Geometrical meaning of the zeroes of a polynomial.

The x-intercepts of a polynomial graph give its zeroes. From the points (−2, 0) and (3, 0), the zeroes are −2 and 3. Their product is calculated directly: (−2) × 3 = −6. Therefore option A is correct. The y-coordinate is zero for both points because they lie on the x-axis, but it is not used when multiplying the zeroes; the zeroes are the x-coordinates. Option B loses the negative sign, option C could arise from adding the zeroes instead of multiplying them because −2 + 3 = 1, and option D is not the product or sum of the given values. The graph information is sufficient without knowing the polynomial’s leading coefficient.

Zeros of a polynomial are the x-values where its graph meets the x-axis (y = 0). The given points (0,0) and (3,0) have x-coordinates 0 and 3, so the zeros are 0 and 3. Distractor C wrongly uses -3 instead of 3 (wrong sign). Options B and D fail to include both intercepts. Exam tip: read the x-coordinate of each x-intercept — its y-value must be zero.

The geometrical meaning of a zero is the x-coordinate of a point where the graph of p(x) meets or crosses the x-axis. On the x-axis, y = 0, so every distinct intersection represents a distinct real solution of p(x) = 0. The graph in the question has three different intersection points; therefore it has three different real zeroes. The fact that the polynomial is cubic is consistent with this result, since a cubic can have at most three real zeroes. Option B and Option C undercount the visible intersections, while Option D would mean that the graph never met the x-axis. Hence Option A is correct.

A root (zero) of a polynomial is an x-value where the polynomial evaluates to 0. For the zero polynomial \(p(x)=0\), the polynomial equals 0 for every real x, so every x is a root. The distractors B and C are incorrect because they restrict roots to only positive or only negative numbers, whereas the zero polynomial has every real number as a root. Exam tip: treat the zero polynomial as a special case — its set of roots is the entire domain (all real numbers).

The value (p(x)=5) is never (0) so it has no zero. Tip: a non-zero constant polynomial has no zero.

Set p(x)=0 to find the root: \(3x=0\) gives \(x=0\). The x-axis intersection is the root written as a point \((x,0)\), so the intersection is \((0,0)\). The nearby distractor \((3,0)\) is incorrect because it corresponds to x=3, whereas \(3x\) vanishes only at x=0. Exam tip: always solve p(x)=0 and express the answer as \((root,0)\).

On the x-axis, \(y=0\). Substituting gives \(0 = x + 2\), so \(x = -2\). Interpretation: the x‑intercept is the value of x that makes the expression equal to zero. The nearest distractor 2 is wrong due to the sign; 0 and 1 do not satisfy the equation. Exam tip: to find an x‑intercept always set \(y=0\) and solve for x.

The zero is the x-value for which y = 0. For the line \(y=5-x\), set \(0=5-x\) which gives \(x=5\). Hence the zero is 5. The closest distractor 0 is wrong because at \(x=0\) the value is \(y=5\), not 0. Exam tip: to find the x-intercept of a line, set \(y=0\) and solve for \(x\).

Only one (x)-axis intersection is visible so there is one real zero. Tip: the count comes from the number of points.

The governing concept is the geometrical meaning of a zero of a polynomial. A number r is a zero of p(x) when p(r) = 0. On a graph, this means that the point (r, 0) lies on the x-axis. Since the stated intersection is (a, 0), substituting its x-coordinate gives p(a) = 0; hence the zero is a. Therefore, option A is the precise answer. The second coordinate, 0, describes the y-coordinate and only confirms that the point lies on the x-axis; it is not the requested zero. Option C simplifies algebraically to a, but option A gives the standard and direct value. Option D changes the sign without any basis.

By definition a number a is a zero (root) of a polynomial iff \(p(a)=0\). Here \(p(2)=3\), which is not zero, so 2 is not a zero. Option C is incorrect because the degree (quadratic or otherwise) does not change the criterion; any degree requires \(p(a)=0\) to be a root. Exam tip: evaluate the polynomial at the given value—if the result is 0, it is a root.

Reason: Interpret the polynomial as the function \(y=p(x)\). \(p(-4)=0\) means at \(x=-4\) the value \(y=0\), so the graph passes through the point \((-4,0)\). The closest distractor (0, −4) comes from swapping coordinates; it would be correct only if \(p(0)=-4\), which is not given. Exam tip: treat \(p(x)\) as the y-coordinate and plot the point \((x,p(x))\).

A zero of a polynomial corresponds to an x‑intercept of its graph, i.e. a point whose y‑coordinate is zero. Thus the representing point must satisfy \(y=0\). The point (9,0) has y = 0, so x = 9 is a zero. The closest distractor (0,9) is incorrect because its y ≠ 0 (it lies on the y‑axis); similarly (9,9) and (0,-9) do not lie on the x‑axis. Exam tip: quickly check the y‑coordinate — zeros appear where y = 0.

The x-intercepts are the zeros of the polynomial; here the zeros are -1 and 1. Their sum is -1 + 1 = 0, so the correct answer is 0. Option D (1) is incorrect because it gives only one zero rather than the sum of both. Tip: symmetric zeros of the form ±a always sum to 0; more generally the sum of zeros relates to the coefficients (negative of the coefficient of x^{n-1} divided by leading coefficient).

The zeroes are (0) and (6) so the product is (0). Tip: any number multiplied by (0) gives (0).

Two real zeroes need two distinct (x)-axis intersections. Tip: (y)-axis intersections do not count as zeroes.

If x = 3 is a zero then p(3) = 0. On the graph this means the point with x‑coordinate 3 lies on the x‑axis, i.e. (3, 0). Options B and C have y ≠ 0 so they cannot be zeros; option D has x = -3, not 3. Exam tip: substitute the given x into p(x) to check if it equals zero, or look for the point (x, 0) on the graph.

A number a is a zero (root) of a polynomial if p(a)=0. If the graph meets the x-axis at (-6,0) then the function value at x=-6 is 0, so p(-6)=0 and -6 is a zero. Option B is wrong because (0,-6) is a y-axis intercept (x=0), not x=-6. Option C gives x=6, the wrong sign. Option D has y=6, so p(-6)=6≠0. Exam tip: a root corresponds to an x-intercept, so check the coordinate is (-6,0) with y=0 and the correct sign for x.

For a polynomial p(x), a real zero is a real number r satisfying p(r) = 0. On the coordinate graph y = p(x), this condition produces the point (r, 0), which lies on the x-axis. Thus real zeroes are represented exactly by x-axis intersections. If no point of the graph lies on that axis, there is no real value of x for which p(x) = 0. Therefore the polynomial has no real zero, as stated in Option A. Options B and C claim intersections that the question explicitly rules out. Option D is also false because a polynomial cannot have every number as a zero unless it is the zero polynomial, whose graph is the x-axis itself.

If the graph crosses the \\(x\\)-axis at \\(x=10\\), the point on the curve is \\( (10,0)\\). The polynomial's value at x equals the y-coordinate, so \\(p(10)=0\\). The closest distractor \\(p(-10)=0\\) is wrong because it refers to x = -10, not x = 10. The options \\(p(0)=10\\) and \\(p(10)=10\\) are inconsistent with crossing the x-axis because y must be 0 there. Exam tip: when told the graph crosses at \\(x=a\\), substitute x = a; a crossing on the x-axis means \\(p(a)=0\\).