Geometrical meaning of the zeroes of a polynomial.

A zero r corresponds to the x-intercept (r,0) because the y-coordinate is zero at an x-axis crossing. Thus zeros -6, 2 and 9 give (-6,0), (2,0) and (9,0). Common mistake: option A swaps coordinates, placing the zero as a y-value instead of x. Exam tip: write each zero as the x-coordinate (r,0) and quickly plot to verify.

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

The zeroes (-8) and (8) are opposite numbers. Tip: opposite zeroes are equally distant from the (y)-axis.

x-axis intersections are the points where \(p(x)=0\). Solve \(16x^2-9=0\). Factor as \((4x-3)(4x+3)=0\), giving \(4x-3=0\) or \(4x+3=0\), hence \(x=\pm\tfrac{3}{4}\). Thus the intersections are \(\left(\tfrac{3}{4},0\right)\) and \(\left(-\tfrac{3}{4},0\right)\). Why other options fail: (C) \(\tfrac{4}{3}\) is the reciprocal error from mishandling \(4x\); (B) \(\pm3\) ignores the coefficient 16. Exam tip: view \(16x^2\) as \((4x)^2\) and use difference of squares to factor quickly.

((-5,0)) is repeated, so the distinct zeroes are (-5) and (4). Tip: count the same (x)-value once.

The x-intercepts give the zeroes of the monic quadratic: -6 and 2. A quadratic with these zeroes can be written as p(x) = (x + 6)(x - 2). Expanding gives x^2 - 2x + 6x - 12 = x^2 + 4x - 12. Comparing this with the standard form x^2 + px + q, the coefficient of x is p = 4 and the constant term is q = -12. Therefore option A is correct. The sum of the zeroes is -4, which equals -p, so p must be 4; confusing the sum directly with p leads to option B. The product is -12, not 12, ruling out option C, and option D has an incorrect coefficient.

Real zeroes are counted from (x)-axis intersections, not from the (y)-axis intercept. Tip: ((0,18)) does not show a zero.

(x^4-81=(x^2-9)(x^2+9)), and the real zeroes are only (\pm3). Tip: (x^2+9) gives no real zero.

For six distinct real zeroes, the degree must be at least (6). Tip: the number of distinct zeroes cannot exceed the degree.

(x^2-2x+10=(x-1)^2+9), so there is no real zero. Tip: an always positive form gives no intersection.

(x^3-8x^2+15x=x(x-3)(x-5)), so the zeroes are (0,3,5). Tip: first take (x) as the common factor.

(12=\frac{6+18}{2}), so it is the midpoint of the two zeroes. Tip: the midpoint need not be a zero.

The relevant principle is that a repeated real zero makes a parabola touch the x-axis, whereas two distinct real zeroes make it cross at two points. Factor the expression: 4x^2 - 24x + 36 = 4(x^2 - 6x + 9) = 4(x - 3)^2. Thus the discriminant is zero and the only zero is x = 3, repeated twice. The graph therefore touches the x-axis at (3, 0). Option A is correct. Option B results from reversing the sign, option C would require a positive discriminant and two different roots, and option D would require a negative discriminant with no real roots.

For (x<-9), both factors are negative and the outside negative makes the value negative. Tip: check factor signs first.

Zeroes are the x-values where the graph meets the x‑axis. From (7,0) and (−4,0) the zeroes are 7 and −4, so their sum is 7 + (−4) = 3. ‘‘Touches’’ indicates an even multiplicity and ‘‘crosses’’ an odd multiplicity, but the zero values remain 7 and −4 — standard exam questions treat these zeroes once each unless multiplicity is explicitly requested. Closest distractor: −3 is simply the wrong sign; 11 and −11 arise from arithmetic errors. Exam tip: count every x‑intercept as a zero; only account for multiplicity if the question asks for sum with multiplicity.

The x-intercepts give the roots, so the zeros are 0 and d. Their product is \(0\times d=0\). Closest distractors: option A (d) is just one root, not the product; option D (d^2) arises from mistakenly multiplying d by d; option C (-d) has the wrong sign. Exam tip: whenever one root is 0, the product of the roots is immediately 0.

Answer: A, 1 unit. A graph meets the x-axis at the zeroes of its polynomial. Solve x²−15x+56=0 by finding two numbers with sum 15 and product 56. These numbers are 7 and 8, so p(x)=(x−7)(x−8). The zeroes are x=7 and x=8, giving points (7,0) and (8,0). Their distance is |8−7|=1 unit. Option A is therefore correct. Option B is only the smaller zero. Option C is the sum of the zeroes, 7+8=15. Option D is their product, 7×8=56. A common mistake is to select the sum or product because they appear in the polynomial; however, the question asks for separation, not a root relationship. Memory cue: first find both roots, then subtract and take the positive magnitude.

For a downward-opening parabola, values outside the zeroes are negative. Tip: opening direction changes sign regions.

For real (x), (x^8\geq0), so (x^8+1>0). Tip: an always positive polynomial does not cut the (x)-axis.