Geometrical meaning of the zeroes of a polynomial.

The first coordinates of intersection points are the zeroes. Tip: choose points whose second coordinate is (0).

The midpoint of two zeros is their average. So the midpoint is \(\frac{-4+6}{2}=1\). Option C (5) is a common mistake — it equals half the distance between the zeros (\(6-(-4)=10\) so half is 5) but not the midpoint itself; to get the midpoint add that half-distance to the left root: \(-4+5=1\). Exam tip: compute the average directly and be careful with signs.

Intersection with the x-axis means \(p(x)=0\). Setting each factor to zero gives \(x+7=0\Rightarrow x=-7\) and \(x-1=0\Rightarrow x=1\). Thus the x-intercepts are \((-7,0)\) and \((1,0)\). Closest distractor: option A has the signs reversed (a common sign error); option B lists y-axis points, not x-axis intercepts. Exam tip: set each factor to zero and verify by substituting into \(p(x)\).

Each distinct x-value where the graph meets the x-axis is a real zero of the polynomial. The graph touches at (2,0), so x=2 is a zero (typically even multiplicity), and crosses at (5,0), so x=5 is another zero (odd multiplicity). These are two distinct real zeros. Why distractors fail: "One" is wrong because there are two different x-values; "Three" and "Zero" are not supported by the given points. Exam tip: count distinct x-coordinates where the graph meets the x-axis; touching vs crossing tells multiplicity, not distinctness.

A zero of a polynomial is an x-value where \(p(x)=0\). The table shows \(p(-5)=0\) and \(p(4)=0\), so \(-5\) and \(4\) are zeros. Option A is incorrect because \(p(-1)=2\) and \(p(6)=9\) are not zero. Exam tip: when given function values in a table, pick the x‑entries whose function value equals 0.

The value p(0) gives the y-intercept of the polynomial's graph. p(0) = -3 means the graph passes through the point (0, -3). x = 0 would be a root only if p(0) = 0, which is not the case here, so B is wrong. The graph crosses the x-axis at x = -3 only if p(-3) = 0, which we are not told, so C is incorrect. Also p(0) ≠ 0 does not imply there are no zeros at all, so D is false. Exam tip: to test whether x = a is a root compute p(a); to get the y-intercept evaluate p(0).

The graph does not meet the (x)-axis, so there is no real zero. Tip: the vertex position can help judge intersections.

Set \(p(x)=0\). Factorizing gives \(x^2-2x=x(x-2)\), so \(x(x-2)=0\) implies \(x=0\) or \(x=2\). Thus the graph meets the x-axis at \(x=0\) and \(x=2\). Closest distractor B is incorrect because \(p(1)=1-2=-1\), not zero. C and D are also wrong since the factorization shows two distinct zeros. Exam tip: factor out the common term first to find zeros quickly.

The x-values where the graph meets the x-axis are the zeros of the polynomial. So the sum = (−2) + 1 + 5 = 4. The nearest distractor 8 is wrong — it might arise from adding absolute values or a calculation error. Exam tip: directly add the x-coordinates of x-intercepts (with their signs) to get the sum of zeros.

Zeros correspond to x‑values where the graph meets the x‑axis, i.e. y=0. Among the given points only (-3,0) and (8,0) have their second coordinate equal to 0, so they represent zeros. Option C is wrong because (0,4) has y≠0; option A is wrong because neither point has y=0; option D is incorrect since not all points have y=0. Exam tip: always check the second coordinate — zeros require y=0.

The zero (root) is the x-value where the graph crosses the x-axis, so set \(y=0\). For \(y=4x+12\), putting \(y=0\) gives \(4x+12=0\) ⇒ \(x=-3\). Option B (3) is the wrong sign; options C (12) and D (−12) confuse the y-intercept (which is 12) with the x-intercept. Exam tip: to find roots of a linear graph, always set \(y=0\).

In \(p(x)=(x-6)^2\) the root \(x=6\) has multiplicity 2 (a repeated root). An even multiplicity causes the graph to touch the x-axis and turn back, so the curve touches at \(x=6\). Option B is wrong due to the wrong sign; option C would require two distinct real roots, not a repeated one; option D is wrong because a real root exists. Exam tip: even multiplicity → touch/turn, odd multiplicity → cross the axis.

For four distinct real zeroes, the degree must be at least (4). Tip: the number of zeroes cannot exceed the degree.

The y-intercept of a polynomial is y = p(0). If the graph passes through (0,5) then the y-intercept is 5, i.e. \(p(0)=5\), so \(p(0)\neq0\) and x=0 is not a root. The other options assert (directly or indirectly) that \(p(0)=0\) (graph through the origin or stated \(p(0)=0\)), which would make x=0 a root, so they are incorrect. Exam tip: evaluate \(p(0)\) — if it equals 0, x=0 is a root; otherwise it is not.

