Geometrical meaning of the zeroes of a polynomial.

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.

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.

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).

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.

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.

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.

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).

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.

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.

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 \((-b/a,0)\).

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.

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.

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.

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.

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.

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

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.

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.

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.

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.

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

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.

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.

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.

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.