Geometrical meaning of the zeroes of a polynomial.

((0,14)) has (y=14), so it is not on the (x)-axis. Tip: a zero point must have second coordinate (0).

In this interval the first two factors are positive and the third is negative, so the product is negative. Tip: check the sign of each factor separately.

In the quadratic, the sum of zeroes is (13), so the other zero is (7). Tip: convert a zero into ((x,0)).

The zeroes are (-12) and (12), so the product is (-144) and the sum is (0). Tip: opposite zeroes have sum (0).

To find x-axis intersections, set \(p(x)=0\). Since \(36x^2-49=(6x-7)(6x+7)\), the equation \((6x-7)(6x+7)=0\) gives \(x=\frac{7}{6}\) or \(x=-\frac{7}{6}\). Therefore, the intercepts are \(\left(\frac{7}{6},0\right)\) and \(\left(-\frac{7}{6},0\right)\). Option C incorrectly uses the reciprocal \(\frac{6}{7}\). Exam tip: factor a difference of squares \(a^2-b^2\) as \((a-b)(a+b)\) before solving.

Repeated points give the same (x)-values, so the distinct zeroes are (-11) and (4). Tip: count the same (x)-value once.

The x-coordinates of the points where a polynomial graph cuts the x-axis are its zeroes. Thus the zeroes here are −8 and 5. Because the coefficient of x² is 1, the polynomial can be reconstructed as (x − (−8))(x − 5) = (x + 8)(x − 5). Expanding gives x² − 5x + 8x − 40 = x² + 3x − 40. Comparing this expression with p(x) = x² + px + q gives p = 3 and q = −40. Therefore option A is correct. The coefficient p is positive because the middle terms combine to +3x, while q is negative because it equals the product of the zeroes, (−8)(5) = −40.

Real zeroes are counted from (x)-axis intersections, not from the (y)-axis intercept. Tip: ((0,25)) does not show a zero.

(x^4-1296=(x^2-36)(x^2+36)), and the real zeroes are only (\pm6). Tip: (x^2+36) gives no real zero.

For eight distinct real zeroes, the degree must be at least (8). Tip: the number of distinct zeroes cannot exceed the degree.

(x^2+16x+80=(x+8)^2+16), so there is no real zero. Tip: an always positive form gives no intersection.

(x^3-12x^2+35x=x(x-5)(x-7)), so the zeroes are (0), (5), (7). Tip: first take (x) as the common factor.

(18=\frac{10+26}{2}), so it is the midpoint of the two zeroes. Tip: the midpoint value need not be a zero.

To determine the contact of a quadratic graph with the x-axis, find its zeroes and check whether they are distinct or repeated. Factor the polynomial: 6x² − 72x + 216 = 6(x² − 12x + 36) = 6(x − 6)². Thus the only zero is x = 6, and it occurs twice. A repeated real zero means that the parabola touches the x-axis at one point and turns back, rather than crossing it at two distinct points. Hence option A is correct. The factor 6 only changes the vertical scale of the graph and does not alter the zero. Therefore x = −6 is incorrect, and the graph does meet the axis, so option D is also incorrect.

For (x<-13), both factors are negative and the outside negative makes the value negative. Tip: check factor signs first.

A touching point is still a zero, so the zeroes are 11 and −6. Their product is 11 × (−6) = −66. The closest distractor 66 fails because it ignores the negative sign. Exam tip: "touches" means the x-value is a root (often with even multiplicity), but for product you simply multiply the zeros with their signs.

For a polynomial graph, an intersection with the x-axis has y-coordinate zero, and its x-coordinate is a zero of the polynomial. The three given intersections therefore represent the zeroes 0, g, and −g. Their sum is 0 + g + (−g) = 0, because g and −g are additive inverses and cancel. Thus option A is correct. The condition g ≠ 0 ensures that the two symbolic nonzero intersections are distinct from the origin and from each other; it does not alter their sum. The expressions g² and −g² involve multiplication, not addition, while g alone represents only one of the three zeroes.

The zeroes of the polynomial are the x-coordinates of the graph's intersections with the x-axis. Factor the quadratic by finding two numbers whose product is 90 and whose sum is 19: x² − 19x + 90 = (x − 9)(x − 10). Therefore the zeroes are 9 and 10. Their distance on the x-axis is the absolute difference |10 − 9| = 1 unit, so option A is correct. The value 9 is only one zero, not the separation between the two intercepts. By the coefficient relationships, 19 is the sum of the zeroes and 90 is their product; neither quantity represents the distance. Using the absolute difference also ensures that distance is treated as positive.

For a downward-opening parabola, values between the two zeroes are positive. Tip: (x=2) lies between the zeroes.

For real (x), (x^{12}\geq0), so (x^{12}+1>0). Tip: an always positive polynomial does not cut the (x)-axis.