Geometrical meaning of the zeroes of a polynomial.

Direct answer: Option A. For an x-axis intersection, y=p(x) must equal 0. Here p(x)=x^2+4. For every real x, x^2 is at least 0, so x^2+4 is at least 4, which is strictly positive. Therefore p(x) can never be 0, and the graph has no real x-intercept. Option A is correct. Option B would require two real solutions of x^2+4=0, but that equation gives x^2=-4, impossible for real x. Option C would require one real solution, but even the smallest value of the polynomial is 4, not 0. Option D is false because the graph is the parabola y=x^2 shifted upward by 4; it is not the x-axis y=0. Complex solutions do not create real graph intersections. Memory cue: x^2+positive number stays above the x-axis.

If the graph passes through (0,0) then by definition the polynomial satisfies \(p(0)=0\), so 0 is a root. This does not imply the polynomial is identically zero (option C is false) — a single zero at x=0 does not make the function zero everywhere. Option B contradicts the given point because a nonzero constant term would give \(p(0)\neq0\). Option D is wrong because "touching" (tangent) the x-axis indicates a repeated (even multiplicity) root, whereas "crossing" indicates the graph passes through the axis; from the fact it crosses we only can be certain that \(p(0)=0\). Exam tip: to check whether \(a\) is a root, substitute \(x=a\) into \(p(x)\); if \(p(a)=0\), then \(a\) is a zero of the polynomial.

Any point on the x-axis has y-coordinate 0. The point (7, 0) therefore means the polynomial's value at x=7 is 0, i.e. \(p(7)=0\). Options B (7) and C (-7) confuse the x-coordinate with the function value; option D (1) is also incorrect. Exam tip: for a point (a, b) on a graph, always use \(p(a)=b\).

Direct answer: Option A, two distinct real zeroes. A graph meets the x-axis when its y-coordinate is zero. Therefore every x-axis intersection gives a real zero of the polynomial. If a quadratic cuts the x-axis at two separate points, the two points have two different x-coordinates, and both coordinates are real. Hence the polynomial has two distinct real zeroes. Option A is correct. Option B describes one repeated zero: the graph touches the axis at one point and turns back. Option C describes a graph that never meets the x-axis, so it has no real zeroes. Option D is impossible for a quadratic with real coefficients in this situation; a quadratic cannot have exactly one real zero and one non-real imaginary zero, because non-real roots occur as a conjugate pair. The practical graph rule is: two crossings mean two different real roots, one touching point means equal real roots, and no meeting means no real roots.

The polynomial is given in factor form: p(x)=2(x-1)(x+4). x-intercepts occur where p(x)=0, so set each factor to zero: x-1=0 → x=1 and x+4=0 → x=-4. Hence the intercepts are (1,0) and (-4,0). The multiplicative constant 2 does not affect the roots. Closest distractor B flips the signs, C wrongly uses 2 instead of 1, and D lists y-intercepts. Exam tip: from factorised form read off roots directly and write intercepts as (root,0).

A zero of a polynomial is an x–value where the polynomial evaluates to zero: if \\(p(a)=0)\\ then \\(x=a)\\ is a zero. Thus \\(x=3)\\ is a zero exactly when \\(p(3)=0)\\. Option C is a common mix-up — a zero at \\(x=3)\\ means the graph passes through \\((3,0)\\), not \\((0,3)\\). Option D is wrong because \\(p(3)=3)\\ gives value 3, not 0. Exam tip: always verify by calculating \\(p(a)\\) and checking whether it equals 0, or look for the x–intercept \\((a,0)\\).

Crossing at \((-5,0)\) means \(x=-5\) is a real root (typically odd multiplicity). Touching at \((2,0)\) means \(x=2\) is also a real root (typically even multiplicity, e.g. multiplicity 2). These are two distinct real zeros, so the answer is two. The option "three" might arise if one counts multiplicities (for example multiplicity 1 at \(-5\) and multiplicity 2 at \(2\) gives three roots counting multiplicity), but the question asks for distinct real zeros. Exam tip: always check whether the question asks for distinct roots or roots counted with multiplicity when interpreting touch vs. cross.

The repeated factor gives equal zero (-2). A quadratic parabola generally touches the (x)-axis at such a point.

If the graph of a polynomial meets the \(x\)-axis at a point, the polynomial's value at that x-coordinate is zero. Hence when the graph crosses the axis at \(a\) and \(b\), we have \(p(a)=0\) and \(p(b)=0\). Even if a root has multiplicity greater than one (the graph just touches the axis), the value at that root is still zero. Option D (“they are not equal”) is incorrect because both values are equal (both zero); option C is also incorrect because the axis-intercept cannot produce 0 at one point and 1 at the other. Exam tip: whenever asked about x-intercepts, immediately set the polynomial value to 0 at those x-values — that gives the roots directly.

The y-intercept gives the value \(p(0)\); here \(p(0)=−8\). Zeros are x-values satisfying \(p(x)=0\), i.e. where the graph meets the x-axis. Knowing only the y-intercept does not determine any x-zero because the y-value (−8) is not an x-coordinate. Option B (zero = −8) is a common confusion that mistakes the y-value for an x-value and is therefore incorrect. Exam tip: to find zeros set \(p(x)=0\) or look for x-axis intersections (y=0).

