Geometrical meaning of the zeroes of a polynomial.

A zero of a polynomial is an x-value at which the polynomial value is 0. Geometrically, p(a) = 0 means that the graph contains the point (a, 0) on the x-axis. Here p(1) = 0, p(2) = 0, and p(3) = 0, so the graph contains (1, 0), (2, 0), and (3, 0). Because the question explicitly states that the three zeroes are distinct, these are three different points. Hence option C is correct. Option A and option B count too few intersections. Option D incorrectly counts both coordinates or treats each zero as producing more than one point. Each distinct x-value paired with y = 0 gives exactly one distinct x-axis point.

The governing concept is the meaning of distinct zeroes. Distinct means that each different numerical value is listed only once. The stated zeroes are 2, 2, and −5. Although 2 occurs twice, both occurrences represent the same x-coordinate and the same x-axis point (2, 0); the repetition may indicate multiplicity, but it does not create a new distinct zero. The value −5 gives another point, (−5, 0). Therefore the distinct real zeroes are 2 and −5, so option A is correct. Option C lists repeated occurrences rather than distinct values. Option B omits −5, and option D omits 2. Thus the repeated entry must be removed only for the purpose of listing distinct zeroes.

Answer: B. The graph meets the x-axis at the points (−5,0) and (5,0). The y-axis is the line x=0. The perpendicular distance of a point (x,y) from the y-axis is |x|. Hence the two distances are |−5|=5 and |5|=5, so the zeroes are equally distant from the y-axis. Option A is false because a point on the y-axis must have x=0, not x=−5 or 5. Option C is false because −5 lies to the left of the y-axis, while 5 lies to its right. Option D is also false for the same reason. The opposite signs show reflection across the y-axis. Memory cue: equal absolute x-values mean equal distance from the y-axis.

The graph meets the x-axis when the polynomial value is zero. In factored form, p(x) = (x + 2)(x − 7), so the product is zero when either factor is zero. From x + 2 = 0, we get x = −2; from x − 7 = 0, we get x = 7. The corresponding x-axis points must have y-coordinate 0, giving (−2, 0) and (7, 0). Therefore option B is correct. Option A reverses the signs of both roots. Option C writes the values as y-coordinates on the y-axis, and option D pairs the roots with nonzero y-coordinates, so neither represents x-axis intersections.

The governing concept is the range of a finite set of real numbers, calculated as the greatest value minus the least value. The zeroes given by the x-intercepts are -12, -3 and 8. The greatest zero is 8 and the least zero is -12. Therefore, range = maximum - minimum = 8 - (-12) = 8 + 12 = 20. Hence option C is correct. The value 11 is the difference between -12 and -3, while 15 is the difference between -3 and 12 only by an incorrect sign interpretation. Option D gives -20, but a range is a non-negative spread and cannot be negative. The middle zero does not affect the maximum-minus-minimum calculation.

For a polynomial graph, an intersection with the x-axis has y-coordinate zero, and its x-coordinate is a zero of the polynomial. The three given intersections therefore represent the zeroes 0, g, and −g. Their sum is 0 + g + (−g) = 0, because g and −g are additive inverses and cancel. Thus option A is correct. The condition g ≠ 0 ensures that the two symbolic nonzero intersections are distinct from the origin and from each other; it does not alter their sum. The expressions g² and −g² involve multiplication, not addition, while g alone represents only one of the three zeroes.

A real zero is an x-value for which p(x) = 0. On a graph, this is exactly an x-coordinate where the curve meets the x-axis. If the entire parabola lies above the x-axis and never touches or crosses it, there is no point on the graph whose y-coordinate is zero. Consequently, it has zero real zeroes, so option A is correct. A parabola tangent to the x-axis at one point has one repeated real zero, while a parabola crossing the x-axis at two points has two distinct real zeroes. Infinitely many zeroes are impossible for a nonzero quadratic polynomial, because a quadratic can have at most two real roots. The graph's position therefore determines the answer directly.

The vertex is the turning point of a parabola. If this vertex lies on the x-axis, its y-coordinate is zero, so the parabola meets the x-axis exactly at its turning point. It does not cross the axis at two separate locations; instead, the contact represents a repeated root. Algebraically, the quadratic can be written in the form a(x − r)², so both real zeroes are r and r. Therefore option A is correct. Distinct real zeroes would place the vertex away from the axis and produce two crossings. No real zero would occur if the parabola stayed completely above or below the axis, and a quadratic cannot have three real zeroes.

A zero of a polynomial is an x-value for which the polynomial equals zero. On the graph, this corresponds to an intersection with the x-axis, whose y-coordinate is 0. The three stated intersections therefore give the x-values −2, 0, and 5, so the zeroes are −2, 0, and 5. Hence option A is correct. The origin (0, 0) lies on the x-axis, so it absolutely can represent the zero x = 0; being the origin does not exclude it. Option B omits a valid zero, option C contradicts the definition of a graph zero, and option D counts only the two nonzero intersections. Since the three points are distinct, they represent three distinct real zeroes.

A solution of p(x) = 0 is an x-value for which the graph has y-coordinate zero. Such points are precisely the intersections of the graph with the x-axis. The stated intersections have x-coordinates -7 and -2, so p(-7) = 0 and p(-2) = 0. Hence the solution set is {-7, -2}, represented by option A. The negative signs must be retained because they are the actual x-coordinates of the intercepts; changing them to 7 and 2 gives different points on the opposite side of the y-axis. Option C keeps only one sign correct, and option D ignores the two explicitly given x-axis intersections. No calculation beyond reading the graph is needed.

Rationalising the denominator gives 1/√2 = (1/√2) × (√2/√2) = √2/2. The number √2 is irrational, and dividing it by the non-zero rational number 2 keeps the result irrational. Therefore 1/√2 is irrational. It is nevertheless a real number because √2 is real and non-zero. It cannot be an integer or a natural number, since those are all rational numbers, whereas the expression has just been shown to be irrational. It is also not zero because the numerator is 1 and the denominator √2 is non-zero. Thus the denominator simplification confirms option A rather than any of the numerical alternatives.

Answer: C, 2 and −3. A zero of p(x) is an x-value for which p(x)=0. The polynomial is already factorised, so use the zero-product property: a product is zero when at least one factor is zero. Set x−2=0, giving x=2. Set x+3=0, giving x=−3. These are the x-intercepts, or the points where the graph meets the x-axis; they can therefore be marked at 2 and −3 on the number line. Checking gives p(2)=0×5=0 and p(−3)=−5×0=0. A changes −3 to +3, B changes 2 to −2, and D changes both signs incorrectly. The sign warning is important: x−a gives zero a, while x+a gives zero −a.