Geometrical meaning of the zeroes of a polynomial.

Polynomial functions are continuous: their graphs have no jumps or breaks. Since p(2) is positive, the graph is above the x-axis at x = 2; since p(5) is negative, it is below the x-axis at x = 5. Moving continuously from x = 2 to x = 5, the graph must pass through y = 0 at least once. Thus it has at least one real zero in the interval (2, 5), so it may cross the x-axis between those values. The wording “may cross” is appropriate because the information does not determine the exact zero or exclude multiple crossings. Options B and C are unsupported, and a polynomial graph cannot be a vertical line parallel to the y-axis.

By definition a number \(a\) is a zero of the polynomial only if \(p(a)=0\). Here \(p(2)=0\) is given, so 2 is definitely a root. The statements \(p(-1)<0\) and \(p(0)<0\) show values that are nonzero (negative), so \(-1\) and \(0\) are not roots. 'None' is incorrect because \(p(2)=0\) explicitly gives a zero. Exam tip: to confirm a zero, look for an explicit equality \(p(a)=0\); sign information alone (positive/negative) does not prove a root.

The graph of y = p(x) intersects the x-axis at a point whose y-coordinate is zero. At (-2, 0), the y-coordinate gives p(-2) = 0. Similarly, at (7, 0), the y-coordinate gives p(7) = 0. Therefore both values are zero, and -2 and 7 are the two zeroes of the quadratic polynomial. The signs of the polynomial between or outside the zeroes do not change these endpoint values.

Distance on the number line is the absolute difference between the coordinates, so it is always nonnegative. Here distance = \\(|9-(-3)| = |12| = 12\\). Option B (9) is just one zero's x-value, not the distance; option D (-12) is negative and therefore not a valid distance. Exam tip: always use |x2 - x1| to find distance between two zeros on the x-axis.

The midpoint of ((-2,0)) and ((6,0)) is ((2,0)). Tip: when (y)-values are equal, average the (x)-values.

The governing concept is that the zeroes of a polynomial are the x-coordinates of the points where its graph meets the x-axis. From the intersections (r, 0) and (s, 0), the two zeroes are r and s. The average of two values is their sum divided by 2, so average = (r + s) / 2. Therefore, option C is correct. Option A is the product of the zeroes, not their average. Option B gives only their sum and omits division by the number of values. Option D uses a difference, which is unrelated to the usual average formula. The y-coordinate 0 is important geometrically because it identifies the points as x-axis intersections, but it does not enter the average calculation.

The x-intercepts give the zeros of the polynomial; here the zeros are -4 and 4. They have equal magnitude and opposite signs, so they are opposites of each other. In fact their sum is zero: \(\alpha+\beta=0\) (with \(\alpha=-4,\beta=4\)). The closest distractor A is wrong because "both equal" would mean the same number (e.g. 4 and 4), which is not the case. Exam tip: read the x-coordinates of intercepts as the zeros and compare their signs and magnitudes directly.

If a graph meets the x-axis at (a,0), then x=a is a root of the polynomial because the polynomial's value is 0 there. Since the graph passes through (0,0), (2,0) and (5,0), the polynomial equals zero at x=0, x=2 and x=5; hence all three are zeros. Closest distractor: option D wrongly omits x=0; options A and C ignore the given x-intercepts. Exam tip: any x-intercept (a,0) directly gives a root x=a — never overlook the origin (0,0).

The zero is the x-value where y = 0. Set \(y=0\) in \(y=-3x-12\): \(-3x-12=0\) ⇒ \(-3x=12\) ⇒ \(x=-4\). So the zero is \(-4\). The distractor \(4\) is wrong due to the sign; \(12\) and \(-12\) do not satisfy the equation. Exam tip: to find the x-intercept of a line set \(y=0\); here the x-intercept is \((-4,0)\).

The direct answer is option A: the graph meets the x-axis at x=0 and x=5. The x-axis has y=0, so we set p(x)=0. Here 0=x(x-5). A product is zero when at least one factor is zero. Therefore x=0 or x-5=0. Solving the second equation gives x=5. The intersections are consequently (0,0) and (5,0). Option A is correct. Option B includes 1, but x=1 gives p(1)=1(-4)=-4, not 0. Option C uses -5 incorrectly; x=-5 gives a nonzero product. Option D is wrong because both factors produce zero at different x-values, so there are two intersections, not only one. Memory cue: for a factored polynomial, set each factor equal to zero.

(-2) and (2) are equally distant from the (y)-axis. Tip: opposite zeroes may indicate symmetry in the graph.

The defining condition for x = a to be a zero of p(x) is p(a) = 0. The graph crosses the x-axis at x = 1 and x = 4, so those values correspond to zero function values. At x = 2, however, the question states p(2) > 0. A positive number is not zero, so the graph point at x = 2 has a positive y-coordinate and lies above the x-axis. Therefore x = 2 is not a zero, making option B correct. Being between two zeroes does not automatically make a number another zero. The fact that 2 itself is positive is irrelevant; the decisive issue is the value of p(2). Option D is false because the graph can certainly contain a point with x = 2 and positive y-coordinate.

