Geometrical meaning of the zeroes of a polynomial.

The zeroes are (-9), (-1), (4) and the greatest is (4). Tip: the value to the right on the number line is greater.

Intersections with the x-axis occur where \(p(x)=0\). Since \(p(-3)=0\) and \(p(8)=0\), the x-intercepts are (-3,0) and (8,0). Because \(p(2)=5\) ≠ 0, the point (2,5) is not an x-intercept. Exam tip: to find x-intercepts from function values, look for input values that give output 0 (roots).

Factorize: \(p(x)=x^2-10x+25=(x-5)^2\). The root \(x=5\) has multiplicity 2 (a repeated zero), so the parabola touches the x-axis at a single point and does not cross it. Closest distractor C is incorrect because cutting at two distinct points requires two distinct real roots (discriminant > 0). Exam tip: if the discriminant \(b^2-4ac=0\), the quadratic has a repeated root and the graph is tangent to the x-axis.

The governing concept is that the real zeroes of p(x) are the x-values where the graph meets the x-axis. For the quadratic p(x) = x² + 6x + 10, complete the square: p(x) = (x + 3)² + 1. Since (x + 3)² is always at least 0, p(x) is always at least 1, so it can never equal 0 for any real x. Equivalently, its discriminant is b² − 4ac = 36 − 40 = −4, which is negative, confirming that there are no real roots. Therefore option C is correct. Option A would require a positive discriminant, option B would require a zero discriminant, and option D is impossible for a nonzero quadratic.

A root (zero) is an x‑value where the function equals zero (the point lies on the x‑axis). At (0, −6) the y‑value is −6, so the point is a y‑intercept (x = 0, y ≠ 0), not an x‑intercept. Hence −6 is the y‑value, not a root. Why other choices are wrong: B is incorrect because sign alone does not prevent a number being a root — roots can be negative, positive or zero. C is incorrect because x = 0 is not automatically a root; only if f(0) = 0. D is false because not every point gives y = 0. Exam tip: always check which coordinate is x and which is y; zeros correspond to x‑coordinates where y = 0 (x‑axis intersections).

The distance is the absolute difference of their x-coordinates. Calculate as \\(|9-(-4)|=|13|=13\\). Options A and B are incorrect because they do not give the correct difference; option D is incorrect because distance cannot be negative. Exam tip: use the absolute value of the difference of coordinates to find distance on an axis.

The (x)-value of the midpoint is (\frac{-2+10}{2}=4). Tip: on the (x)-axis both points keep (y=0).

The axis of symmetry passes through the average (x=\frac{1+7}{2}=4). Tip: the middle of two zeroes is useful in a parabola.

A non-zero constant multiplier does not change zeroes. Tip: zeroes come from factors that make the value zero.

Factorize: \(x^3-4x = x(x^2-4)=x(x-2)(x+2)\). The x-intercepts (zeros) are values making any factor zero, so \(x=-2,0,2\). Distractor B is wrong because 4 is not a root (\(4^3-4\cdot4=64-16=48\neq0\)). Exam tip: always factor out the common factor \(x\) first, then factor the remaining quadratic as a difference of squares.

A number a is a real zero of a polynomial p(x) if and only if \(p(a)=0\). Therefore 0 is a zero exactly when \(p(0)=0\). On the graph this means the curve must pass through the origin (0,0). The point (5,0) would indicate 5 is a zero, not 0; the point (0,5) has x=0 but y≠0 so 0 is not a root there. Exam tip: compute \(p(0)\) or check the y‑coordinate at x=0 on the graph to decide quickly.

For five distinct real zeroes the degree must be at least (5). Tip: the number of zeroes cannot exceed the degree.

The direct answer is option B: at x = −5. A zero is found by setting the polynomial equal to zero: −2(x + 5)² = 0. The non-zero factor −2 can be divided out, leaving (x + 5)² = 0. A square is zero only when its base is zero, so x + 5 = 0 and x = −5. Therefore the graph meets the x-axis at the point (−5, 0). The squared factor means the parabola touches the x-axis there rather than crossing it, although the question only asks where it meets. Option A is wrong because x = 5 gives −2(10)², not 0. Option B is correct because substituting x = −5 gives −2(0)² = 0. Option C is wrong because x = 2 gives a non-zero value. Option D is wrong because there is one real zero, namely −5. The outside negative factor changes the opening direction of the parabola but does not change where its value is zero. Memory cue: ignore any non-zero multiplier when finding roots; solve the factor equal to zero.

