Geometrical meaning of the zeroes of a polynomial.

Correct answer: B. -4

Explanation: The zeroes of a polynomial are the x-coordinates where its graph meets or cuts the x-axis. Here, the zeroes are -12 and -4. On the number line, -4 lies to the right of -12, so -4 is greater. Equivalently, among negative numbers, the number with the smaller absolute value is greater. Therefore, option B is correct; 0 and 12 are not the given zeroes.

Correct answer: A. (9, 0)

Explanation: If x = 9 is a zero of p(x), then p(9) = 0. On the graph this corresponds to a point on the x‑axis with y = 0, namely (9, 0). Option D has y = 0 but x = -9, so it does not represent x = 9. Options B and C have y ≠ 0, so they cannot be zeros. Exam tip: For a zero x = a of a polynomial, the graph passes through (a, 0) — check the x‑coordinate equals the given zero and y = 0.

Correct answer: C. 5

Explanation: Point \\((5,0)\\) has y=0, so it's an x-intercept and therefore \(x=5\) is a zero of the polynomial/function. The point \\((2,3)\\) has y=3 (y\neq0), so \(x=2\) is not a zero. Exam tip: an x-value is a zero only when the corresponding y-coordinate equals 0 (i.e., the graph meets the x-axis).

Correct answer: A. 2 and −3

Explanation: The governing concept is that zeroes are the values of x that make the polynomial equal to zero. Factor the expression by finding two numbers whose product is −6 and whose sum is 1: these are 3 and −2. Thus x² + x − 6 = (x + 3)(x − 2). Applying the zero-product property, either x + 3 = 0, giving x = −3, or x − 2 = 0, giving x = 2. Therefore the real zeroes are 2 and −3, so option A is correct. Option B would result from changing both signs, option C confuses the constant and coefficient with roots, and option D is false because two real solutions have been obtained directly by factorisation.

Correct answer: B. \(0\)

Explanation: The x-intercepts are the polynomial's zeroes, here \(a,\;0,\;-a\). Their sum is \(a+0+(-a)=0\). The distractor \(2a\) might come from adding absolute values incorrectly; \(-a\) is only one root, not the sum. Exam tip: opposite numbers cancel, so pairs like \(a\) and \(-a\) give zero when summed.

Correct answer: B. (4)

Explanation: The direct answer is B: x=4. A graph meets the x-axis exactly where its y-value, p(x), is zero. The given values are p(3)<0, p(4)=0, and p(5)<0. Only the middle statement gives an output of zero, so the point (4,0) lies on the x-axis. Option B is correct. Option A is wrong because p(3) is negative, not zero; its point lies below the x-axis. Option C is wrong for the same reason: p(5)<0 means the point lies below the x-axis. Option D is wrong because one of the listed values, p(4), does equal zero. A negative value is below the x-axis, while a zero value is on it. Do not confuse “less than zero” with “equal to zero”; only equality identifies a zero.

Correct answer: A. There are two distinct real zeroes

Explanation: If the graph crosses the x-axis at two distinct points, each crossing corresponds to a different real root of the polynomial. Crossing typically indicates an odd multiplicity (often 1) at that root. Choice B is incorrect because a repeated root refers to the same x-value occurring multiple times (leading to touching or flattened behavior at one point), not two separate intersection points. C and D are clearly wrong: no real roots would give no intersections, and every point being a root would mean the polynomial is identically zero. Exam tip: count distinct intersection x-values for number of distinct real roots; check whether the graph crosses (odd multiplicity) or merely touches (even multiplicity).

Correct answer: A. Because (x^2+1) is always positive

Explanation: The direct answer is option A: x^2+1 is always positive for real x, so the graph never reaches the x-axis. For every real x, x^2 is at least 0. Adding 1 gives x^2+1 at least 1, which is strictly greater than 0. Therefore p(x) can never equal 0, so there is no real zero and no x-axis intersection. Option A is correct. Option B is false because x^2 is never negative for real x. Option C is false: substituting x=1 gives p(1)=2, not 0, so 1 is not a zero. Option D is false because x^2+1 is a quadratic polynomial, not a linear one. The graph is an upward-opening parabola whose lowest y-value is 1. Memory cue: a graph cuts the x-axis only when its y-value becomes exactly zero.

