Geometrical meaning of the zeroes of a polynomial.

Correct answer: A. (0,0)

Explanation: Set p(x)=0 to find the root: \(3x=0\) gives \(x=0\). The x-axis intersection is the root written as a point \((x,0)\), so the intersection is \((0,0)\). The nearby distractor \((3,0)\) is incorrect because it corresponds to x=3, whereas \(3x\) vanishes only at x=0. Exam tip: always solve p(x)=0 and express the answer as \((root,0)\).

Correct answer: A. -2

Explanation: On the x-axis, \(y=0\). Substituting gives \(0 = x + 2\), so \(x = -2\). Interpretation: the x‑intercept is the value of x that makes the expression equal to zero. The nearest distractor 2 is wrong due to the sign; 0 and 1 do not satisfy the equation. Exam tip: to find an x‑intercept always set \(y=0\) and solve for x.

Correct answer: A. 5

Explanation: The zero is the x-value for which y = 0. For the line \(y=5-x\), set \(0=5-x\) which gives \(x=5\). Hence the zero is 5. The closest distractor 0 is wrong because at \(x=0\) the value is \(y=5\), not 0. Exam tip: to find the x-intercept of a line, set \(y=0\) and solve for \(x\).

Correct answer: A. One

Explanation: Only one (x)-axis intersection is visible so there is one real zero. Tip: the count comes from the number of points.

Correct answer: A. a

Explanation: The governing concept is the geometrical meaning of a zero of a polynomial. A number r is a zero of p(x) when p(r) = 0. On a graph, this means that the point (r, 0) lies on the x-axis. Since the stated intersection is (a, 0), substituting its x-coordinate gives p(a) = 0; hence the zero is a. Therefore, option A is the precise answer. The second coordinate, 0, describes the y-coordinate and only confirms that the point lies on the x-axis; it is not the requested zero. Option C simplifies algebraically to a, but option A gives the standard and direct value. Option D changes the sign without any basis.

Correct answer: A. No

Explanation: By definition a number a is a zero (root) of a polynomial iff \(p(a)=0\). Here \(p(2)=3\), which is not zero, so 2 is not a zero. Option C is incorrect because the degree (quadratic or otherwise) does not change the criterion; any degree requires \(p(a)=0\) to be a root. Exam tip: evaluate the polynomial at the given value—if the result is 0, it is a root.

Correct answer: A. (-4, 0)

Explanation: Reason: Interpret the polynomial as the function \(y=p(x)\). \(p(-4)=0\) means at \(x=-4\) the value \(y=0\), so the graph passes through the point \((-4,0)\). The closest distractor (0, −4) comes from swapping coordinates; it would be correct only if \(p(0)=-4\), which is not given. Exam tip: treat \(p(x)\) as the y-coordinate and plot the point \((x,p(x))\).

Correct answer: A. (9,0)

Explanation: A zero of a polynomial corresponds to an x‑intercept of its graph, i.e. a point whose y‑coordinate is zero. Thus the representing point must satisfy \(y=0\). The point (9,0) has y = 0, so x = 9 is a zero. The closest distractor (0,9) is incorrect because its y ≠ 0 (it lies on the y‑axis); similarly (9,9) and (0,-9) do not lie on the x‑axis. Exam tip: quickly check the y‑coordinate — zeros appear where y = 0.

Correct answer: A. 0

Explanation: The x-intercepts are the zeros of the polynomial; here the zeros are -1 and 1. Their sum is -1 + 1 = 0, so the correct answer is 0. Option D (1) is incorrect because it gives only one zero rather than the sum of both. Tip: symmetric zeros of the form ±a always sum to 0; more generally the sum of zeros relates to the coefficients (negative of the coefficient of x^{n-1} divided by leading coefficient).

Correct answer: A. (0)

Explanation: The direct answer is A: 0. The zeroes of a polynomial are the x-coordinates where its graph meets the x-axis. The intersections \((0,0)\) and \((6,0)\) therefore give zeroes 0 and 6. Their product is \(0\times6=0\). Option A is correct. Option B, 6, is one of the zeroes, but the question asks for their product, not one zero. Option C, -6, would be the product only if one zero were positive and the other negative, such as -1 and 6; that is not the case. Option D, 36, could result from using \(6\times6\), but the first zero is 0, not 6. The zero at the origin makes the product immediately zero. Exam cue: whenever one factor in a product is zero, the whole product is zero.

