Geometrical meaning of the zeroes of a polynomial.

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

(x^2+4) is always positive so (y=0) never occurs. Count a zero only when the graph meets the 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 \(p(a)=b\).

Two separate intersections give two distinct real zeroes. Different (x)-intercepts mean different zeroes.

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.

The vertex is the turning point of a parabola. If this vertex lies on the x-axis, its y-coordinate is zero, so the parabola meets the x-axis exactly at its turning point. It does not cross the axis at two separate locations; instead, the contact represents a repeated root. Algebraically, the quadratic can be written in the form a(x − r)², so both real zeroes are r and r. Therefore option A is correct. Distinct real zeroes would place the vertex away from the axis and produce two crossings. No real zero would occur if the parabola stayed completely above or below the axis, and a quadratic cannot have three real zeroes.

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.

A zero of a polynomial is an x-value for which the polynomial equals zero. On the graph, this corresponds to an intersection with the x-axis, whose y-coordinate is 0. The three stated intersections therefore give the x-values −2, 0, and 5, so the zeroes are −2, 0, and 5. Hence option A is correct. The origin (0, 0) lies on the x-axis, so it absolutely can represent the zero x = 0; being the origin does not exclude it. Option B omits a valid zero, option C contradicts the definition of a graph zero, and option D counts only the two nonzero intersections. Since the three points are distinct, they represent three distinct 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).