The governing concept is the geometrical meaning of a zero: a real number r is a zero of p(x) exactly when the graph y = p(x) passes through (r, 0). Thus the two points (a, 0) and (b, 0) represent the zeroes a and b, respectively. The condition a < b tells us their order on the number line, so a is the smaller value. Therefore option A is correct. The value b is the larger zero, while 0 is merely the common y-coordinate of both x-axis points and is not automatically a zero. The sum a + b combines the two zeroes but does not identify the smaller one. This reasoning applies regardless of the actual numerical values of a and b.

Zeros of the polynomial are the x-values where p(x)=0. From the table p(-1)=0 and p(3)=0, so the zeros are -1 and 3. Their product is (-1)×3 = −3. Option B (3) ignores the sign; options C and D are incorrect because p(6) and p(-4) are not zero (p(-4)=−2, p(6)=5). Exam tip: only use x-values that satisfy p(x)=0 when identifying roots.

An intersection with the x-axis at \((r,0)\) means the x-coordinate \(r\) is a zero (root) of the graph's equation. Since \(r>0\), the zero is positive. Option A is wrong because \(r\) is not negative; C is wrong because the zero would be 0 only if \(r=0\); D is wrong because an x-axis intersection implies a zero exists. Exam tip: the x-intercept's x-coordinate gives the zero directly.

The governing concept is that x-axis intersections occur where p(x) = 0, and their coordinates are (zero, 0). Factor the polynomial: x² − 3x − 10 = (x − 5)(x + 2), because the two numbers 5 and −2 have product −10 and sum −3. Setting each factor equal to zero gives x − 5 = 0, so x = 5, and x + 2 = 0, so x = −2. Consequently, the graph intersects the x-axis at (5, 0) and (−2, 0), making option A correct. Option B reverses both signs, option C gives y-axis points rather than x-axis points, and option D incorrectly uses the constant and linear coefficients as roots.

The governing concept is the graphical interpretation of zeroes. A real zero is an x-value for which the function value is zero, so it is represented by a point where the graph meets the x-axis. If the graph cuts the x-axis at two different points, the two points have different x-coordinates and therefore give exactly two distinct real zeroes. Hence option A is correct. A graph staying above the x-axis has no real zero, while a graph touching the x-axis at only one point has one distinct real zero, often a repeated root for a quadratic. Intersections with the y-axis do not determine zeroes; they show the value p(0), and an ordinary function can meet the y-axis at only one point.

Both have the same (x)-value (3), so there is one distinct zero. Tip: count a repeated value once for distinct count.

The direct answer is option B: the zeroes are 0 and -6. A zero makes the polynomial value equal to zero, so solve x(x+6)=0. By the zero-product property, x=0 or x+6=0. The second equation gives x=-6. Thus the x-axis intersections are (0,0) and (-6,0). Option A is wrong because it changes -6 to +6; x=6 gives 6(12), not zero. Option B is correct because both values make one factor vanish. Option C is wrong because 1 is not obtained from either equation, and x=6 again has the wrong sign. Option D is incomplete: -6 is a zero, but 0 is also a zero. A plus sign inside a factor gives a negative root: x+a=0 means x=-a.

The graph does not meet the (x)-axis, so there is no real solution. Tip: (p(x)=0) means an (x)-axis intersection on the graph.

The zeroes are the x-coordinates of the x-intercepts. Here the zeroes are -3, 0 and 6, so their sum is -3 + 0 + 6 = 3. The closest distractor B (6) is wrong because it corresponds to omitting the -3 (i.e., adding only 0 and 6). Exam tip: always include every x-intercept's x-value and keep track of signs when summing roots.

x-intercepts occur where the polynomial equals zero. Factorizing gives \(p(x)=x^2-49=(x-7)(x+7)\). Setting \(p(x)=0\) yields \(x=\pm7\), so the graph meets the x-axis at \((7,0)\) and \((-7,0)\). Option C is a common confusion: \((0,7)\) and \((0,-7)\) are y-intercepts, not x-intercepts. Option B would require zeros at \(x=\pm49\) which correspond to \(x^2-2401\), not our polynomial. Exam tip: to find x-intercepts set \(p(x)=0\) and factor; recognize difference of squares quickly.

When listing distinct zeroes, repeated roots are counted only once. Here 5 is repeated, so the distinct real zeroes are 5 and −1. Option C is incorrect because it lists 5 twice; options B and D are incorrect because each omits one of the values. Exam tip: write the values and remove duplicates (think of a set) to list distinct roots quickly.

The vertex lies on the (x)-axis, so the parabola touches at ((-2,0)). Tip: the opening direction does not change the touching (x)-value.