Correct answer: A. One that cuts the (x)-axis at two distinct points

Explanation: The direct answer is A: a graph that cuts the x-axis at two distinct points. A zero means p(x)=0, and graphically this is exactly a point whose y-coordinate is 0, that is, a point on the x-axis. If the graph has two different x-axis intersections, their two x-coordinates are two real zeroes. Option A is correct because it directly describes these two intersections. Option B is wrong because a graph cannot normally cut the same y-axis at two different points if it represents a function, and, in any case, y-axis intersections do not define zeroes. Option C is not enough: a horizontal graph may have no zero or may coincide with the x-axis, so it does not guarantee two distinct zeroes. Option D is wrong because touching the y-axis concerns x=0, not p(x)=0. Exam cue: zeroes are read from x-axis intersections.

Correct answer: A. The graph passes through the point (3, 0)

Explanation: If x = 3 is a zero then p(3) = 0. On the graph this means the point with x‑coordinate 3 lies on the x‑axis, i.e. (3, 0). Options B and C have y ≠ 0 so they cannot be zeros; option D has x = -3, not 3. Exam tip: substitute the given x into p(x) to check if it equals zero, or look for the point (x, 0) on the graph.

Correct answer: A. The graph cuts the x-axis at (-6,0)

Explanation: A number a is a zero (root) of a polynomial if p(a)=0. If the graph meets the x-axis at (-6,0) then the function value at x=-6 is 0, so p(-6)=0 and -6 is a zero. Option B is wrong because (0,-6) is a y-axis intercept (x=0), not x=-6. Option C gives x=6, the wrong sign. Option D has y=6, so p(-6)=6≠0. Exam tip: a root corresponds to an x-intercept, so check the coordinate is (-6,0) with y=0 and the correct sign for x.

Correct answer: A. It has no real zero

Explanation: For a polynomial p(x), a real zero is a real number r satisfying p(r) = 0. On the coordinate graph y = p(x), this condition produces the point (r, 0), which lies on the x-axis. Thus real zeroes are represented exactly by x-axis intersections. If no point of the graph lies on that axis, there is no real value of x for which p(x) = 0. Therefore the polynomial has no real zero, as stated in Option A. Options B and C claim intersections that the question explicitly rules out. Option D is also false because a polynomial cannot have every number as a zero unless it is the zero polynomial, whose graph is the x-axis itself.

Correct answer: A. \\(p(10)=0\\)

Explanation: If the graph crosses the \\(x\\)-axis at \\(x=10\\), the point on the curve is \\( (10,0)\\). The polynomial's value at x equals the y-coordinate, so \\(p(10)=0\\). The closest distractor \\(p(-10)=0\\) is wrong because it refers to x = -10, not x = 10. The options \\(p(0)=10\\) and \\(p(10)=10\\) are inconsistent with crossing the x-axis because y must be 0 there. Exam tip: when told the graph crosses at \\(x=a\\), substitute x = a; a crossing on the x-axis means \\(p(a)=0\\).

Correct answer: A. 5 and −2

Explanation: The governing concept is the relationship between polynomial zeroes and x-axis intersections. Whenever p(a) = 0, the point (a, 0) belongs to the graph, so x = a is an x-axis intersection. Here p(5) = 0 gives the point (5, 0), and p(−2) = 0 gives the point (−2, 0). Thus the graph cuts the x-axis at x-values 5 and −2, making option A correct. Option B introduces x = 0 without any information about p(0). Option C changes 5 to −5 and −2 to 2, reversing the signs incorrectly. Option D again replaces −2 with 2. The two given function values directly determine the two requested x-values.

Correct answer: A. It is a y-axis intercept with y ≠ 0, so 7 is not a zero

Explanation: A zero (root) corresponds to an x-coordinate where the graph meets the x-axis, so the y-coordinate must be 0. The point (0, 7) lies on the y-axis with y = 7, not on the x-axis, therefore 7 is not a zero. The closest distractor (C) is wrong because a y-intercept has x = 0 but its y need not be 0. Exam tip: always check that the second coordinate (y) equals 0 before calling an x-value a zero.

