Geometrical meaning of the zeroes of a polynomial.

The opening direction does not change the count, there are two intersections. Tip: decide zeroes by (x)-axis intersections.

An x-intercept occurs where the function value is zero, i.e. \(p(x)=0\). Here \(p(-3)=0\) and \(p(4)=0\), so the graph meets the x-axis at \((-3,0)\) and \((4,0)\). Option A is the closest distractor because it incorrectly treats \(p(0)=5\) as an x-intercept by listing \((0,5)\), but y=5≠0 so it is not on the x-axis. Exam tip: whenever you have \(p(a)=0\), record the intercept as \((a,0)\).

Real zeroes are counted only from (x)-axis intersections. Tip: do not add the (y)-axis intercept to zeroes.

A (y)-axis intercept does not give a zero because (p(x)\neq0). Tip: meeting the (x)-axis is necessary for a zero.

Zeroes of a polynomial are the x‑values for which \(p(x)=0\). The table shows \(p(1)=0\) and \(p(5)=0\), so x=1 and x=5 are zeroes. Option A is wrong because \(p(-2)=6\) and \(p(3)=-4\) are not zero. Option C lists output values (y‑values), not x‑values. Exam tip: scan the table for entries where \(p(x)=0\), not where the function value is nonzero.

Both crossing and touching correspond to roots: crossing at \(x=-2\) indicates an odd-multiplicity root (the sign of \(p(x)\) changes), while touching at \(x=3\) indicates an even-multiplicity root (no sign change). Therefore the distinct real zeros are \(x=-2\) and \(x=3\). The closest distractor 'Only -2' is incorrect because touching at 3 still gives \(p(3)=0\). Exam tip: list distinct zeros as the x-values (comma-separated) and note multiplicities if required.

A zero corresponds to an x-value where the graph meets the x-axis. If a graph touches the x-axis at exactly one point (is tangent there), that x-location is a single real zero — although the root may have multiplicity >1, it still gives only one distinct real solution, so B is correct. Option A gives two distinct intersections → two real zeros. Option C never meets the x-axis → no real zeros. Option D (graph coincides with the x-axis) yields infinitely many zeros. Exam tip: check whether the graph crosses the x-axis (gives distinct zeros) or merely touches it (gives one distinct zero with even multiplicity).

X-intercepts occur where the polynomial equals zero, i.e. set \(p(x)=0\). Solving \(x^2-16=0\) gives \(x^2=16\) so \(x=\pm4\). Therefore the x-intercepts are \((4,0)\) and \((-4,0)\). Option A lists y-axis-like points that are not zeros here; options C and D are incorrect because substituting those x-values does not make \(p(x)=0\) (for example \(p(16)=256-16\neq0\)). Exam tip: always solve \(p(x)=0\) to find zeros and report them as \((x,0)\).

Reason: For real x, \(x^2\ge0\), so \(x^2+9\ge9>0\). Thus \(x^2+9=0\) has no real solution and the parabola never meets the x-axis. Using the discriminant: \(D=b^2-4ac=0^2-4\cdot1\cdot9=-36<0\), confirming no real roots. The closest distractor “Two times” is wrong because a quadratic has two x-intercepts only when \(D>0\). Exam tip: check the discriminant or the vertex/minimum value to decide intersection with the x-axis quickly.

((x+1)^2=0) gives only (x=-1), and it is a repeated zero. Tip: a squared factor often indicates touching.

The zeroes are (-5) and (1), and (1) is greater. Tip: the value to the right on the number line is greater.

The zeroes are (m) and (n), so the product is (mn). Tip: the first coordinate of the intersection point is the zero.

At (3) and (5), (p(x)) is (4) and (-1). Tip: for a zero, the function value must be exactly (0).

A degree (4) polynomial can have at most (4) real zeroes. Tip: the number of real zeroes cannot exceed the degree.

A line parallel to and above the (x)-axis does not meet the (x)-axis. Tip: such a graph behaves like a non-zero constant 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.