Geometrical meaning of the zeroes of a polynomial.

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.

(x^2-9x-52=(x-13)(x+4)), so the zeroes are (13) and (-4). Tip: write intersection points from factors.

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.

Answer: A, p=5 and q=−36. The x-axis intersections give the roots −9 and 4. Since the coefficient of x² is 1, form the polynomial by multiplying the corresponding factors: (x−(−9))(x−4)=(x+9)(x−4). Expand: x²−4x+9x−36=x²+5x−36. Comparing this with x²+px+q gives p=5 and q=−36. Option A is correct. Option B has the wrong sign for p. Option C uses 13 and 36 without respecting the sum and product signs. Option D incorrectly treats the root sum as p and also gives the wrong sign for q. Check by Vieta’s relations: root sum = −p, so −5=−p and p=5; root product=q, so (−9)(4)=−36. Memory cue: for x²+px+q, sum of roots is −p and product is q.

Real zeroes are counted from (x)-axis intersections, not from the (y)-axis intercept. Tip: ((0,-20)) does not show a zero.

(x^4-625=(x^2-25)(x^2+25)), and the real zeroes are only (\pm5). Tip: (x^2+25) gives no real zero.

For seven distinct real zeroes, the degree must be at least (7). Tip: the number of distinct zeroes cannot exceed the degree.