Geometrical meaning of the zeroes of a polynomial.

The zeroes of a polynomial are the x-coordinates of its intersections with the x-axis. The point \((0,0)\) therefore gives the zero 0, while the point \((b,0)\) gives the zero b. The condition \(b\ne0\) ensures that these are two distinct points, so both zeroes must be included in the sum.

Adding the two x-coordinates gives \(0+b=b\). Thus the sum of the zeroes is b, which is option B. The value \(b^2\) would be related to a product in some contexts, not to this sum, and \(-b\) has the wrong sign. The origin should not be ignored: although its x-coordinate is 0, it contributes 0 to the sum while still being an important zero.

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

For real (x), (x^6\geq0), so (x^6+1>0). Tip: an always positive polynomial does not cut the (x)-axis.

A zero of the polynomial is an x‑value where the function equals 0. Here \(p(-7)=0\), \(p(1)=0\) and \(p(4)=0\), so there are three zeroes. The closest incorrect choice "four" is wrong because \(p(-3)=2\), not 0. Exam tip: verify by substituting the x‑value into \(p(x)\) and confirm the result is exactly 0 before counting it as a zero.

Factor: \(p(x)=x^2-bx= x(x-b)\). x-intercepts occur where \(p(x)=0\), so \(x(x-b)=0\) gives \(x=0\) or \(x=b\). Thus the intercepts are \((0,0)\) and \((b,0)\). The closest distractor (B) is incorrect because \(p(-b)=(-b)^2-b(-b)=2b^2\), which is not zero for general \(b\ne0\). Option (D) is wrong since \((0,b)\) is on the y-axis, not the x-axis. Exam tip: set \(p(x)=0\) and factor out the common \(x\) to find x-intercepts quickly.

For equal distance from the (y)-axis, zeroes should be opposites, so (4) is needed with (-4). Tip: symmetric zeroes are (a) and (-a).

The zeroes are (-2) and (7), and ((x+2)^2) causes touching at (-2). Tip: the outside (5) does not change the zeroes.

In this interval the signs are (+), (-), (-), so the product is positive. Tip: an odd number of negative factors gives a negative value.

The discriminant is (c^2-4c^2=-3c^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 ((4,0)). Tip: if the vertex has (y=0), there is one distinct zero.

The discriminant is (64-88=-24), so there are no real zeroes. Tip: with negative discriminant a parabola does not meet the (x)-axis.

It is ((x-a)^2-9), so (x-a=\pm3) and the zeroes are (a-3), (a+3). Tip: use difference of squares.

Mean \(=\dfrac{-1+3+7}{3}=3\). Explanation: The x-coordinates of the intercepts are the zeros; their sum is 9 and dividing by 3 gives the mean 3. Option 9 is incorrect because it is the sum, not the average (division was omitted). Option -3 results from a sign error, and 7 is just one of the zeros. Exam tip: read off the x-values from intercepts first, then compute average = (sum)/(count).

The average of the two zeroes is (4), so the other zero is (10). Tip: connect the axis of symmetry with the midpoint of zeroes.

The even-power factor ((x+6)^2) gives touching and the single factor (x-2) gives crossing. Tip: identify graph behavior from the power of the factor.

(x=-1) lies between the zeroes and an upward-opening parabola is below the axis there. Tip: check the sign region between zeroes.

In this interval the first factor is positive and the second is negative, so the outside negative makes the value positive. Tip: check each factor's sign separately.

Factorizing gives \(p(x)=x^2-6kx+9k^2=(x-3k)^2\), so there is a repeated root at \(x=3k\) and the parabola touches the x-axis at that x-value. Alternatively, the discriminant \(D=(-6k)^2-4\cdot1\cdot9k^2=0\) shows a double root. Note: when \(k=0\) all choices collapse to 0, but for a general (nonzero) k the unique touching point is \(x=3k\). Exam tip: spot a perfect square trinomial or check discriminant = 0 to identify a tangent to the x-axis quickly.

Direct answer: Option A, (u,0) and (v,0). To find x-axis intersections, set p(x)=0. We have x^2-(u+v)x+uv=(x-u)(x-v), because multiplying gives x^2-vx-ux+uv=x^2-(u+v)x+uv. Thus (x-u)(x-v)=0, so x=u or x=v. Each root becomes a point on the x-axis with y=0: (u,0) and (v,0). Option A is correct. Option B gives points on the y-axis, not generally x-axis intersections. Option C changes the signs and is not obtained from the factors. Option D gives general plane points whose y-coordinates need not be zero. If u=v, the two listed points coincide, but the stated locations remain correct. Memory cue: factors (x-a)(x-b) give x-intercepts (a,0) and (b,0).

The average of the two zeroes is (3), so the other zero is (-5). Tip: set (\frac{a+b}{2}) equal to the axis of symmetry.

A zero is an input x for which the polynomial value is exactly zero. Positive or negative values do not represent zeroes; they only show that the graph lies above or below the x-axis at those inputs. Therefore, in a list of function values, we select only the statements written with equality to 0.

Here \(p(-8)=0\), so -8 is a zero, and \(p(3)=0\), so 3 is another zero. The statements \(p(-2)>0\) and \(p(9)<0\) do not add zeroes. Their sum is \((-8)+3=-5\). Therefore option A is correct. A common mistake is to use the signs or to include every listed input instead of checking which function values are exactly zero.

To find where the graph meets the x-axis, solve \(p(x)=0\). For the given polynomial, \(x^3-25x=0\). Taking the common factor x gives \(x(x^2-25)=0\), and the difference of squares factors the second part as \(x(x-5)(x+5)=0\). A product is zero when at least one factor is zero, so the possible x-values are \(0\), 5, and -5.

These three values are distinct, so the graph has three different x-axis intersection points: \((0,0)\), \((5,0)\), and \((-5,0)\). Repeated factors, if present, would affect multiplicity but not create an additional distinct point. Since the question asks for distinct points, the answer is three, which is option C.

Both factors have even powers, so the graph touches at both places. Tip: an even-power zero usually gives touching, not crossing.

The midpoint is \(\left(\frac{-11+5}{2},0\right)=(-3,0)\). Tip: on the (x)-axis the midpoint has \(y=0\).

(x^2-8x-33=(x-11)(x+3)), so the zeroes are (11) and (-3). Tip: form ((x,0)) points from factors.