Geometrical meaning of the zeroes of a polynomial.

The average of the two zeroes is (2), so the other zero is (7). Tip: in a parabola the axis of symmetry passes through the midpoint of the zeroes.

The graph meets the (x)-axis at three distinct (x)-values. Tip: both touching and crossing give zeroes.

The squared factor ((x+4)^2) gives touching, and the single factor (x-3) gives crossing. Tip: an even-power factor usually shows touching.

For an upward-opening parabola, the graph lies below the (x)-axis between the two zeroes. Tip: identify the sign region between zeroes.

Between them ((x-1)) is positive and ((x-5)) is negative, and the outside negative makes the value positive. Tip: graph position is decided by the sign of (p(x)).

The governing concept is that every x-axis intersection (r, 0) identifies r as a real zero of the polynomial. From the three given points, the zeroes are −4, 2, and 9. Their product is (−4) × 2 × 9 = −8 × 9 = −72. Thus option B is correct. The result must be negative because exactly one of the three factors is negative; multiplying one negative factor by positive factors gives a negative product. Option A has the right absolute value but loses the sign. Option C, 7, is the sum −4 + 2 + 9, not the product. Option D, −11, is not obtained by the required multiplication. The second coordinate, 0, only confirms that the points lie on the x-axis and is not multiplied.

The vertex is below the (x)-axis and the graph opens upward, so it cuts the (x)-axis twice. Tip: check vertex height and opening direction together.

The vertex lies on the (x)-axis, so the graph touches there. Tip: the opening direction does not change the touching point.

Factor the polynomial: \(p(x)=x^2-2kx+k^2=(x-k)^2\). Hence the root is repeated at \(x=k\), so the graph touches (but does not cross) the x-axis at \(x=k\). The distractor \(x=-k\) is incorrect because it is not a root of \((x-k)^2\). Exam tip: recognize perfect-square trinomials quickly, or check the discriminant \(b^2-4ac=0\) to identify a repeated root.

The polynomial equals ((x-a)(x-b)), so the zeroes are (a) and (b). Tip: connect factor form with graph intersections.

The average of the two zeroes is (-1), so the other zero is (-6). Tip: set the average equal to the axis of symmetry.

A value is a zero of \(p(x)\) only when the function value at that input is exactly 0. Statements such as \(p(-5)>0\) and \(p(1)<0\) describe points above and below the x-axis, respectively; they do not make -5 or 1 zeroes. The equalities \(p(x)=0\), on the other hand, directly identify zeroes.

From \(p(-2)=0\), the value \(-2\) is a zero. From \(p(6)=0\), the value 6 is another zero. Their sum is \((-2)+6=4\). Hence option A is correct. It would be an error to add all four listed x-values, because the inequality signs show only the position relative to the axis, not an x-axis intersection.

If a polynomial's graph either crosses or merely touches the x-axis at a value of x, the polynomial equals zero at that x (\\(p(x)=0\\)). Therefore both x = -3 (crossing) and x = 2 (touching) are real zeroes. Geometrically, crossing usually indicates an odd multiplicity (often 1) and touching indicates an even multiplicity (\\(\ge2\\)), but both imply roots. The closest distractor B is wrong because 'touching' still means the polynomial vanishes there. Exam tip: check whether the graph crosses or touches the axis — that tells you the presence of a root and suggests whether its multiplicity is odd or even.

(x^3-9x=x(x-3)(x+3)), so there are three distinct zeroes. Tip: take the common factor and use difference of squares.

Both factors have even powers, so the graph touches at both points. Tip: at an even power the graph usually turns back.

To find the zeroes of a factored polynomial, set each factor equal to zero. A product is zero whenever at least one of its factors is zero. In this expression, the factor \(x+1\) gives one zero, while the factor \((x-4)^3\) gives another zero. The exponent 3 shows that the second zero occurs repeatedly, but it does not create three different zeroes.

Solving \(x+1=0\) gives \(x=-1\). Solving \(x-4=0\) gives \(x=4\). Thus the zeroes are \(-1\) and \(4\), and the distinct zeroes are exactly these two values. Option C lists the repeated value three times, so it describes multiplicity rather than distinct zeroes. Therefore option A is correct.

The midpoint is \(\left(\frac{-7+1}{2},0\right)=(-3,0)\). Tip: on the (x)-axis the midpoint has \(y=0\).

The range is the difference between greatest and smallest zero, (4-(-8)=12). Tip: range is always non-negative.

(x^2+2x-15=(x+5)(x-3)), so the zeroes are (-5) and (3). Tip: form points ((x,0)) from factors.

A root (zero) means the x-value for which the polynomial equals zero, i.e. the point has y=0. The origin (0,0) has x=0 and y=0, so if the graph passes through (0,0) then \(p(0)=0\) and x=0 is a root. Why other choices fail: A and C confuse being on the y-axis with not being a root; D is wrong because the information (0,0) is sufficient to conclude \(p(0)=0\). Exam tip: remember roots are x-values where y=0 — check coordinates to see which axis values are zero.