Correct answer: A. One

Explanation: For a repeated zero the only distinct (x)-value is (2). Tip: do not count the same point twice for distinct zeroes.

Correct answer: A. 1 and 4

Explanation: Zeroes are the x-values where the graph meets the x-axis, i.e. where y = 0. Among the given points \\((1,0)\\) and \\((4,0)\\) have y = 0, so the zeroes are 1 and 4. Option B is incorrect because \\((0,6)\\) has y = 6 (an y-intercept), not y = 0. Exam tip: always check the y-coordinate — only points with y = 0 give zeroes.

Correct answer: A. (-3)

Explanation: The direct answer is A: -3. The graph cuts the x-axis at \((-7,0)\) and \((-3,0)\), so the zeroes are -7 and -3, obtained from the first coordinates. To compare negative numbers, imagine the number line: numbers farther to the right are greater. Since -3 lies to the right of -7, \(-3>-7\). Option A is correct. Option B, -7, is the smaller zero, not the greater one. Option C, 0, is not listed as an x-intercept, so it is not a zero from the given information. Option D, 7, changes the sign of -7 and is not a given zero. A common mistake is thinking that the negative number with the larger absolute value is greater; in fact, -7 is less than -3. Remember: among negative numbers, the one closer to zero is greater.

Correct answer: A. (-3, 0) and (3, 0)

Explanation: X-intercepts occur where p(x)=0. Solve \\(x^2-9=0\\). Factor as \\(x^2-9=(x-3)(x+3)\\), giving \\x=\pm3\\. Therefore the x-intercepts are \\(-3,0\\) and \\(3,0\\). Option B is incorrect: (0,-9) is the y-intercept since \\(p(0)=-9\\), and \\(p(9)=81-9=72\\) so (9,0) is not a root. Exam tip: set p(x)=0 and use difference of squares to factor quickly.

Correct answer: A. It touches the x-axis at the origin \(x=0\)

Explanation: Solve \(x^2=0\) to find zeros: the only root is \(x=0\), with multiplicity 2. A repeated root of even multiplicity means the parabola touches (is tangent to) the x-axis at that point and does not cross it. Option B is incorrect because two distinct real roots would be required to cut the axis at two points (discriminant > 0), which is not the case here. Option D is wrong since \(x=1\) is not a root. Exam tip: check factorization or discriminant and note root multiplicities to determine whether the graph crosses or merely touches the x-axis.

Correct answer: A. (2, 0) and (-5, 0)

Explanation: The zeros come from setting each linear factor to zero: x-2=0 gives x=2 and x+5=0 gives x=-5. x-axis intercepts have coordinates (x,0), so the graph crosses at (2,0) and (-5,0). Option B simply reverses the signs; options C and D are not x-axis intercepts. Exam tip: set each factor equal to 0 and write the intercepts as (x,0).

Correct answer: A. To know the number of real zeroes

Explanation: The x-axis is the set of points whose y-coordinate is zero. Therefore, when a graph y = p(x) intersects the x-axis, the corresponding x-coordinate satisfies p(x) = 0 and is a real zero. Counting distinct x-axis intersections consequently gives the number of distinct real zeroes visible from the graph. It does not always determine the degree: a polynomial may have a higher degree than the number of real intersections. It also does not immediately give the constant term, which is related to the y-intercept p(0), not the x-intercepts. Colour has no mathematical role here. Thus Option A states the correct use.

Correct answer: A. It has one real zero, −8

Explanation: A real zero is read from the x-coordinate of an x-axis intersection. The graph has exactly one such intersection, so it has exactly one distinct real zero. Because the intersection occurs at x = −8, the zero is −8, not 8. The minus sign is part of the coordinate and cannot be discarded. Option B confuses the numerical magnitude 8 with the number of zeroes; one intersection does not mean eight zeroes. Option D contradicts the stated intersection. Therefore Option A is correct. This conclusion concerns distinct real zeroes visible on the graph and does not claim anything about possible complex zeroes.