Geometrical meaning of the zeroes of a polynomial.

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.

The same (x)-value (4) is repeated, so there is one distinct zero. Tip: do not count repetition for distinct zeroes.

Zeros (roots) occur where \(p(x)=0\). Solving \(x^2-1=0\) gives \(x=\pm1\), so the points \((-1,0)\) and \((1,0)\) are zeros on the graph. The point \((0,-1)\) has \(y=-1\) (i.e. \(y\neq0\)), so it is not a root. The closest distractor D listing both points is incorrect because those two points do represent zeros. Exam tip: roots correspond to x‑intercepts where \(y=0\).

On the number line a number located to the right is greater. -2 lies to the right of -9, so -2 is the greater zero. -9 is incorrect because it is further left (smaller) than -2. Exam tip: among negative numbers the one with the smaller absolute value is the greater number (e.g. |−2|=2 < |−9|=9).

The zeros (roots) are the x-coordinates where the graph meets the x-axis: here they are \(-10\) and \(0\). Thus the product is \((-10)\times 0 = 0\). Option A (-10) is incorrect because it uses only one root; option D (100) is wrong — it seems to confuse absolute values or squaring. Exam tip: if one root is \(0\), the product of the roots is always \(0\).

Core idea: \(p(x)=5\) is a nonzero constant polynomial, so its graph is the horizontal line \(y=5\). This line never meets the \(x\)-axis, hence it is parallel to the \(x\)-axis and does not cut it. Why the closest distractor is wrong: option A (cuts once) would apply to a linear polynomial with a real root; here \(p(x)\) never equals 0, so no intersection. Exam tip: a zero of a polynomial is an \(x\) with \(p(x)=0\); nonzero constant polynomials have no real zeros.

The zero polynomial satisfies p(x)=0 for every real x, so the graph consists of all points (x,0) — i.e. the x-axis. B is wrong because the y-axis is the vertical line x=0, not y=0 for all x. C is wrong because "above the x-axis" means y>0, but here y=0. D is wrong because the graph does touch the x-axis at every point (it is the x-axis). Exam tip: check p(x) for an arbitrary x; if p(x)=0 always then the graph is y=0. Note: the degree of the zero polynomial is not normally defined (sometimes taken as −∞).

If \(p(a)=0\) then the graph passes through the point \((a,0)\). Since the zeros 2, 5 and 9 are distinct, there are three distinct x‑intercepts. The closest distractor 'Two' would only be correct if two of the zeros coincided; 'Nine' confuses the value 9 with the count of intercepts. Exam tip: always check whether zeros are distinct — repeated roots (multiplicity) still produce a single x‑intercept.

The axis of symmetry passes through the midpoint of zeroes, (\frac{-2+6}{2}=2). Tip: the average of two zeroes is useful for a parabola.

It is ((x+1)^2+4), which is always positive. Tip: if the value never becomes (0), there is no (x)-axis intersection.

A number \(a\) is a zero (root) of a polynomial if and only if \(p(a)=0\). Here \(p(3)=0\), so 3 is a zero and the graph meets the x-axis at x=3. Since \(p(7)=4\neq0\), 7 is not a zero. Closest distractors (B and C) are wrong because they assume \(p(7)=0\). Exam tip: always check the function value equals exactly 0 before calling a point a root.

Zeros are the x-values that make the polynomial zero. Solve \(2x^2-8=0\). Divide both sides by 2 to get \(x^2=4\), so \(x=\pm2\). Hence the zeros are (2, -2). The closest distractor (4, -4) would arise from the mistake of not dividing by 2 and assuming \(x^2=16\). Exam tip: factor first: \(2(x^2-4)=2(x-2)(x+2)\) to read off the zeros quickly.

The first coordinate (r) of the intersection is the zero, and (r<0). Tip: the same rule works in symbolic questions.

The negative sign does not change the zero, and ((x+3)^2=0) gives (x=-3). Tip: an outside multiplier does not change the zero.

Points where the graph meets the x-axis are the polynomial's zeros, so the product is the product of the x-values: (-1)×2×4 = -8. Option A (8) would be the result if the negative sign were ignored; options C and D come from incorrect arithmetic. Exam tip: multiply the x-intercepts directly and pay attention to signs (negative × positive = negative).