Geometrical meaning of the zeroes of a polynomial.

The geometrical rule for polynomial zeroes is that every point (r, 0) on the graph corresponds to p(r) = 0, so r is a real zero. The three listed points have different x-coordinates: −4, 1, and 6. Thus they represent three distinct real zeroes. The y-coordinate is 0 in each case because all three points lie on the x-axis. Option A counts only one point, Option B misses one of the three intersections, and Option D adds an unsupported fourth zero. The degree of the polynomial is not needed to answer the question; simply counting the distinct x-axis intersections gives three. Hence Option C is correct.

The touching point gives only one distinct zero (3). Tip: count a repeated zero once when distinct zeroes are asked.

When the graph does not meet the (x)-axis, (p(x)=0) never occurs. Tip: a real zero is seen on the (x)-axis.

If \(p(a)=0\), the graph has y-coordinate zero at x=a, so the point is \((a,0)\); similarly \(p(b)=0\) gives \((b,0)\). Closest distractor: \((0,a)\) and \((0,b)\) would be points on the y-axis (x=0), not x-intercepts. Exam tip: a zero (root) of a polynomial is the x-value where the graph crosses the x-axis, so represent it as \((\text{root},0)\).

The direct answer is B: 2 and 5. A polynomial zero is an x-value at which the output y=p(x) equals 0. Therefore inspect only the points with second coordinate 0. The points (2,0) and (5,0) have y=0, so the zeroes are x=2 and x=5. Option B is correct. Option A, 0 and 7, is wrong: the point (0,-5) has a nonzero y-coordinate, while (7,3) also has a nonzero y-coordinate. Option C, -5 and 3, incorrectly uses the y-coordinates from those two non-axis points; they are outputs, not zeroes. Option D, all x-values, is wrong because only x-values paired with y=0 qualify. The point (0,-5) is a y-axis intercept, not an x-axis intercept. Remember: read zeroes as the x-coordinates of points on the x-axis.

A root (zero) of a polynomial is an x-value for which the polynomial evaluates to 0, i.e. the graph meets the x-axis. At (0,8) the value is 8, not 0, so this point does not indicate a zero. Option B is misleading because x=0 alone doesn't guarantee a root—only if f(0)=0. Exam tip: to identify roots check that the y-coordinate (or f(x)) equals 0, not merely the x- or y-intercept label.

An x-axis intersection occurs when y = p(x) = 0. For p(x) = (x − 4)(x + 2), the product is zero when either factor is zero. Thus x − 4 = 0 gives x = 4, and x + 2 = 0 gives x = −2. Each zero x-coordinate is paired with y = 0, so the intersections are (4, 0) and (−2, 0). Option B reverses the signs and therefore gives the wrong zeroes. Option C lists y-axis points because their first coordinate is 0, and Option D uses the two numbers as y-coordinates rather than forming points on the x-axis. Hence Option A is correct.

Every point on the x-axis has y-coordinate 0. Therefore, to find the intersection of y = 2x − 10 with the x-axis, set y = 0: 0 = 2x − 10. Adding 10 to both sides gives 2x = 10, and dividing by 2 gives x = 5. The intersection point is consequently (5, 0), so option B is correct. Option A comes from an incorrect sign while solving the equation. Option C is the y-intercept, found by setting x = 0, not the x-intercept. Option D does not work because x = 10 gives y = 20 − 10 = 10 rather than 0. The value 5 is also the zero of the corresponding linear polynomial 2x − 10.

The zeroes of the parabola are the x-coordinates of its x-intercepts. Here the zeroes are \\(-1\\) and \\(4\\), so their sum is \\(-1 + 4 = 3\\). A common trap (option C: 4) is to pick one root instead of the sum. Exam tip: read the x-values from intercept points to get the zeroes directly.

The direct answer is C: -12. The x-axis points are (-6,0) and (2,0). Their x-coordinates, -6 and 2, are the zeroes. To find their product, multiply step by step: 8? No; use (-6)(2)=-12. Thus the product is -12. Option C is correct. Option A, -8, is the sum (-6)+2, not the product. Option B, 12, results from ignoring the negative sign or multiplying the absolute values 6 and 2. Option D, 8, does not represent the product of the zeroes; it may arise from an incorrect subtraction or sign error. The y-coordinates are both 0 and are not used as the zeroes. Always identify x-values first, then perform the requested operation, preserving signs.

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 direct answer is option C: the greater zero is 1. A zero of a polynomial is an x-value where its graph meets the x-axis, whose y-coordinate is 0. The two given intersections have x-coordinates -5 and 1, so the zeroes are -5 and 1. On a number line, numbers farther to the right are greater; therefore 1 is greater than -5. Option A, -5, is a zero but it is the smaller one. Option B, 0, is not one of the stated x-intercepts. Option C, 1, is correct because the point (1,0) lies on the x-axis. Option D, 5, is not an x-coordinate of either intersection. Remember: read the x-coordinates of x-axis intersections, then compare them on the number line.

The direct answer is option B: mn. The graph meets the x-axis at (m, 0) and (n, 0). At an x-axis point, the first coordinate is the x-value and therefore the zero of the polynomial. Thus the two zeroes are m and n. Their product is m × n, written as mn. Option A, m + n, is the sum of the zeroes, not their product. Option B is correct because multiplication of the two zeroes gives mn. Option C, 0, is not generally correct; the product would be zero only if m or n were zero, which is not stated. Option D, m − n, is their difference, not their product. The second coordinates are both 0 because the points lie on the x-axis, but those 0s are not the zeroes being multiplied. Memory cue: read the first coordinate of each x-axis intersection, then apply the operation asked—product means multiply.

The direct answer is option B: 3 and 5 are not zeroes because their y-values are not 0. A polynomial zero is an x-value a for which p(a)=0. From the graph points, p(2)=0, p(3)=4, p(4)=0, and p(5)=-1. Thus 2 and 4 are zeroes, because their points lie on the x-axis. At x=3 the graph has height 4, and at x=5 it has height -1; neither height equals 0. Option A is wrong because being positive does not prevent a number from being a zero; 2 and 4 are positive zeroes. Option B is correct. Option C is wrong because these points are not on the y-axis; their x-values are 3 and 5. Option D is wrong because a polynomial graph is formed and is explicitly given. Exam cue: check p(x)=0, not merely whether x is positive or negative.

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.