Geometrical meaning of the zeroes of a polynomial.

Correct answer: A. (x=b-4)

Explanation: It is ((x-(b-4))^2), so the repeated zero is (b-4). Tip: a perfect square form shows the zero quickly.

Correct answer: A. ((n,0)) and ((n+3,0))

Explanation: The direct answer is option A: the intersections are \\(n,0\\) and \\(n+3,0\\). To find an x-axis intersection, put the y-value, here \\(p(x)\\), equal to zero. Factor the polynomial: \\(x^2-(2n+3)x+n(n+3)=(x-n)(x-(n+3))\\), because the roots add to \\(2n+3\\) and multiply to \\(n(n+3)\\). Thus \\(x=n\\) or \\(x=n+3\\). Points on the x-axis have y-coordinate zero, so the points are \\( (n,0)\\) and \\( (n+3,0)\\). Option A is correct. Option B gives points on the y-axis, because their x-coordinate is zero, so it does not show x-axis intersections. Option C uses negative roots, which do not come from the factorisation. Option D gives points whose y-coordinates are generally nonzero, so they are not on the x-axis. Memory cue: zero of a polynomial gives an x-coordinate, and always write the graph point as \\( (zero,0)\\).

Correct answer: A. (11)

Explanation: Direct answer: Option A, 11. A zero is identified by p(x)=0. The given equalities show p(-7)=0 and p(4)=0, so the zeroes are -7 and 4. The statements p(-3)<0 and p(8)>0 do not identify zeroes; they only tell us that the polynomial has negative and positive values at those inputs. The distance between two points on the number line is the absolute difference: |4-(-7)|=|4+7|=11. Option A is correct. Option B, 3, is only the distance from -7 to -3, not between the two zeroes. Option C, 15, is the distance from -7 to 8, not the requested pair. Option D, -11, cannot be a distance because distance is non-negative. Exam cue: first select only p(x)=0, then subtract and take the absolute value.

Correct answer: A. Two, touches at both

Explanation: There are two distinct zeroes (-2) and (5), and both have even powers. Tip: at an even-power zero the graph usually touches.

Correct answer: A. (c) and (-d)

Explanation: The zeroes of a factored polynomial are found by setting each factor equal to zero. A factor raised to a power still contributes only one distinct zero; its exponent gives the multiplicity. This distinction is important because the question asks for distinct zeroes rather than all zeroes counted with repetition. The given inequality ensures that the two values do not coincide.

From \(x-c=0\), we obtain \(x=c\). From \(x+d=0\), we obtain \(x=-d\). The powers 5 and 2 mean that these zeroes have multiplicities 5 and 2, respectively, but the distinct list is only \(c\) and \(-d\). Since \(c\ne-d\), they are genuinely different. Therefore option A is correct; option C repeats \(-d\) unnecessarily.

Correct answer: A. ((-4,0))

Explanation: The midpoint is \(\left(\frac{-17+9}{2},0\right)=(-4,0)\). Tip: on the (x)-axis the midpoint has \(y=0\).

Correct answer: C. (31)

Explanation: The range is the difference between the greatest and smallest zero, (13-(-18)=31). Tip: range is always non-negative.

Correct answer: A. ((17,0)) and ((-4,0))

Explanation: (x^2-13x-68=(x-17)(x+4)), so the zeroes are (17) and (-4). Tip: write intersection points from factors.

Correct answer: A. (0) and (-9)

Explanation: The origin is also on the (x)-axis, and (x=-9) is another (x)-axis intersection. Tip: count ((0,0)) as zero (0).

Correct answer: A. ((-10,0))

Explanation: (x^2+20x+100=(x+10)^2), so the touching point is ((-10,0)). Tip: change the sign in a perfect square to get the zero.

Correct answer: A. (s+\frac{4}{3})

Explanation: The mean is (\frac{(s-4)+(s+1)+(s+7)}{3}=s+\frac{4}{3}). Tip: take the average even for symbolic zeroes.

Correct answer: A. Because it is ((x-7)^2+4)

Explanation: ((x-7)^2+4) is always positive, so (p(x)=0) will not occur. Tip: adding a positive number to a square gives no real intersection.

Correct answer: A. The graph will cross the (x)-axis

Explanation: ((x-3)^5) is an odd-power factor, so the graph crosses at (x=3). Tip: at an odd power the graph usually crosses the axis.

Correct answer: A. (x=q-2)

Explanation: The axis of symmetry is at the average of the zeroes, (\frac{(q-11)+(q+7)}{2}=q-2). Tip: take the midpoint even with symbols.

Correct answer: C. ((0,14))

Explanation: ((0,14)) has (y=14), so it is not on the (x)-axis. Tip: a zero point must have second coordinate (0).

Correct answer: B. Negative

Explanation: 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.

Correct answer: A. Other (7), intersections ((6,0)), ((7,0))

Explanation: In the quadratic, the sum of zeroes is (13), so the other zero is (7). Tip: convert a zero into ((x,0)).

Correct answer: A. Product (-144), sum (0)

Explanation: The zeroes are (-12) and (12), so the product is (-144) and the sum is (0). Tip: opposite zeroes have sum (0).

Correct answer: A. \(\left(\frac{7}{6},0\right),\ \left(-\frac{7}{6},0\right)\)

Explanation: To find x-axis intersections, set \(p(x)=0\). Since \(36x^2-49=(6x-7)(6x+7)\), the equation \((6x-7)(6x+7)=0\) gives \(x=\frac{7}{6}\) or \(x=-\frac{7}{6}\). Therefore, the intercepts are \(\left(\frac{7}{6},0\right)\) and \(\left(-\frac{7}{6},0\right)\). Option C incorrectly uses the reciprocal \(\frac{6}{7}\). Exam tip: factor a difference of squares \(a^2-b^2\) as \((a-b)(a+b)\) before solving.

Correct answer: B. Two

Explanation: Repeated points give the same (x)-values, so the distinct zeroes are (-11) and (4). Tip: count the same (x)-value once.

Correct answer: B. Two

Explanation: Real zeroes are counted from (x)-axis intersections, not from the (y)-axis intercept. Tip: ((0,25)) does not show a zero.

Correct answer: A. ((-6,0)) and ((6,0))

Explanation: (x^4-1296=(x^2-36)(x^2+36)), and the real zeroes are only (\pm6). Tip: (x^2+36) gives no real zero.

Correct answer: C. (8)

Explanation: For eight distinct real zeroes, the degree must be at least (8). Tip: the number of distinct zeroes cannot exceed the degree.

Correct answer: C. The graph does not cut the (x)-axis

Explanation: (x^2+16x+80=(x+8)^2+16), so there is no real zero. Tip: an always positive form gives no intersection.

Correct answer: A. (0,5,7)

Explanation: (x^3-12x^2+35x=x(x-5)(x-7)), so the zeroes are (0), (5), (7). Tip: first take (x) as the common factor.