Geometrical meaning of the zeroes of a polynomial.

(x^2+10x+29=(x+5)^2+4), so there is no real zero. Tip: an always positive form gives no intersection.

(x^3-6x^2+5x=x(x-1)(x-5)), so the zeroes are (0,1,5). Tip: first take (x) as the common factor.

(8=\frac{4+12}{2}), so it is the midpoint of the two zeroes. Tip: the midpoint need not be a zero.

The governing concept is the geometrical meaning of zeroes: a zero is an x-coordinate where the graph meets the x-axis. Factor the polynomial: p(x) = 3x^2 - 18x + 27 = 3(x^2 - 6x + 9) = 3(x - 3)^2. Thus the only zero is x = 3, and it is repeated. A quadratic with a repeated real zero touches the x-axis at that point instead of crossing it. Therefore, option A is correct. Option B has the wrong sign, option C would require two distinct zeroes, and option D would apply only when there are no real zeroes.

For (x<-8), both factors are negative and the outside negative makes the value negative. Tip: check factor signs first.

The x-intercepts \\((5,0)\\) and \\((-3,0)\\) correspond to zeroes 5 and −3. Their sum is \\(5+(-3)=2\\). Note: a touch-point is still a zero (usually even multiplicity) and a cross-point is a zero with odd multiplicity, but multiplicity does not change the numerical values of the zeroes. Exam tip: use the x-coordinates of the given intercepts as the roots directly.

The zeroes of a polynomial are the x-coordinates of its intersections with the x-axis. The point \((0,0)\) therefore gives the zero 0, while the point \((b,0)\) gives the zero b. The condition \(b\ne0\) ensures that these are two distinct points, so both zeroes must be included in the sum.

Adding the two x-coordinates gives \(0+b=b\). Thus the sum of the zeroes is b, which is option B. The value \(b^2\) would be related to a product in some contexts, not to this sum, and \(-b\) has the wrong sign. The origin should not be ignored: although its x-coordinate is 0, it contributes 0 to the sum while still being an important zero.

Answer: A, 1 unit. The zeroes are the x-values where the graph meets the x-axis. Set p(x)=0 and factor the quadratic. We need two numbers with product 42 and sum 13: 6 and 7. Thus x²−13x+42=(x−6)(x−7), so the zeroes are 6 and 7. The corresponding points are (6,0) and (7,0). Their distance on the x-axis is the absolute difference |7−6|=1 unit. Option A is correct. Option B is one zero, not the separation. Option C is the sum of the zeroes, since 6+7=13. Option D is their product, since 6×7=42. Those relationships help with factorisation but do not answer the distance question. Memory cue: distance between two x-axis points is the absolute difference of their x-coordinates.

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

For real (x), (x^6\geq0), so (x^6+1>0). Tip: an always positive polynomial does not cut the (x)-axis.

A zero of the polynomial is an x‑value where the function equals 0. Here \(p(-7)=0\), \(p(1)=0\) and \(p(4)=0\), so there are three zeroes. The closest incorrect choice "four" is wrong because \(p(-3)=2\), not 0. Exam tip: verify by substituting the x‑value into \(p(x)\) and confirm the result is exactly 0 before counting it as a zero.

Factor: \(p(x)=x^2-bx= x(x-b)\). x-intercepts occur where \(p(x)=0\), so \(x(x-b)=0\) gives \(x=0\) or \(x=b\). Thus the intercepts are \((0,0)\) and \((b,0)\). The closest distractor (B) is incorrect because \(p(-b)=(-b)^2-b(-b)=2b^2\), which is not zero for general \(b\ne0\). Option (D) is wrong since \((0,b)\) is on the y-axis, not the x-axis. Exam tip: set \(p(x)=0\) and factor out the common \(x\) to find x-intercepts quickly.

For equal distance from the (y)-axis, zeroes should be opposites, so (4) is needed with (-4). Tip: symmetric zeroes are (a) and (-a).

The zeroes are (-2) and (7), and ((x+2)^2) causes touching at (-2). Tip: the outside (5) does not change the zeroes.

In this interval the signs are (+), (-), (-), so the product is positive. Tip: an odd number of negative factors gives a negative value.

The discriminant is (c^2-4c^2=-3c^2<0), so there are no real zeroes. Tip: a negative discriminant means no (x)-axis intersection.

The vertex lies on the (x)-axis, so the parabola touches at ((4,0)). Tip: if the vertex has (y=0), there is one distinct zero.

The discriminant is (64-88=-24), so there are no real zeroes. Tip: with negative discriminant a parabola does not meet the (x)-axis.

It is ((x-a)^2-9), so (x-a=\pm3) and the zeroes are (a-3), (a+3). Tip: use difference of squares.

Mean \(=\dfrac{-1+3+7}{3}=3\). Explanation: The x-coordinates of the intercepts are the zeros; their sum is 9 and dividing by 3 gives the mean 3. Option 9 is incorrect because it is the sum, not the average (division was omitted). Option -3 results from a sign error, and 7 is just one of the zeros. Exam tip: read off the x-values from intercepts first, then compute average = (sum)/(count).