Geometrical meaning of the zeroes of a polynomial.

(18=\frac{10+26}{2}), so it is the midpoint of the two zeroes. Tip: the midpoint value need not be a zero.

For (x<-13), both factors are negative and the outside negative makes the value negative. Tip: check factor signs first.

A touching point is still a zero, so the zeroes are 11 and −6. Their product is 11 × (−6) = −66. The closest distractor 66 fails because it ignores the negative sign. Exam tip: "touches" means the x-value is a root (often with even multiplicity), but for product you simply multiply the zeros with their signs.

For a downward-opening parabola, values between the two zeroes are positive. Tip: (x=2) lies between the zeroes.

For real (x), (x^{12}\geq0), so (x^{12}+1>0). Tip: an always positive polynomial does not cut the (x)-axis.

A root (zero) is an x-value where \(p(x)=0\). Here \(p(-12)=0,\; p(4)=0,\; p(15)=0\), so there are three zeros. Since \(p(-2)=6\), \(-2\) is not a zero. The closest distractor (Four) is incorrect because it would wrongly include \(-2\) despite \(p(-2)\neq0\). Exam tip: count only those x for which the function value is exactly 0 — check each given pair carefully.

Factor: \(p(x)=x^2-hx=x(x-h)\). X-intercepts occur when \(p(x)=0\), so \(x=0\) or \(x=h\). Thus the intercepts are (0,0) and (h,0). Choice B is wrong due to sign (root is \(h\), not \(-h\)); choice D is invalid because (0,h) is not on the x-axis (y≠0). Exam tip: set \(p(x)=0\) or factor out the common \(x\); the roots give the x-coordinates of intercepts.

For equal distance from the (y)-axis, zeroes should be opposites, so (10) is needed with (-10). Tip: symmetric zeroes are (a) and (-a).

The zeroes are (-5) and (14), and ((x+5)^2) causes touching at (-5). Tip: the outside (11) does not change the zeroes.

In this interval the signs are (+), (-), (-), so the product is positive. Tip: the product of two negative factors is positive.

The discriminant is (g^2-4g^2=-3g^2<0), so there are no real zeroes. Tip: a negative discriminant means no (x)-axis intersection.

The vertex lies on the (x)-axis, so the parabola touches at ((-14,0)). Tip: if the vertex has (y=0), there is one distinct zero.

The discriminant is (400-460=-60), so there are no real zeroes. Tip: with negative discriminant a parabola does not meet the (x)-axis.

It is ((x-d)^2-36), so (x-d=\pm6) and the zeroes are (d-6), (d+6). Tip: use difference of squares.

The zeros are the x-values -4, 6 and 16. Mean = \(\dfrac{-4+6+16}{3}=\dfrac{18}{3}=6\). Option B (18) is the sum of the zeros, not the average. Option D (16) is just one root, not the mean. Exam tip: read the x-coordinates of intercepts first and then compute their average.

A zero of a polynomial is an x-value at which its graph meets the x-axis, because the y-value there is \(p(x)=0\). The question says that the graph cuts the x-axis at two different positions, \(x=-2\) and \(x=5\). Each position gives one zero, so the quadratic polynomial has two zeroes. These are distinct real zeroes because the two x-values are different.

This can also be understood from the factor form: a quadratic with these zeroes would be proportional to \((x+2)(x-5)\). The two factors become zero at \(-2\) and \(5\), respectively. Therefore the number of zeroes is two, and option A is correct. The answer is not one, because the graph has two separate x-intercepts; it is not three, because a quadratic polynomial can have at most two zeroes. “No zeroes” would apply only if the graph did not meet the x-axis.

If a polynomial's graph touches the x-axis at x=a without crossing, the root at a has even multiplicity. For example, (x-a)^2 touches but does not cross. For even multiplicity the sign of f(x) on both sides of a is the same, so the curve does not pass through the axis. Option B is incorrect because an odd multiplicity root (e.g., 1 or 3) causes the graph to cross the axis. Exam tip: factor or check derivatives — if f(a)=0 and f'(a)=0 (and higher derivatives as needed), the root is likely repeated (even multiplicity).

Such a line never meets the (x)-axis so it has no zero. First check the intercept from the graph.

Set p(x)=0: \(x^2-9=0\). Factor: \((x-3)(x+3)=0\) so \(x=\pm3\). Zeros of the polynomial appear on the x-axis as points \((x,0)\), therefore \((-3,0)\) and \((3,0)\) are the intercepts. Options B and D list points on the y-axis (x=0), not x-axis intercepts; option C mistakes the roots as ±9 instead of ±3. Exam tip: either factor the quadratic or take square roots (\(x^2=9\) gives \(x=\pm3\)) to find intercepts quickly.

When a quadratic graph touches at one point its two zeroes are equal. Treat it as a repeated zero in exams.

p(2)=0 means the function value at x=2 is y=p(2)=0, so the point (2,0) lies on the graph — i.e., the graph crosses the x-axis at (2,0). Option B is wrong because the y-axis is at x=0, not x=2. Options C and D are also incorrect: C asserts a vertex at (0,2) without justification, and D implies the graph is a vertical line parallel to x=2, which is not true for a typical polynomial function. Exam tip: p(a)=0 indicates a is a root and corresponds to the x-intercept (a,0) on the graph.

The distinct zeroes are only (-1) and (2). A touching zero may be repeated but its distinct value is counted once.

An x‑intercept is the point where the function value is zero (y=0). Set \(p(x)=0\): \(3x+6=0\) gives \(x=-2\), so the intercept is \((-2,0)\). Option (2,0) is a sign error; (0,6) is the y‑intercept, and (0,-2) is incorrect for both coordinates. Exam tip: find x‑intercepts by solving the polynomial equal to zero.

A zero (root) of a polynomial is the x-coordinate where the graph meets the x-axis (y=0). From the intercepts (-4,0), (1,0), (6,0) the x-values are -4, 1 and 6, so the set of zeroes is {-4,1,6}. The closest distractor C mixes a y-value (0) into the set in place of 1; remember 0 in the ordered pair is the y-coordinate, not a root by itself. Exam tip: read off only the x-coordinates of x-intercepts to list zeroes.

The number of real zeroes cannot exceed the degree of the polynomial. Three crossings need minimum degree (3).