There is only one intersection, so there is one real zero. Tip: a cubic can have at most three real zeroes, not necessarily three.

The graph does not meet the (x)-axis, so (p(x)=0) does not occur. Tip: staying above or below without meeting gives no intersection.

If x = 7 is a zero of the polynomial p(x), then p(7) = 0. Points on the graph are of the form (x, p(x)), so at x = 7 the y-value is 0 and the point (7,0) must lie on the graph. The distractor (7,7) is wrong because it gives y ≠ 0; (0,7) is wrong due to reversed coordinates; (-7,0) is wrong because the x-coordinate is -7, not 7. Exam tip: put the zero as the x-coordinate and set y = 0 to get the point.

An x-axis intercept (x,0) means the function value is zero at that x, so the x-coordinates a, b, c of (a,0), (b,0), (c,0) are the zeroes. Option A is wrong because (0,0,0) does not represent the three x-values a, b, c. Option C is false since (a+0)=a (and similarly for b, c), so writing (a+0,b+0,c+0) is equivalent to (a,b,c). Option D is wrong because all three a, b and c are zeroes, not just c. Exam tip: whenever you see an intercept (p,0), the zero of the polynomial/function is p (the x-coordinate).

((x-3)^2=0) gives only (x=3), and the square causes touching. Tip: the outside (2) does not change the zero.

Set p(x)=0 to find zeros:
\(-x^2+25=0\) gives \(x^2=25\), so \(x=\pm5\). Thus the real zeros are 5 and −5. Option B is incorrect because it confuses the value of \(x^2\) (25) with the values of x; the roots are the square roots, \(\pm5\). Option C is wrong since p(0)=25, not 0. Option D is wrong because real zeros do exist. Exam tip: always solve p(x)=0 (or set y=0) and then solve the resulting equation step by step.

Both points lie on the x-axis, so their distance is found by taking the absolute difference of their x-coordinates. Thus distance = |6 - (-2)| = |6 + 2| = 8 units. A distance cannot be negative, so the sign of the subtraction must be handled using the absolute value. Therefore, the correct answer is 8 units; the other numerical choices do not represent the separation of the two points.

The direct answer is option A: the graph passes through (0, 0) and (5, 0). The statement p(0) = 0 means that when x = 0, the y-value p(x) is also 0. Therefore the graph contains the point (0, 0), the origin. Similarly, p(5) = 0 means that when x = 5, the y-value is 0, so the graph contains (5, 0). Both points lie on the x-axis, so they represent real zeroes 0 and 5. Option A is correct. Option B is wrong because (0, 5) would mean p(0) = 5, and (5, 5) would mean p(5) = 5, opposite to the given values. Option C is wrong because the graph does meet the x-axis at the stated points. Option D is wrong because it meets the x-axis at x = 5 as well as meeting the y-axis at the origin; it is not limited to the y-axis. Strictly, the information proves that the graph passes through these points; whether it crosses or merely touches at either point is not needed. Memory cue: p(a) = 0 always gives the graph point (a, 0).

Crossing the x-axis at a point indicates a root with odd multiplicity (a simple real zero); crossing at (2,0) gives the root x=2. Touching the axis indicates a root with even multiplicity; touching at (−1,0) gives the root x=−1. These are two distinct x-values, so there are 2 distinct real zeros. Why "three" is wrong: touching does not create an extra distinct x-value, it only implies even multiplicity at x=−1. Exam tip: count distinct x-values of intercepts; use multiplicity only to determine crossing vs touching behavior.

The graph does not meet the (x)-axis and the given values are not zero. Tip: do not call a value a zero unless the function value is (0).

The direct answer is A: -3. The x-axis intersections give the zeroes directly: -3, 2, and 8. To find the smallest number, compare them on the number line. Every negative number is less than every positive number, so -3 is less than 2 and 8. Option A is correct. Option B, 2, is a zero but not the smallest because it is greater than -3. Option C, 8, is also a zero but is the greatest of the three, not the smallest. Option D, 0, is not an x-coordinate of any given intersection, so it is not a zero identified by the data. A common mistake is to choose the number with the smallest-looking positive size or to ignore the negative sign. On a number line, values farther left are smaller. Thus the order is -3, 2, 8.

In both touching and crossing, the graph meets the (x)-axis. Tip: for a zero, meeting the axis is enough, crossing is not necessary.

An intersection with the x-axis means y=0, so the x-coordinate satisfies \(p(x)=0\); hence x-axis intercepts are the polynomial's zeros (roots). An intersection with the y-axis occurs at x=0 and the y-value is \(p(0)\), which is not necessarily zero. Thus B is incorrect because it wrongly claims the y-axis always gives \(p(x)=0\). C is only true in the special case that \(p(0)=0\) (i.e. zero is a root); D is false because an x-axis intersection does indicate a zero. Exam tip: set y=0 and solve \(p(x)=0\) for roots; set x=0 to find the y-intercept \(p(0)\).