x‑intercept \((a,0)\), immediately record \(p(a)=0\) — that a is a root of the polynomial.

Set the quadratic equal to zero to find x-intercepts (roots).

Factor: \(x^2+2x-15=(x+5)(x-3)\).
Setting \((x+5)(x-3)=0\) gives x=-5 or x=3, so the graph meets the x-axis at x=3 and x=-5.

The closest distractor (-3, 5) is wrong because the signs of the roots are incorrect — the constant term -15 forces the product of roots to be -15, so one root must be negative.

Exam tip: Check constant term and middle coefficient quickly — product of roots = constant term, sum of roots = - (coefficient of x).

p(1)=0 directly shows x=1 is a root — this is explicit. The facts p(-3)<0 and p(4)>0 only give signs of p at those x-values; they do not make -3 or 4 roots. (By continuity of polynomials, a sign change between -3 and 4 guarantees at least one root in (-3,4), but it does not identify the endpoint values as roots.) Therefore the only certain root from the given information is x=1. Exam tip: look for statements of the form p(a)=0 for a guaranteed root; sign changes indicate existence of a root in an interval, not at a specific endpoint.)

The geometrical meaning of a zero is tied specifically to the x-intercept. A number r is a zero of p(x) when p(r) = 0, so the point (r, 0) lies on the graph. The line y = 2 is not the x-axis; every point on it has ordinate 2, not 0. Therefore, its two intersections do not represent zeroes. Since the graph is stated not to cut the x-axis, the supplied information displays no zero. It does not prove that the polynomial has no real zero anywhere unless the whole graph is known, but among the given data, option A is the only certain statement. Options B and C incorrectly treat y = 2 intersections as x-intercepts, while D has no basis.

To find the x-intercept set p(x)=0. With p(x)=4x we get 4x=0 ⇒ x=0, so y=p(0)=0 and the intersection point is (0,0). Option (4,0) is incorrect because p(4)=16 ≠ 0; (0,4) is impossible for an x-intercept since y must be 0. Exam tip: always solve p(x)=0 to get x-intercepts of a polynomial.

For zeroes (m) and (n), the factors are ((x-m)) and ((x-n)). A non-zero multiplier does not change zeroes.

The root (zero) of a polynomial is the x‑value where its graph meets the x‑axis. Hence the zero is the number 2.5. Option B (2.5, 0) denotes the coordinate point on the plane — a point, not the numeric root. Option C is incorrect because x=0 is not where the graph crosses; Option D is the point on the y‑axis and irrelevant. Exam tip: when asked for a root or zero give the x‑value (a single number), not the ordered pair of the point.

Points where the graph meets the x-axis are the zeros of \(p(x)\), so set \(p(x)=0\). Factor: \(x^2-2x-8=(x-4)(x+2)\). Therefore \(x=4\) and \(x=-2\), giving points \((4,0)\) and \((-2,0)\). Option B has signs swapped (gives wrong x-values); option C lists points on the y-axis, not x-axis; option D is just incorrect roots. Exam tip: set the quadratic equal to zero and factor or use the quadratic formula; then report roots as (x,0).

Three distinct (x)-intercepts give three zeroes. Count by intersections not by the type of numbers.

A zero can be a fraction and touching is enough. The key point is \(p\left(\frac{1}{2}\right)=0\).

A zero (root) of a polynomial is the x‑value where its graph meets the x‑axis — i.e. the x‑coordinate of the intercept. The given intercept \((-\sqrt{3},0)\) has x‑coordinate \(-\sqrt{3}\), so that is the zero. Option D is the full point (ordered pair), not the zero itself; option C is the y‑coordinate (0); option B has the wrong sign. Exam tip: from an x‑axis intercept (a,b) pick the first component a as the root.

A zero is related to the (x)-axis not a general (y)-intercept. A (y)-intercept gives a zero only if it is the origin.

Completing the square gives \(x^2+6x+10=(x+3)^2+1\). Since \((x+3)^2\ge0\), the minimum value is 1, so the parabola never reaches 0 and does not meet the x-axis. Equivalently, the discriminant is \(b^2-4ac=36-40=-4<0\), so there are no real roots. The closest distractor (touch once) is wrong because that requires discriminant zero. Exam tip: use the discriminant or complete the square to quickly decide if real x-intercepts exist.

The graph meets the x-axis where \(p(x)=0\). Solve \(2x^2-8=0\) to get \(x^2=4\), so \(x=\pm 2\). Equivalently factor as \(2(x-2)(x+2)\), giving zeros \(-2\) and \(2\). Choice B would follow from the wrong step \(x^2=16\); C is wrong because \(p(0)=-8\) (not zero); D is wrong because there are two zeros, both \(2\) and \(-2\), not only \(4\). Exam tip: set \(p(x)=0\) first and factor out common constants to simplify solving.

A solution of p(x) = 0 is an x-value for which the graph has y-coordinate zero. Such points are precisely the intersections of the graph with the x-axis. The stated intersections have x-coordinates -7 and -2, so p(-7) = 0 and p(-2) = 0. Hence the solution set is {-7, -2}, represented by option A. The negative signs must be retained because they are the actual x-coordinates of the intercepts; changing them to 7 and 2 gives different points on the opposite side of the y-axis. Option C keeps only one sign correct, and option D ignores the two explicitly given x-axis intersections. No calculation beyond reading the graph is needed.

A real zero of a polynomial is a real number r for which p(r) = 0. On the graph y = p(x), this means that the graph must have a point (r, 0), so it must intersect or touch the x-axis. The statement says that the quadratic graph neither touches nor intersects the x-axis. Consequently, there is no real x-coordinate at which p(x) becomes zero, and the number of distinct real zeroes is zero. In discriminant language, a quadratic with no real x-axis contact has discriminant less than zero, although calculating the discriminant is not necessary here. One or two would require contact with the axis; infinitely many is impossible for a nonzero quadratic.

The x-axis intersections occur where p(x) = 0. Since 5 is non-zero, it cannot itself create a zero; the product is zero when at least one factor is zero. Thus x + 1 = 0 gives x = -1, x - 2 = 0 gives x = 2, and x - 4 = 0 gives x = 4. These are three different real values, so the graph has three distinct x-intercepts: (-1, 0), (2, 0), and (4, 0). Therefore option A is correct. The degree is three, so three distinct zeroes are possible and occur here. Option B or C would require repeated or missing roots, while option D wrongly counts the constant coefficient 5 as an additional zero.

The point (9,0) lies on the x-axis, so for the polynomial p(x) we have \(p(9)=0\); hence x=9 is a root (zero). The fact that the graph goes from above to below shows the curve crosses the axis at that point, which indicates a root of odd order (often a simple root). Option B is wrong because y=9 is not implied by a point on the x-axis; option C is wrong because a nonzero constant polynomial does not cross the x-axis; option D contradicts the given point lying on the x-axis. Exam tip: A crossing of the x-axis implies a zero with odd multiplicity; touching-and-returning implies even multiplicity.

Points on the x-axis have y-coordinate zero, so set \(p(x)=0\): \(x^2-11x+30=0\). Factorising gives \(x^2-11x+30=(x-5)(x-6)\), so the roots are \(x=5\) and \(x=6\). Thus the graph meets the x-axis at (5,0) and (6,0). The closest distractor (B) has the signs reversed; option C lists y-axis intercepts, and option D confuses coefficients with roots. Exam tip: set the polynomial equal to zero and factorise (or use the quadratic formula) to find x-intercepts quickly.

Geometrically each (x)-axis intersection gives one real zero. A quadratic may have (0), (1), or (2) real zeroes.