Zeros are the x-values for which \(p(x)=0\). A multiplicative constant (the leading '-' here) does not change the roots. Set each factor equal to zero: \(x-3=0\) gives \(x=3\), and \(x+1=0\) gives \(x=-1\). Thus the zeros are 3 and -1. Why distractors fail: option D mistakenly uses +1 instead of -1; option B flips signs of both roots. Exam tip: Factor the polynomial and set each linear factor to zero; ignore overall constant factors when finding zeros.

The opening direction alone does not decide the number of zeroes. Two intersections clearly give two real zeroes.

(x^2-6x+9=(x-3)^2), so (3) is a repeated zero. A repeated zero appears as touching on the graph.

Four distinct real zeroes need degree at least four. The number of zeroes cannot exceed the degree.

To find zeros set \(p(x)=0\). For \(p(x)=5\) this gives \(5=0\), which is impossible, so there are no real roots. Geometrically the graph \(y=5\) is a horizontal line that does not meet the x-axis. Note the contrast: the zero polynomial \(p(x)=0\) has every real number as a root (infinitely many). Exam tip: always solve \(p(x)=0\) to determine number of zeros and check if the polynomial is the zero polynomial first.

The zero polynomial \(p(x)=0\) yields \(y=p(x)=0\) for every real x. Thus the graph consists of all points whose y-coordinate is zero — the entire line \(y=0\), i.e. the x-axis. Option D is incorrect because the graph is not just the origin; it contains infinitely many points (every (x,0)). Option B is wrong since the y-axis is the vertical line \(x=0\), not \(y=0\). Option C is false because a value exists for every x. Exam tip: Remember the graph of a polynomial is the set of points \((x,p(x))\); plug in the definition to see the shape quickly.

A zero (root) is an x-value where the function's value is 0, i.e., the graph meets the x-axis. The graph meets the x-axis only at x = -6, so the root is -6. Option B (6) is incorrect because the graph does not intersect at x = 6; option C (0) is wrong unless the graph intersects at x = 0, which it does not here; option D (none) is false because an intersection at x = -6 is given. Exam tip: read the x-coordinate of the point where the graph crosses or touches the x-axis—any intersection gives a root (crossing usually indicates odd multiplicity, touching indicates even multiplicity).

The point \((r,0)\) has x‑coordinate \(r\) and y‑coordinate 0. For the polynomial \(p(x)\), the y‑value at x=\(r\) is \(p(r)\); therefore \(p(r)=0\). Option B (\(p(0)=r\)) incorrectly relates the value at x=0 to r and need not hold. Option C (\(r=0\)) is only true if the intercept happens at the origin, not in general. Option D (\(p(r)=r\)) would mean the function value equals the x‑coordinate at that point, which contradicts y=0 here. Exam tip: Whenever you see an x‑intercept \((a,0)\), immediately record \(p(a)=0\) — that a is a root of the polynomial.

Set the quadratic equal to zero to find x-intercepts (roots).

Factor: \(x^2+2x-15=(x+5)(x-3)\).
Setting \((x+5)(x-3)=0\) gives x=-5 or x=3, so the graph meets the x-axis at x=3 and x=-5.

The closest distractor (-3, 5) is wrong because the signs of the roots are incorrect — the constant term -15 forces the product of roots to be -15, so one root must be negative.

Exam tip: Check constant term and middle coefficient quickly — product of roots = constant term, sum of roots = - (coefficient of x).

p(1)=0 directly shows x=1 is a root — this is explicit. The facts p(-3)<0 and p(4)>0 only give signs of p at those x-values; they do not make -3 or 4 roots. (By continuity of polynomials, a sign change between -3 and 4 guarantees at least one root in (-3,4), but it does not identify the endpoint values as roots.) Therefore the only certain root from the given information is x=1. Exam tip: look for statements of the form p(a)=0 for a guaranteed root; sign changes indicate existence of a root in an interval, not at a specific endpoint.)

To find the x-intercept set p(x)=0. With p(x)=4x we get 4x=0 ⇒ x=0, so y=p(0)=0 and the intersection point is (0,0). Option (4,0) is incorrect because p(4)=16 ≠ 0; (0,4) is impossible for an x-intercept since y must be 0. Exam tip: always solve p(x)=0 to get x-intercepts of a polynomial.

For zeroes (m) and (n), the factors are ((x-m)) and ((x-n)). A non-zero multiplier does not change zeroes.

The root (zero) of a polynomial is the x‑value where its graph meets the x‑axis. Hence the zero is the number 2.5. Option B (2.5, 0) denotes the coordinate point on the plane — a point, not the numeric root. Option C is incorrect because x=0 is not where the graph crosses; Option D is the point on the y‑axis and irrelevant. Exam tip: when asked for a root or zero give the x‑value (a single number), not the ordered pair of the point.

Points where the graph meets the x-axis are the zeros of \(p(x)\), so set \(p(x)=0\). Factor: \(x^2-2x-8=(x-4)(x+2)\). Therefore \(x=4\) and \(x=-2\), giving points \((4,0)\) and \((-2,0)\). Option B has signs swapped (gives wrong x-values); option C lists points on the y-axis, not x-axis; option D is just incorrect roots. Exam tip: set the quadratic equal to zero and factor or use the quadratic formula; then report roots as (x,0).

Three distinct (x)-intercepts give three zeroes. Count by intersections not by the type of numbers.