Geometrical meaning of the zeroes of a polynomial.

The direct answer is A: one distinct zero. A distinct zero means a different x-value, not the number of times that value is written or the multiplicity of a root. Both listed points are exactly (4,0), so both refer to the same location and the same x-value x=4. Therefore there is only one distinct real zero. Option A is correct. Option B, two, incorrectly counts the repeated writing as two different zeroes. Option C, four, has no mathematical basis and may confuse the coordinate value 4 with the number of zeroes. Option D, zero, is wrong because (4,0) lies on the x-axis, so p(4)=0. The word distinct is important: repeated information is counted once. A graph may touch the x-axis at x=4 without crossing it, and it is still a zero.

Zeros (roots) occur where \(p(x)=0\). Solving \(x^2-1=0\) gives \(x=\pm1\), so the points \((-1,0)\) and \((1,0)\) are zeros on the graph. The point \((0,-1)\) has \(y=-1\) (i.e. \(y\neq0\)), so it is not a root. The closest distractor D listing both points is incorrect because those two points do represent zeros. Exam tip: roots correspond to x‑intercepts where \(y=0\).

On the number line a number located to the right is greater. -2 lies to the right of -9, so -2 is the greater zero. -9 is incorrect because it is further left (smaller) than -2. Exam tip: among negative numbers the one with the smaller absolute value is the greater number (e.g. |−2|=2 < |−9|=9).

The zeros (roots) are the x-coordinates where the graph meets the x-axis: here they are \(-10\) and \(0\). Thus the product is \((-10)\times 0 = 0\). Option A (-10) is incorrect because it uses only one root; option D (100) is wrong — it seems to confuse absolute values or squaring. Exam tip: if one root is \(0\), the product of the roots is always \(0\).

Core idea: \(p(x)=5\) is a nonzero constant polynomial, so its graph is the horizontal line \(y=5\). This line never meets the \(x\)-axis, hence it is parallel to the \(x\)-axis and does not cut it. Why the closest distractor is wrong: option A (cuts once) would apply to a linear polynomial with a real root; here \(p(x)\) never equals 0, so no intersection. Exam tip: a zero of a polynomial is an \(x\) with \(p(x)=0\); nonzero constant polynomials have no real zeros.

The zero polynomial satisfies p(x)=0 for every real x, so the graph consists of all points (x,0) — i.e. the x-axis. B is wrong because the y-axis is the vertical line x=0, not y=0 for all x. C is wrong because "above the x-axis" means y>0, but here y=0. D is wrong because the graph does touch the x-axis at every point (it is the x-axis). Exam tip: check p(x) for an arbitrary x; if p(x)=0 always then the graph is y=0. Note: the degree of the zero polynomial is not normally defined (sometimes taken as −∞).

If \(p(a)=0\) then the graph passes through the point \((a,0)\). Since the zeros 2, 5 and 9 are distinct, there are three distinct x‑intercepts. The closest distractor 'Two' would only be correct if two of the zeros coincided; 'Nine' confuses the value 9 with the count of intercepts. Exam tip: always check whether zeros are distinct — repeated roots (multiplicity) still produce a single x‑intercept.

The axis of symmetry passes through the midpoint of zeroes, (\frac{-2+6}{2}=2). Tip: the average of two zeroes is useful for a parabola.

The direct answer is option C: it has no real zeroes. Rewrite the polynomial by completing the square: x² + 2x + 5 = (x + 1)² + 4. For every real x, (x + 1)² is at least 0. Therefore (x + 1)² + 4 is at least 4, so it is always positive and can never equal 0. Hence the graph never meets the x-axis and there are no real zeroes. Option A is wrong because two real zeroes would require two x-values making the expression 0. Option B is wrong because even one real zero is impossible; the expression is always at least 4. Option C is correct. Option D is wrong because not every number is a zero; in fact, no real number is a zero. The graph is an upward-opening parabola with its lowest point at x = −1 and y = 4, safely above the x-axis. Memory cue: completing the square and obtaining a positive constant shows no real x-axis intersection.

A number \(a\) is a zero (root) of a polynomial if and only if \(p(a)=0\). Here \(p(3)=0\), so 3 is a zero and the graph meets the x-axis at x=3. Since \(p(7)=4\neq0\), 7 is not a zero. Closest distractors (B and C) are wrong because they assume \(p(7)=0\). Exam tip: always check the function value equals exactly 0 before calling a point a root.

Zeros are the x-values that make the polynomial zero. Solve \(2x^2-8=0\). Divide both sides by 2 to get \(x^2=4\), so \(x=\pm2\). Hence the zeros are (2, -2). The closest distractor (4, -4) would arise from the mistake of not dividing by 2 and assuming \(x^2=16\). Exam tip: factor first: \(2(x^2-4)=2(x-2)(x+2)\) to read off the zeros quickly.

The first coordinate (r) of the intersection is the zero, and (r<0). Tip: the same rule works in symbolic questions.

The direct answer is A: the graph touches the x-axis at x=-3. To find a zero, set p(x) equal to zero: -(x+3)^2=0. Multiplying by -1 does not change the solutions, so (x+3)^2=0. A square is zero only when its base is zero; hence x+3=0 and x=-3. Because the factor is squared, the graph touches the x-axis there rather than crossing it in the usual quadratic shape. Option A is correct. Option B, x=3, has the wrong sign; substituting 3 gives -36, not 0. Option C is wrong because there is only one distinct solution, even though it is repeated algebraically. Option D is wrong because p(-3)=0, so the graph does meet the x-axis. The outside negative sign changes the opening direction, not the zero location.

Points where the graph meets the x-axis are the polynomial's zeros, so the product is the product of the x-values: (-1)×2×4 = -8. Option A (8) would be the result if the negative sign were ignored; options C and D come from incorrect arithmetic. Exam tip: multiply the x-intercepts directly and pay attention to signs (negative × positive = negative).

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.

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.

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.

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.