Geometrical meaning of the zeroes of a polynomial.

The point (0,0) means that when \(x=0\), \(y=p(0)=0\). Roots of a polynomial are the x-values for which \(p(x)=0\). Hence \(x=0\) is indeed a root. The closest distractor is C — claiming it is only a y-intercept — which is incorrect here because (0,0) lies on the x-axis as well, so it gives an x-intercept. Exam tip: substitute the x-coordinate of the given point into \(p(x)\); if it yields zero, that x is a root.

(x^2+14x+49=(x+7)^2), so the touching point is ((-7,0)). Tip: in a perfect square, change the sign to get the zero.

The zeroes are (m), (n), (r), so the mean is (\frac{m+n+r}{3}). Tip: take the first coordinate even in symbolic points.

((x-4)^2+4) is always positive, so (p(x)=0) will not occur. Tip: adding a positive number to a square can prevent intersection.

((x+4)^2) is an even-power factor, so the graph touches at (x=-4). Tip: power (2) shows a repeated zero.

The axis of symmetry is at the average of the zeroes, (\frac{(c-7)+(c+3)}{2}=c-2). Tip: the average rule also works with symbols.

A zero r corresponds to the x-intercept (r,0) because the y-coordinate is zero at an x-axis crossing. Thus zeros -6, 2 and 9 give (-6,0), (2,0) and (9,0). Common mistake: option A swaps coordinates, placing the zero as a y-value instead of x. Exam tip: write each zero as the x-coordinate (r,0) and quickly plot to verify.

In this interval the first two factors are positive and the third is negative, so the product is negative. Tip: check factor signs separately.

In the quadratic, the sum of zeroes is (9), so the other zero is (5). Tip: quickly convert a zero to ((x,0)).

Direct answer: Option B. The x-axis crossings give zeroes x=-8 and x=8. These numbers are opposites because 8=-(-8), and their sum is (-8)+8=0. Option A is wrong: the two numbers are not equal. Option B is correct: they are opposite zeroes and their sum is zero. Option C is wrong because 8 is positive, so both cannot be negative. Option D is wrong because the product is (-8)(8)=-64, not +64. Geometrically, the two intercepts are equally far from the y-axis, at distances 8 on opposite sides. The word “cuts” confirms that the graph meets the x-axis, so both listed x-values are zeroes. Exam cue: for roots -a and a, add them to get zero; do not forget the negative sign when multiplying.

x-axis intersections are the points where \(p(x)=0\). Solve \(16x^2-9=0\). Factor as \((4x-3)(4x+3)=0\), giving \(4x-3=0\) or \(4x+3=0\), hence \(x=\pm\tfrac{3}{4}\). Thus the intersections are \(\left(\tfrac{3}{4},0\right)\) and \(\left(-\tfrac{3}{4},0\right)\). Why other options fail: (C) \(\tfrac{4}{3}\) is the reciprocal error from mishandling \(4x\); (B) \(\pm3\) ignores the coefficient 16. Exam tip: view \(16x^2\) as \((4x)^2\) and use difference of squares to factor quickly.

The direct answer is B: two distinct real zeroes. A zero is an x-value where the graph has y=0. The listed points are (-5,0), (-5,0), and (4,0). The x-value -5 is repeated, so it represents one distinct zero, not two. The other x-value is 4. Thus the distinct zero set is {-5,4}, containing two values. Option B is correct. Option A, one, ignores the different value 4. Option C, three, counts the three written entries rather than distinct x-values. Option D, four, is unrelated and may result from confusing the coordinate value 4 with the number of roots. The y-coordinate 0 only confirms that each point is on the x-axis. When a question says distinct, remove duplicates before counting.

The x-intercepts give the zeroes of the monic quadratic: -6 and 2. A quadratic with these zeroes can be written as p(x) = (x + 6)(x - 2). Expanding gives x^2 - 2x + 6x - 12 = x^2 + 4x - 12. Comparing this with the standard form x^2 + px + q, the coefficient of x is p = 4 and the constant term is q = -12. Therefore option A is correct. The sum of the zeroes is -4, which equals -p, so p must be 4; confusing the sum directly with p leads to option B. The product is -12, not 12, ruling out option C, and option D has an incorrect coefficient.