Correct answer: C. -25

Explanation: The x-coordinates of the x-intercepts are the zeroes of the polynomial, here −5 and 5. Their product is (−5)×5 = −25. Option B (25) ignores the negative sign; A (0) would be correct only if one root were 0; D (10) is not the product. Exam tip: always account for signs when multiplying roots — opposite signs give a negative product.

Correct answer: B. (5,0)

Explanation: The x-intercept occurs where the polynomial equals zero, i.e. \(p(x)=0\). Solving \(3x-15=0\) gives \(x=5\), so the intercept is \((5,0)\). Option \((0,-15)\) is the y-intercept (when \(x=0\), \(p(0)=-15\)), not the x-intercept. Options \((3,0)\) and \((-5,0)\) are simple arithmetic/sign errors — \((3,0)\) would arise from an incorrect solution and \((-5,0)\) flips the sign. Exam tip: for a linear polynomial \(ax+b\), the x-intercept is \((-b/a,0)\).

Correct answer: B. One

Explanation: A touch at (m,0) means the polynomial satisfies \(p(m)=0\). If the graph only touches and does not cross the x-axis at that point, the root at x = m has even multiplicity (2, 4, ...). Regardless of that multiplicity, it represents a single distinct real zero — x = m. The distractor 'two' is incorrect because two distinct real zeros would require the graph to meet the x-axis at two different x-values. Exam tip: check \(p(m)=0\) and use root multiplicity (even → touch, odd → cross) to decide crossing vs touching.

Correct answer: C. Three

Explanation: The direct answer is option C: there are three distinct x-axis intersections. To find them, set p(x)=0: (x-2)(x-5)(x-8)=0. A product is zero if at least one factor is zero, so x-2=0, x-5=0, or x-8=0. These give x=2, 5, and 8. The corresponding points are (2,0), (5,0), and (8,0), all different. Option A is wrong because there are not just one but three roots. Option B is wrong because no root is repeated and no two values coincide. Option C is correct. Option D is wrong because the degree is 3 and the polynomial has three distinct real roots, not six intersections. A repeated root would still represent only one distinct point. Exam cue: count distinct solutions of p(x)=0, not the total number of factors written with repetition.

Correct answer: A. 0 is also a zero

Explanation: The governing concept is the relationship between an intercept and a polynomial value. The y-axis intercept is obtained by putting x = 0, so a graph passing through (0, 0) means p(0) = 0. By definition, this makes 0 a zero of the polynomial. The same point is also on the x-axis, because every point with y-coordinate 0 lies on that axis; therefore it is an x-axis intersection as well. Hence option A is correct. The conclusion does not say that every number is a zero, since only x = 0 has been established. It also cannot imply that there is no zero or that only 1 is a zero. Other zeroes may exist, but they cannot be determined from this single intercept alone.

Correct answer: A. 6

Explanation: The governing concept is that repeated roots are counted once when the question asks for distinct zeroes. Recognise the perfect-square trinomial: p(x) = x² − 12x + 36 = (x − 6)². Setting p(x) equal to zero gives (x − 6)² = 0, so x − 6 = 0 and x = 6. Both algebraic roots coincide, meaning the polynomial has one distinct real zero, 6, with multiplicity two. Graphically, the parabola touches the x-axis at the single point (6, 0) rather than crossing it at two different points. Therefore option A is correct. The value −6 has the wrong sign, 12 is the coefficient magnitude rather than the root, and option D is false because x = 6 is a valid real solution.

Correct answer: B. −42

Explanation: The governing concept is that the x-coordinate of every x-axis intersection is a zero of the polynomial. The three zeroes are therefore −2, 3, and 7. Their product is calculated using only the x-values: (−2) × 3 × 7 = (−6) × 7 = −42. Thus option B is correct. The negative sign is essential because there is one negative factor and two positive factors, so the final product is negative. Option A has the correct magnitude but the wrong sign. Option C is the sum of the three zeroes, since −2 + 3 + 7 = 8, not their product. Option D does not result from multiplying the listed zeroes. The y-coordinate 0 is not included in the product.

Correct answer: C. There is no real solution

Explanation: (p(x)=4) is never (0), so the graph does not cut the (x)-axis. Tip: a non-zero constant polynomial has no zero.

