Geometrical meaning of the zeroes of a polynomial.

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.

Direct answer: Option A, (m,0) and (m-1,0). Factor the polynomial: x^2-(2m-1)x+m(m-1)=(x-m)(x-(m-1)). Indeed, the sum of the roots is m+(m-1)=2m-1 and their product is m(m-1), matching the polynomial. Setting p(x)=0 gives x=m or x=m-1. Since an x-axis point has y=0, the intersections are (m,0) and (m-1,0). Option A is correct. Option B has x=0 and therefore describes y-axis points. Option C uses negative roots, which would come from factors (x+m) and (x+m-1), not these factors. Option D gives points whose second coordinates are not zero. If m has a special value causing equality, the two locations may coincide, but the stated root expressions remain valid. Memory cue: compare the factor form with (x-root).

Direct answer: Option B, the distance is 11. A zero is identified by \(p(x)=0\). The statements \(p(-6)=0\) and \(p(5)=0\) therefore give the two zeroes -6 and 5. The values \(p(-2)<0\) and \(p(9)>0\) describe signs of the polynomial at other points; they are not zeroes and must not be used as endpoints of the requested distance. Distance on a number line is non-negative, so subtract the smaller value from the larger: \(5-(-6)=5+6=11\). Option A, 1, is unrelated. Option B is correct because it is the distance between -6 and 5. Option C, 14, is obtained by an incorrect combination such as 5+9 or -6 to 8, not by the given zeroes. Option D, -11, cannot be a distance because distance is never negative. Memory cue: select only values with p(x)=0, then use larger minus smaller.

(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.

A zero of a polynomial is a value of the variable that makes the polynomial equal to zero. In a product, the polynomial becomes zero whenever at least one factor becomes zero. The exponents show multiplicity, or how many times a zero is repeated, but they do not create different zero values. The condition ensures that the two values obtained here are distinct.

Set the first factor equal to zero: \(x+a=0\), so \(x=-a\). Set the second factor equal to zero: \(x-b=0\), so \(x=b\). The powers 3 and 2 indicate repeated zeroes, but the distinct zeroes are counted only once each. Since \(a\ne-b\), these values are different. Hence option A, \(-a\) and \(b\), is correct.

The midpoint is \(\left(\frac{-13+7}{2},0\right)=(-3,0)\). Tip: on the (x)-axis the midpoint has \(y=0\).

The range is the difference between the greatest and smallest zero, (11-(-15)=26). Tip: range is always non-negative.

Direct answer: Option A, the intersections are \((13,0)\) and \((-4,0)\). At an x-axis intersection, y=0, so solve \(p(x)=0\): \(x^2-9x-52=0\). We need two numbers whose product is -52 and whose difference is 9: 13 and -4. Hence \(x^2-9x-52=(x-13)(x+4)\). Setting factors to zero gives x=13 or x=-4. The corresponding points must have y=0, so they are \((13,0)\) and \((-4,0)\). Option A is correct. Option B reverses both signs and does not satisfy the equation. Option C gives y-axis points because its x-coordinate is 0, not x-axis points. Option D uses the constant and coefficient as if they were roots; they are not. Exam cue: for x-axis intersections, find zeroes and attach y=0.

The origin is also on the (x)-axis, and (x=6) is another (x)-axis intersection. Tip: count ((0,0)) as zero (0).

(x^2-18x+81=(x-9)^2), so the touching point is ((9,0)). Tip: change the sign in a perfect square to get the zero.

The zeroes (x-values where the graph meets the x-axis) are \(r-1,\ r+2,\ r+5\). Mean = \(\dfrac{(r-1)+(r+2)+(r+5)}{3}=\dfrac{3r+6}{3}=r+2\). The closest distractor \(r+1\) arises from arithmetic mistake; the correct procedure is sum all roots then divide by their count. Exam tip: always add the symbolic roots first, then divide by the number of roots to avoid sign or division errors.

((x-5)^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-7)^4) is an even-power factor, so the graph touches at (x=7). Tip: at an even power the graph usually turns back.

The axis of symmetry is at the average of the zeroes, (\frac{(t-9)+(t+5)}{2}=t-2). Tip: take the midpoint even with symbols.

((0,12)) has (y=12), so it is not on the (x)-axis. Tip: a zero point must have second coordinate (0).

In this interval the first two factors are positive and the third is negative, so the product is negative. Tip: check the sign of each factor separately.

In the quadratic, the sum of zeroes is (11), so the other zero is (7). Tip: convert a zero into ((x,0)).

The zeroes are (-10) and (10), so the product is (-100) and the sum is (0). Tip: opposite zeroes have sum (0).

x-axis intersections occur where \(p(x)=0\). So solve \(25x^2-36=0\). Recognize a difference of squares: \((5x)^2-6^2=0\), hence \((5x-6)(5x+6)=0\). Solving gives \(x=\pm\tfrac{6}{5}\). Thus the intersections are \(\left(\tfrac{6}{5},0\right)\) and \(\left(-\tfrac{6}{5},0\right)\). Distractor B errs by effectively taking \(\sqrt{36}=6\) without accounting for the factor 25 on \(x^2\). Exam tip: either factor as a difference of squares or divide the equation by 25 first to simplify.

Repeated points give the same (x)-values, so the distinct zeroes are (-7) and (2). Tip: count the same (x)-value once.