Real zeroes are counted from (x)-axis intersections, not from the (y)-axis intercept. Tip: ((0,18)) does not show a zero.

The direct answer is option A: the real x-axis intersections are (-3,0) and (3,0). Set x^4-81=0 and use difference of squares: x^4-81=(x^2-9)(x^2+9)=(x-3)(x+3)(x^2+9). The first two factors give x=3 and x=-3. The last factor gives x^2=-9, which has no real solution. Hence the real intersections are exactly (-3,0) and (3,0). Option A is correct. Option B incorrectly uses ±9 instead of the square roots ±3. Option C wrongly includes (0,0); p(0)=-81, so the origin is not on the graph. Option D is wrong because two real roots do exist. The word real is important: complex roots from x^2+9=0 are not real x-axis points. Memory cue: for x^4=a^2, real solutions are x=±√a when the factor permits them.

For six distinct real zeroes, the degree must be at least (6). Tip: the number of distinct zeroes cannot exceed the degree.

Direct answer: Option C, the graph does not cut the x-axis. A zero of a polynomial is an x-value for which the y-value is zero, so it gives a point where the graph meets the x-axis. Rewrite the polynomial: \(p(x)=x^2-2x+10=(x-1)^2+9\). Since \((x-1)^2\geq0\), the smallest possible value of p(x) is 9, which is above zero. Therefore \(p(x)=0\) has no real solution and the parabola has no x-intercept. Option A is wrong because two crossings would require two distinct real zeroes. Option B is wrong because touching once would require one repeated real zero, usually when the minimum is exactly zero. Option C is correct because the whole graph lies above the x-axis. Option D is wrong because the graph is a parabola, not the y-axis. Memory cue: rewrite a quadratic as a square plus a positive number; if it is always positive, there is no x-axis intersection.

Direct answer: Option A, the zeroes are 0, 3 and 5. A zero is an x-value that makes the polynomial equal to zero. First take the common factor x: \(p(x)=x^3-8x^2+15x=x(x^2-8x+15)\). Now factor the quadratic: \(x^2-8x+15=(x-3)(x-5)\), because 3+5=8 and 3 times 5=15. Thus \(p(x)=x(x-3)(x-5)\). A product is zero when at least one factor is zero, so x=0, x=3 or x=5. These are also the x-coordinates of the graph’s x-axis intersections. Option A is correct. Option B changes the signs and does not give the original factorisation. Option C lists 8, which is a coefficient, not a zero; it also omits 0. Option D wrongly keeps only one zero and ignores the other factors. Exam cue: factor completely and set each factor equal to zero.

(12=\frac{6+18}{2}), so it is the midpoint of the two zeroes. Tip: the midpoint need not be a zero.

For (x<-9), both factors are negative and the outside negative makes the value negative. Tip: check factor signs first.

Zeroes are the x-values where the graph meets the x‑axis. From (7,0) and (−4,0) the zeroes are 7 and −4, so their sum is 7 + (−4) = 3. ‘‘Touches’’ indicates an even multiplicity and ‘‘crosses’’ an odd multiplicity, but the zero values remain 7 and −4 — standard exam questions treat these zeroes once each unless multiplicity is explicitly requested. Closest distractor: −3 is simply the wrong sign; 11 and −11 arise from arithmetic errors. Exam tip: count every x‑intercept as a zero; only account for multiplicity if the question asks for sum with multiplicity.

The x-intercepts give the roots, so the zeros are 0 and d. Their product is \(0\times d=0\). Closest distractors: option A (d) is just one root, not the product; option D (d^2) arises from mistakenly multiplying d by d; option C (-d) has the wrong sign. Exam tip: whenever one root is 0, the product of the roots is immediately 0.

For a downward-opening parabola, values outside the zeroes are negative. Tip: opening direction changes sign regions.

For real (x), (x^8\geq0), so (x^8+1>0). Tip: an always positive polynomial does not cut the (x)-axis.

A zero (root) of a polynomial is an x-value where \(p(x)=0\). Here \(p(-9)=0\), \(p(2)=0\) and \(p(6)=0\), so there are three zeros among the given x-values. \(p(-4)=2\) is not zero, so it is not a root. Exam tip: always check the function value equals zero — other values (like 2) do not count as zeros.