Geometrical meaning of the zeroes of a polynomial.

A zero (root) of a polynomial is an x-value where \(p(x)=0\). Here \(p(-9)=0\), \(p(2)=0\) and \(p(6)=0\), so there are three zeros among the given x-values. \(p(-4)=2\) is not zero, so it is not a root. Exam tip: always check the function value equals zero — other values (like 2) do not count as zeros.

We have \(p(x)=x^2-cx\). Factor the expression: \(p(x)=x(x-c)\). Setting \(p(x)=0\) gives \(x=0\) or \(x=c\). Therefore the x-intercepts are \((0,0)\) and \((c,0)\). Why others are wrong: option B uses \(-c\) which is the wrong sign; option C omits the zero at \(x=0\); option D includes \((0,c)\), a point on the y-axis, not an x-intercept. Exam tip: set \(p(x)=0\) and factor out the common \(x\) to find zeros quickly.

For equal distance from the (y)-axis, zeroes should be opposites, so (6) is needed with (-6). Tip: symmetric zeroes are (a) and (-a).

The zeroes are (-3) and (10), and ((x+3)^2) causes touching at (-3). Tip: the outside (7) does not change the zeroes.

In this interval the signs are (+), (-), (-), so the product is positive. Tip: the product of two negative factors is positive.

The discriminant is (d^2-4d^2=-3d^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 ((-5,0)). Tip: if the vertex has (y=0), there is one distinct zero.

The discriminant is (144-204=-60), so there are no real zeroes. Tip: with negative discriminant a parabola does not meet the (x)-axis.

It is ((x-b)^2-16), so (x-b=\pm4) and the zeroes are (b-4), (b+4). Tip: use difference of squares.

Zeroes are the x‑values where the graph meets the x‑axis; here they are -2, 4 and 10. The mean is the sum divided by the count: \(\frac{-2+4+10}{3}=4\). The closest distractor B (12) is the sum of the zeroes, not the mean. Exam tip: read only the x‑coordinates from intercepts and divide by the number of zeroes.

The average of the two zeroes is (-2), so the other zero is (-9). Tip: connect the axis of symmetry with the midpoint of zeroes.

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

(0) lies between the two zeroes and an upward 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-(a+3))^2), so the repeated zero is (a+3). Tip: a perfect square form shows the zero quickly.

The polynomial is ((x-m)(x-(m-1))), so the zeroes are (m) and (m-1). Tip: write zeroes as ((x,0)).

Answer: A, x=−2. For a parabola with zeroes r1 and r2, the axis of symmetry passes through the midpoint of the two x-intercepts, so x=(r1+r2)/2. The first zero is −8. The second is 12 more than −8, so r2=−8+12=4. Their midpoint is (−8+4)/2=−4/2=−2. Therefore the axis is the vertical line x=−2. Option A is correct. Option B results from a sign error in the midpoint. Option C is merely the second zero, not the midpoint. Option D is the sum of the zeroes without dividing by 2. Memory cue: the symmetry axis lies halfway between the two roots, not at either root.

The given zeroes are (-6) and (5), so the distance is (5-(-6)=11). Tip: take only (x)-values where (p(x)=0).

(p(x)=x(x-5)(x-2)), so the zeroes are (0), (2), (5), and the mean is (\frac{7}{3}). Tip: factor first.

There are two distinct zeroes (1) and (-4), and both have even powers. Tip: at an even-power zero the graph usually touches.