Geometrical meaning of the zeroes of a polynomial.

The governing concept is that a zero occurs when at least one factor of a factored polynomial equals zero. From x−2=0, we obtain x=2. From (x+6)^3=0, the base must be zero, so x+6=0 and x=−6. The exponent 3 shows that −6 has multiplicity three, meaning it is repeated three times as a root, but it is still only one distinct zero value. Therefore the distinct zeroes are 2 and −6, making option A correct. Option C lists repetition rather than distinct values, option B changes both signs, and option D omits the zero from the first factor.

The midpoint is \(\left(\frac{-9+3}{2},0\right)=(-3,0)\). Tip: on the (x)-axis the midpoint has \(y=0\).

The relevant concept is the range of a set of real numbers, which is the greatest value minus the least value. The zeroes are -10, -4 and 6. The greatest zero is 6 and the least zero is -10. Therefore the range is 6-(-10)=6+10=16. The middle value -4 does not determine the range. Hence Option C is correct. Option A is the magnitude of the smallest zero, and Option B can arise from comparing only -4 and 8 or from an unrelated difference; it is not the full spread. Option D reverses the subtraction and gives a negative value, but a range is never negative.

(x^2-4x-21=(x-7)(x+3)), so the zeroes are (7) and (-3). Tip: form ((x,0)) points from factors.

A zero (root) is an x-value a for which the function value is zero, i.e. the graph passes through (a,0). If the graph meets the y-axis at (0,0), that point lies on the x-axis as well, so \(p(0)=0\) and x=0 is a root. The closest distractor (C) is wrong because although a generic y-intercept need not be a root, the specific y-intercept here is the origin, which is also an x-intercept. Exam tip: any point of the form (a,0) on the graph immediately tells you x=a is a zero of the polynomial.

(x^2-12x+36=(x-6)^2), so the touching point is ((6,0)). Tip: in a perfect square, change the sign to get the zero.

The zeroes are (a), (b), (c), so their product is (abc). Tip: even in symbolic points, the first coordinate is the zero.

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

((x-3)^2) is an even-power factor, so the graph touches at (x=3). Tip: power (2) shows a repeated zero.

The axis of symmetry is at the average of the zeroes, (\frac{(b-5)+(b+1)}{2}=b-2). Tip: the average rule also works with symbols.

A zero r corresponds to the point (r,0) on the x-axis because at x=r the function value y=0. Therefore for zeros -4, 1 and 8 the x-axis intersections are (-4,0), (1,0) and (8,0), which is option B. Option A is a common error: it places the zeros as y-coordinates, giving points on the y-axis. Exam tip: always treat a zero as the x-coordinate (write it as (x,0)) and keep the given order unless asked otherwise.

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

In the quadratic, the sum of zeroes is (7), so the other zero is (4). Tip: quickly convert a zero to ((x,0)).

The zeroes (-6) and (6) are opposite numbers. Tip: opposite zeroes are equally distant from the (y)-axis.

Set p(x)=0 to find x-intercepts: 9x^2-16=0 ⇒ 9x^2=16 ⇒ x^2=16/9 ⇒ x=±4/3. Alternatively factor as a difference of squares: 9x^2-16=(3x-4)(3x+4), giving x=4/3 and x=-4/3. Distractor C (±3/4) is the reciprocal mistake; B (±4) would be from x^2-16=0. Exam tip: either factor as (3x-4)(3x+4) or take square roots after isolating x^2 to avoid arithmetic slips.

The governing concept is the geometrical meaning of a zero: the x-coordinate of a point where the polynomial graph meets the x-axis. The listed x-coordinates are 2, 2 and 9. The repeated point (2,0) represents the same x-value twice, so it contributes only one distinct real zero. The other distinct x-value is 9. Thus the set of distinct zeroes is {2,9}, containing two elements, and Option B is correct. Option A ignores 9, while Option C counts the repeated coordinate twice rather than counting distinct values. Option D is not supported because only two different x-coordinates have been given.

The governing concept is the relationship between the zeroes and the factorised form of a monic quadratic. Since the graph meets the x-axis at x=-4 and x=3, these are the zeroes. Therefore p(x)=(x+4)(x-3). Expanding gives x^2-3x+4x-12=x^2+x-12. Comparing this with x^2+px+q, the coefficient of x is p=1 and the constant term is q=-12. Hence Option A is correct. Option B has the wrong sign for p, Option C uses the sum with an incorrect sign and the wrong constant sign, and Option D also has the wrong coefficient of x. The leading coefficient is 1, so no additional scaling is needed.

Real zeroes are counted from (x)-axis intersections, not from the (y)-axis intercept. Tip: ((0,-15)) does not show a zero.

(x^4-16=(x^2-4)(x^2+4)), and the real zeroes are only (\pm2). Tip: (x^2+4) gives no real zero.

For four distinct real zeroes, the degree must be at least (4). Tip: the number of distinct zeroes cannot exceed the degree.