Geometrical meaning of the zeroes of a polynomial.

(x^2+8x+16=(x+4)^2), so the touching point is ((-4,0)). Tip: in a perfect square the sign changes to get the zero.

The zeroes are the first coordinates (r), (s), (t). Tip: read the first coordinate even in symbolic points.

((x-3)^2+4) is always positive, so (p(x)=0) will not occur. Tip: a positive number added to a square can prevent intersection.

Distinct zeroes are counted from distinct meeting points with the (x)-axis. Tip: degree gives the maximum, but the actual count is read from the graph.

((x-2)^2) is an even-power factor, so the graph touches at (x=2). Tip: power (2) shows a repeated zero.

The axis of symmetry is at the average of zeroes, (\frac{(a-2)+(a+4)}{2}=a+1). Tip: take the average even for symbolic zeroes.

If a polynomial has a zero a, the graph meets the x-axis at (a,0) because the polynomial's value is zero at x = a. For zeros −2, 3 and 7 the x-intercepts are (−2,0), (3,0) and (7,0), which is option B. Distractor A incorrectly treats zeros as y-intercepts (0,a); C and D alter the order or signs of coordinates, so they are wrong. Exam tip: always write a zero a as the point (a,0) on the x-axis.

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 (5), so the other zero is (3). Tip: immediately convert a zero to ((x,0)).

The zeroes are (-5) and (5), which are opposite numbers. Tip: such zeroes are equally distant from the (y)-axis.

x-intercepts occur where \(p(x)=0\). From \(4x^2-25=0\) we get \(4x^2=25\), so \(x^2=\tfrac{25}{4}\) and hence \(x=\pm\tfrac{5}{2}\). Therefore the intercepts are \(\left(\tfrac{5}{2},0\right)\) and \(\left(-\tfrac{5}{2},0\right)\). Alternatively factor \(4x^2-25=(2x-5)(2x+5)\) to read off the roots. The closest distractor (B) reflects a common slip of using 5 instead of \(\tfrac{5}{2}\). Exam tip: treat \(4x^2\) as \((2x)^2\) or take square roots carefully to avoid fraction errors.

The governing concept is that a real zero of a polynomial is an x-coordinate at which its graph meets the x-axis. The listed points have x-coordinates 1, 1, and 4. Although the point (1,0) appears twice, repetition does not create a new distinct zero; it only indicates that the same zero may have repeated multiplicity. Therefore the distinct x-values are 1 and 4, giving exactly two distinct real zeroes. Option A is incorrect because there are two different x-values, while Option C incorrectly counts the repeated listing separately. Option D has no basis in the data.

The zeroes are (-2) and (5), so the polynomial is ((x+2)(x-5)=x^2-3x-10). Tip: form factors from graph intersections.

Real zeroes are counted from (x)-axis intersections, not from the (y)-axis intercept. Tip: ((0,12)) is not a zero.

(x^4-1=(x^2-1)(x^2+1)), and the real zeroes are only (\pm1). Tip: (x^2+1) gives no real zero.

For three distinct real zeroes, the degree must be at least (3). Tip: the number of distinct zeroes cannot exceed the degree.

(x^2-2x+5=(x-1)^2+4), so it cannot be zero for real (x). Tip: an always positive form gives no real intersection.

(x^3-4x^2-5x=x(x-5)(x+1)), so the zeroes are (0), (5), (-1). Tip: first take (x) as the common factor.

The relevant concept is the midpoint of two numbers on the x-number line. The two x-intercepts have x-coordinates 2 and 8. Their midpoint is calculated by averaging them: (2+8)/2 = 10/2 = 5. Thus x=5 lies exactly halfway between the two given zeroes. This calculation only identifies its position; it does not prove that p(5)=0, because a polynomial may take a nonzero value between its zeroes. Therefore Option A is correct. Option B confuses a midpoint with a zero, Option C refers to a point on the y-axis where x=0, and Option D is false because 8, not 5, is the greater zero.

The governing idea is that the real zeroes determine where a polynomial graph meets the x-axis, and a repeated zero generally means the graph touches the axis without crossing it. Factor the polynomial: 2x^2+12x+18 = 2(x^2+6x+9) = 2(x+3)^2. Hence the only zero is x=-3, with multiplicity two. Since the square is never negative and the leading factor is positive, the graph has its minimum value 0 at x=-3 and touches the x-axis there. Option A is correct. Option B has the wrong sign, Option C would require two distinct real zeroes, and Option D ignores the real repeated zero.