Correct answer: B. Every real x is a zero

Explanation: A zero of p(x) is a value of x for which p(x) equals 0. For the zero polynomial, p(x) is identically 0, so substituting any real number gives p(x) = 0. Thus every real number is a zero, and the polynomial has infinitely many real zeroes. It is not restricted to 0 or to positive values, so option B is correct.

Correct answer: B. 8

Explanation: The zeroes are the x-coordinates where the graph meets the x-axis, so take the average of those x-values. Average = \(\dfrac{2+14}{2}=8\). Option A (6) is incorrect because 6 would be the average of 2 and 10, not of 2 and 14. Exam tip: for x-intercepts, average their x-coordinates: add them and divide by 2.

Correct answer: A. (1) and (-4)

Explanation: Core idea: zeros occur when the product equals zero, so at least one factor must be zero. From \(-(x-1)(x+4)=0\) the leading minus sign does not change the roots; it only flips the overall sign. Set each factor to zero: \(x-1=0\) gives \(x=1\), and \(x+4=0\) gives \(x=-4\). Choice B is wrong because the signs are reversed; C and D are not obtained from the given factors. Exam tip: to find x-intercepts of a factored polynomial, set each factor equal to zero and solve.

Correct answer: A. The graph will pass through (−6, 0), (0, 0), and (5, 0)

Explanation: The governing concept is the coordinate form of a function graph: every input a is represented by the point (a, p(a)). If p(a) = 0, that point becomes (a, 0), which lies on the x-axis and shows that a is a zero. Applying this rule gives p(−6) = 0 → (−6, 0), p(0) = 0 → (0, 0), and p(5) = 0 → (5, 0). Therefore option A is correct. Option B places the given inputs in the second coordinate and incorrectly treats them as y-values. Option C contradicts the three stated zeroes, while option D ignores −6 and 0. The statements establish at least these three x-axis points, whether or not additional zeroes exist.

Correct answer: B. Two

Explanation: The zeroes are (-3) and (1), so there are two distinct meeting points. Tip: count the repeated zero (1) only once for distinct points.

Correct answer: B. Two

Explanation: The governing concept is that a real zero of a polynomial is an x-coordinate at which its graph meets the x-axis. The listed points have x-coordinates 1, 1, and 4. Although the point (1,0) appears twice, repetition does not create a new distinct zero; it only indicates that the same zero may have repeated multiplicity. Therefore the distinct x-values are 1 and 4, giving exactly two distinct real zeroes. Option A is incorrect because there are two different x-values, while Option C incorrectly counts the repeated listing separately. Option D has no basis in the data.

Correct answer: A. It is the midpoint of the two zeroes

Explanation: The relevant concept is the midpoint of two numbers on the x-number line. The two x-intercepts have x-coordinates 2 and 8. Their midpoint is calculated by averaging them: (2+8)/2 = 10/2 = 5. Thus x=5 lies exactly halfway between the two given zeroes. This calculation only identifies its position; it does not prove that p(5)=0, because a polynomial may take a nonzero value between its zeroes. Therefore Option A is correct. Option B confuses a midpoint with a zero, Option C refers to a point on the y-axis where x=0, and Option D is false because 8, not 5, is the greater zero.

Correct answer: A. It will touch at x=-3

Explanation: The governing idea is that the real zeroes determine where a polynomial graph meets the x-axis, and a repeated zero generally means the graph touches the axis without crossing it. Factor the polynomial: 2x^2+12x+18 = 2(x^2+6x+9) = 2(x+3)^2. Hence the only zero is x=-3, with multiplicity two. Since the square is never negative and the leading factor is positive, the graph has its minimum value 0 at x=-3 and touches the x-axis there. Option A is correct. Option B has the wrong sign, Option C would require two distinct real zeroes, and Option D ignores the real repeated zero.

Correct answer: A. 1

Explanation: The governing concept is that polynomial zeroes are the x-coordinates where the graph intersects the x-axis. Factor the quadratic by finding two numbers with product 30 and sum 11: x²−11x+30=(x−5)(x−6). Thus the zeroes are 5 and 6, corresponding to points (5,0) and (6,0). Because both points lie on the x-axis, their distance is the absolute difference of their x-coordinates: |6−5|=1 unit. Therefore option A is correct. The value 5 is one zero, 11 is the sum of the zeroes, and 30 is their product; these are related quantities but none is the requested distance.