Geometrical meaning of the zeroes of a polynomial.

The average of the two zeroes is (5), so the other zero is (11). Tip: the axis of symmetry passes through the midpoint of zeroes.

An even-power zero gives touching and an odd-power zero gives crossing. Tip: identify graph behavior from the power of the factor.

(x=-4) lies between the two zeroes and an upward-opening parabola stays below there. Tip: check the sign region between zeroes.

In this interval the factor signs are (+), (+), (-), and the outside negative makes the value positive. Tip: apply the outside sign at the end.

It is ((x-(b-4))^2), so the repeated zero is (b-4). Tip: a perfect square form shows the zero quickly.

The polynomial is ((x-n)(x-(n+3))), so the zeroes are (n) and (n+3). Tip: write zeroes as ((x,0)).

The zeroes of a quadratic polynomial are the x-coordinates where its parabola meets the x-axis. The axis of symmetry of a parabola with two real zeroes lies exactly halfway between those zeroes. The first zero is −10, and the second is 18 greater, so the second zero is −10 + 18 = 8. Their midpoint is (−10 + 8) ÷ 2 = −2 ÷ 2 = −1. Therefore the vertical axis of symmetry is x = −1, making option A correct. The value x = 8 is merely the second zero, not the midpoint. The values x = 1 and x = −8 are neither the average of the zeroes nor the location of the symmetry axis.

The given zeroes are (-7) and (4), so the distance is (4-(-7)=11). Tip: take only (x)-values where (p(x)=0).

The zeroes of a polynomial are the x-values at which its graph meets the x-axis. Factor the given cubic: p(x) = x³ − 9x² + 18x = x(x² − 9x + 18) = x(x − 3)(x − 6). Hence the three zeroes are 0, 3, and 6. Their mean is the sum divided by the number of zeroes: (0 + 3 + 6) ÷ 3 = 9 ÷ 3 = 3. Therefore option A is correct. The value 6 is one zero, while 9 is the sum of all three zeroes rather than their mean. The original option 18 ÷ 3 was mathematically equal to 6 and could duplicate option B, so it has been corrected to 18, which is the constant term and still remains an incorrect distractor.

There are two distinct zeroes (-2) and (5), and both have even powers. Tip: at an even-power zero the graph usually touches.

The zeroes of a factored polynomial are found by setting each factor equal to zero. A factor raised to a power still contributes only one distinct zero; its exponent gives the multiplicity. This distinction is important because the question asks for distinct zeroes rather than all zeroes counted with repetition. The given inequality ensures that the two values do not coincide.

From \(x-c=0\), we obtain \(x=c\). From \(x+d=0\), we obtain \(x=-d\). The powers 5 and 2 mean that these zeroes have multiplicities 5 and 2, respectively, but the distinct list is only \(c\) and \(-d\). Since \(c\ne-d\), they are genuinely different. Therefore option A is correct; option C repeats \(-d\) unnecessarily.

The midpoint is \(\left(\frac{-17+9}{2},0\right)=(-4,0)\). Tip: on the (x)-axis the midpoint has \(y=0\).

The range is the difference between the greatest and smallest zero, (13-(-18)=31). Tip: range is always non-negative.

(x^2-13x-68=(x-17)(x+4)), so the zeroes are (17) and (-4). Tip: write intersection points from factors.

The origin is also on the (x)-axis, and (x=-9) is another (x)-axis intersection. Tip: count ((0,0)) as zero (0).

(x^2+20x+100=(x+10)^2), so the touching point is ((-10,0)). Tip: change the sign in a perfect square to get the zero.

The mean is (\frac{(s-4)+(s+1)+(s+7)}{3}=s+\frac{4}{3}). Tip: take the average even for symbolic zeroes.

((x-7)^2+4) is always positive, so (p(x)=0) will not occur. Tip: adding a positive number to a square gives no real intersection.

((x-3)^5) is an odd-power factor, so the graph crosses at (x=3). Tip: at an odd power the graph usually crosses the axis.

The axis of symmetry is at the average of the zeroes, (\frac{(q-11)+(q+7)}{2}=q-2). Tip: take the midpoint even with symbols.