Geometrical meaning of the zeroes of a polynomial.

An intersection with the x-axis means y=0, so the x-coordinate satisfies \(p(x)=0\); hence x-axis intercepts are the polynomial's zeros (roots). An intersection with the y-axis occurs at x=0 and the y-value is \(p(0)\), which is not necessarily zero. Thus B is incorrect because it wrongly claims the y-axis always gives \(p(x)=0\). C is only true in the special case that \(p(0)=0\) (i.e. zero is a root); D is false because an x-axis intersection does indicate a zero. Exam tip: set y=0 and solve \(p(x)=0\) for roots; set x=0 to find the y-intercept \(p(0)\).

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.