Geometrical meaning of the zeroes of a polynomial.

Zeros (roots) are the x-coordinates where the graph meets the x-axis, here they are \(a\) and \(-a\). Their sum is \(a + (-a) = 0\). The nearest distractor \(2a\) is wrong because it corresponds to adding magnitudes \(a + a\), not \(a + (-a)\). Exam tip: when roots are opposites (symmetric about origin), their sum is always 0 — spot symmetry to answer quickly.

The governing concept is the meaning of distinct zeroes. Distinct means that each different numerical value is listed only once. The stated zeroes are 2, 2, and −5. Although 2 occurs twice, both occurrences represent the same x-coordinate and the same x-axis point (2, 0); the repetition may indicate multiplicity, but it does not create a new distinct zero. The value −5 gives another point, (−5, 0). Therefore the distinct real zeroes are 2 and −5, so option A is correct. Option C lists repeated occurrences rather than distinct values. Option B omits −5, and option D omits 2. Thus the repeated entry must be removed only for the purpose of listing distinct zeroes.

Answer: A, −1, 0, and 1. A graph meets the x-axis where its y-coordinate is zero. Since y = p(x), set p(x) equal to zero: x^3 − x = 0. Take x common to obtain x(x^2 − 1) = 0. Now use the difference of squares: x^2 − 1 = (x − 1)(x + 1). Thus x(x − 1)(x + 1) = 0, so x = 0, x = 1, or x = −1. The intersection points are therefore (−1,0), (0,0), and (1,0). Option A lists all three values. Option B wrongly includes 3, which is not a zero. Option C changes 1 and −1 to 3 and −3. Option D ignores two valid zeroes. Memory cue: x-axis means y=0, so solve p(x)=0.

If a graph touches the x‑axis at a point, the function value there is zero, so the corresponding x is a real root. Therefore x=11 is a real zero (usually with even multiplicity). Option B is wrong because touching implies f(11)=0; options C and D are incorrect because they give different x‑values. Exam tip: touching the x‑axis at x=a means x=a is a root (often of even multiplicity).

Answer: A, (3,0) and (4,0). At an x-axis intersection, y=0, so solve p(x)=0: x^2 − 7x + 12 = 0. We need two numbers whose product is 12 and whose sum is 7; they are 3 and 4. Therefore x^2 − 7x + 12 = (x−3)(x−4). Setting each factor to zero gives x=3 or x=4. The corresponding points are (3,0) and (4,0). Option A is correct. Option B uses the wrong signs: substituting −3 or −4 does not make the polynomial zero. Option C has x=0, so those are points on the y-axis, not the x-axis. Option D confuses the coefficient 7 and constant term 12 with the roots. Memory cue: x-axis points always have the form (root,0).

The zeros are the x-coordinates of the x-intercepts, namely 3 and 12. The average of two numbers is half their sum, so \\(\frac{3+12}{2}=7.5\\). Option A (15) is the sum, not the average. Option B (9) is an incorrect calculation. Option D (4.5) equals half the distance between the roots \\(\frac{12-3}{2}\\), not their midpoint. Exam tip: the average of two roots equals the midpoint of their x-coordinates on the x-axis.

Answer: B. The graph meets the x-axis at the points (−5,0) and (5,0). The y-axis is the line x=0. The perpendicular distance of a point (x,y) from the y-axis is |x|. Hence the two distances are |−5|=5 and |5|=5, so the zeroes are equally distant from the y-axis. Option A is false because a point on the y-axis must have x=0, not x=−5 or 5. Option C is false because −5 lies to the left of the y-axis, while 5 lies to its right. Option D is also false for the same reason. The opposite signs show reflection across the y-axis. Memory cue: equal absolute x-values mean equal distance from the y-axis.

A point \((a,0)\) lies on the x-axis exactly when \(p(a)=0\), because the y-value is zero there. Given \(p(2)=0\), the point \((2,0)\) is on the x-axis. The conditions \(p(1)<0\) and \(p(3)>0\) only give the sign of the function at those x-values, not roots, so \((1,0)\) and \((3,0)\) are not x-intercepts. \((0,2)\) has y=2, not zero. Exam tip: verify roots by checking where \(p(x)=0\) rather than relying on inequalities indicating sign changes.

The zeroes are (-6), (-2), (4), and (4) is the greatest. Tip: the value to the right on the number line is greater.