The graph represents \(y=p(x)\). If \(p(a)=0\), the point \((a,0)\) lies on the graph because the function value (y-coordinate) is zero. Given \(p(-2)=0\) and \(p(3)=0\), the required points are \((-2,0)\) and \((3,0)\). Distractor (B) would be true only if \(p(0)=-2\) and \(p(0)=3\), which is a different (and inconsistent) statement. Exam tip: remember that the zero of a function gives the x-coordinate of the x-intercept — write \(p(a)=0\) as \((a,0)\).

The polynomial factors as \(x^2+4x+4=(x+2)^2\). Thus the only root is \(x=-2\), but it has multiplicity 2 — so there is one distinct real zero (a repeated root). Using the discriminant: \(\Delta=b^2-4ac=4^2-4\cdot1\cdot4=0\), which indicates a repeated real root. The closest distractor (A) is incorrect because \(\Delta\) is not positive, so there are not two distinct real zeros. Exam tip: check for a perfect square trinomial or compute \(\Delta\) to decide distinct/repeated/no real roots quickly.

The x-intercepts are the zeros of the polynomial; here the zeros are -7 and 2. Product = (-7) × 2 = -14. The closest incorrect choice 14 ignores the negative sign; always track signs when multiplying zeros. Exam tip: use the x-coordinates of the intercepts as the zeros and include their signs when computing product or sum.

The x-coordinate of an x-axis intersection gives the real zero. For the point (-5, 0) the x-value is -5, so the real zero is -5. The phrase 'graph crosses there' indicates an odd multiplicity (often 1) but the asked quantity is simply the x-value. The closest distractor 5 is wrong because it flips the sign; 0 is wrong because the intersection's x-coordinate is not 0; 'No real zero' is wrong because an intersection exists. Exam tip: read the ordered pair carefully and take the x-coordinate (including its sign).

Zeros correspond to x-coordinates on the number line, so the distance between them is the absolute difference of their x-values. Distance = \(|10-2|=8\). Common mistake: adding the coordinates to get \(2+10=12\), which explains why option B is incorrect. Exam tip: visualize the points on a number line and use the absolute difference to get a positive distance.

Zeros are values of x that make the polynomial equal to zero. For \(p(x)=3(x+4)(x-2)\), setting \(p(x)=0\) gives \(3(x+4)(x-2)=0\). Since 3\neq0, this is equivalent to \((x+4)(x-2)=0\), so the zeros are \(x=-4\) and \(x=2\). A non-zero constant factor does not change the zeros. Option A is incorrect because multiplying the zeros would require the factor to act on x (e.g. a factor inside the variable like \(3x\)); a constant multiplier does not do that. Options C and D are wrong because there is no factor that introduces \(x=3\) or cancels the existing roots. Exam tip: when finding zeros, set the polynomial equal to zero and factor; you can ignore any non-zero constant multiplier.

The vertex is on the (x)-axis, so the parabola touches there. Tip: if the vertex has (y=0), check for touching.

Why correct: A polynomial is continuous everywhere. If a continuous function takes opposite signs at two points (here \(p(-1)>0\) and \(p(4)<0\)), the Intermediate Value Theorem guarantees at least one point in between where \(p(x)=0\). Hence there is at least one x-axis intersection in \((-1,4)\).
Why other options are wrong: A is wrong because positive at \(-1\) and negative at \(4\) does not mean those points themselves are zeros. C contradicts the sign change — a root must exist. D is irrelevant: cutting the y-axis is unrelated and not implied by the given values.
Exam tip: For polynomials check values at two x's — a sign change implies a root between them by the Intermediate Value Theorem.

If the graph touches the x-axis at (a,0), the polynomial evaluates to zero at x=a, i.e. \(p(a)=0\); thus x=a is a real zero. Typically an even multiplicity zero causes the graph to touch (not cross) the axis, while an odd multiplicity causes crossing. Option A is incorrect because touching still implies a zero. Option D is wrong because (a,0) lies on the x-axis, not the y-axis. Exam tip: substitute x=a into the polynomial to verify \(p(a)=0\), or factor the polynomial to check the multiplicity of the root.

To find where the graph touches the x-axis, set p(x) equal to zero because every x-axis point has y-coordinate 0. Factor the polynomial: x² − 6x + 9 = (x − 3)². Thus (x − 3)² = 0 gives x = 3 as a repeated zero. A repeated zero explains why the parabola touches the x-axis at one point instead of crossing it. Since the y-coordinate there is 0, the contact point is (3, 0), so option B is correct. Option A has the wrong sign. Option C confuses the constant term 9 with the zero. Option D is the y-intercept, obtained by putting x = 0, not the x-axis contact point. The factorisation and coordinate condition together confirm the answer.

Every point where a polynomial graph meets the x-axis has y-coordinate zero, so its x-coordinate is a real zero of the polynomial. A crossing and a touching are both types of x-axis contact; the difference concerns the behaviour of the graph near the zero, not whether the value is zero. The graph crosses at three points and touches at one more point. Since the question describes these as separate contacts, there are 3 + 1 = 4 distinct x-axis points and therefore four distinct real zeroes. Option B is correct. Counting only crossings gives three and wrongly ignores the touching point; one and zero do not account for the stated contacts.