The three different factors give three distinct zeroes (1), (4), (-2). Tip: each linear factor gives a possible intersection.

Real zeros of a polynomial are the x-values where the graph meets the x-axis, i.e. where \(y=0\). Among the given points only \((-4,0)\) and \((2,0)\) have \(y=0\), so the zeros are \(-4\) and \(2\). Closest distractor B is wrong because \((0,3)\) has \(y=3\) and \((5,-1)\) has \(y=-1\); neither lies on the x-axis. Exam tip: pick x-coordinates only from points whose y-coordinate equals zero.

The zeroes of the quadratic polynomial are -5 and 3. Their average is calculated by adding the two zeroes and dividing by 2: (-5 + 3)/2 = -2/2 = -1. Geometrically, this average is the x-coordinate of the midpoint of the two x-intercepts and the axis of symmetry of the parabola. Therefore, -1 is correct; 1, -2, and 8 result from incorrect arithmetic or from using the difference instead of the average.

The graph meets the x-axis when the polynomial value is zero. In factored form, p(x) = (x + 2)(x − 7), so the product is zero when either factor is zero. From x + 2 = 0, we get x = −2; from x − 7 = 0, we get x = 7. The corresponding x-axis points must have y-coordinate 0, giving (−2, 0) and (7, 0). Therefore option B is correct. Option A reverses the signs of both roots. Option C writes the values as y-coordinates on the y-axis, and option D pairs the roots with nonzero y-coordinates, so neither represents x-axis intersections.

A crossing at a point means the zero has odd multiplicity and a touch means even multiplicity, but in both cases the polynomial satisfies \(p(x)=0\). The x‑values given are \(x=1\) and \(x=6\), so there are two distinct real zeros. The distractor “three” is incorrect because multiplicities (odd/even) do not create additional distinct x‑values. Exam tip: count distinct x‑values where the graph meets the axis; use crossing vs touching only to infer multiplicity, not the number of distinct zeros.

The zeroes are (-9), (-1), (4) and the greatest is (4). Tip: the value to the right on the number line is greater.

Intersections with the x-axis occur where \(p(x)=0\). Since \(p(-3)=0\) and \(p(8)=0\), the x-intercepts are (-3,0) and (8,0). Because \(p(2)=5\) ≠ 0, the point (2,5) is not an x-intercept. Exam tip: to find x-intercepts from function values, look for input values that give output 0 (roots).

Factorize: \(p(x)=x^2-10x+25=(x-5)^2\). The root \(x=5\) has multiplicity 2 (a repeated zero), so the parabola touches the x-axis at a single point and does not cross it. Closest distractor C is incorrect because cutting at two distinct points requires two distinct real roots (discriminant > 0). Exam tip: if the discriminant \(b^2-4ac=0\), the quadratic has a repeated root and the graph is tangent to the x-axis.

The governing concept is that the real zeroes of p(x) are the x-values where the graph meets the x-axis. For the quadratic p(x) = x² + 6x + 10, complete the square: p(x) = (x + 3)² + 1. Since (x + 3)² is always at least 0, p(x) is always at least 1, so it can never equal 0 for any real x. Equivalently, its discriminant is b² − 4ac = 36 − 40 = −4, which is negative, confirming that there are no real roots. Therefore option C is correct. Option A would require a positive discriminant, option B would require a zero discriminant, and option D is impossible for a nonzero quadratic.

A root (zero) is an x‑value where the function equals zero (the point lies on the x‑axis). At (0, −6) the y‑value is −6, so the point is a y‑intercept (x = 0, y ≠ 0), not an x‑intercept. Hence −6 is the y‑value, not a root. Why other choices are wrong: B is incorrect because sign alone does not prevent a number being a root — roots can be negative, positive or zero. C is incorrect because x = 0 is not automatically a root; only if f(0) = 0. D is false because not every point gives y = 0. Exam tip: always check which coordinate is x and which is y; zeros correspond to x‑coordinates where y = 0 (x‑axis intersections).

The distance is the absolute difference of their x-coordinates. Calculate as \\(|9-(-4)|=|13|=13\\). Options A and B are incorrect because they do not give the correct difference; option D is incorrect because distance cannot be negative. Exam tip: use the absolute value of the difference of coordinates to find distance on an axis.