Geometrical meaning of the zeroes of a polynomial.

For (x<-6), both factors are negative and the outside negative makes the value negative. Tip: first check factor signs and then apply the outside sign.

A touch at (4,0) means x=4 is a zero and a crossing at (-2,0) means x=-2 is a zero. Thus the sum of zeros is \(4+(-2)=2\). Closest distractor: 6 would result from incorrectly adding absolute values (4+2) and ignoring the sign; that is incorrect. Exam tip: a touch-point is still a root (often with even multiplicity), and you must include its value when summing zeros.

The x-intercepts give the zeros of the polynomial: here the zeros are \(0\) and \(a\). Their product is \(0\times a=0\). The closest distractor, \(a\), incorrectly treats the product as the nonzero root alone; but a single zero root forces the whole product to be zero. Exam tip: whenever one root is 0, the product of the roots is 0 immediately.

The governing concept is that polynomial zeroes are the x-coordinates where the graph intersects the x-axis. Factor the quadratic by finding two numbers with product 30 and sum 11: x²−11x+30=(x−5)(x−6). Thus the zeroes are 5 and 6, corresponding to points (5,0) and (6,0). Because both points lie on the x-axis, their distance is the absolute difference of their x-coordinates: |6−5|=1 unit. Therefore option A is correct. The value 5 is one zero, 11 is the sum of the zeroes, and 30 is their product; these are related quantities but none is the requested distance.

For a downward-opening parabola, values outside the zeroes are negative. Tip: when the direction changes, sign regions also change.

For real (x), (x^4\geq0), so (x^4+1>0). Tip: an always positive polynomial does not cut the (x)-axis.

A zero (root) is an x‑value where the function value equals 0. Here \(p(-4)=0\), \(p(2)=0\), and \(p(5)=0\), so there are three zeros. Since \(p(0)=3\) is not zero, x=0 is not a root. Exam tip: always verify that the function value is 0 at a given x before counting it as a root.

Set \(p(x)=0\). Factor: \(p(x)=x^2-ax=x(x-a)\). Thus the roots are \(x=0\) and \(x=a\), so the x-intercepts are \((0,0)\) and \((a,0)\). The closest distractor \((0,0),(-a,0)\) is wrong because it flips the sign of the second root; sign errors are common when factoring or solving. Exam tip: always factor and set each factor equal to zero; intercepts have y-coordinate 0, so give points of the form \((\text{root},0)\).

For equal distance from the (y)-axis, zeroes should be opposites, so (3) is needed with (-3). Tip: symmetric zeroes are (a) and (-a).

The zeroes are (2) and (-1), and ((x-2)^2) causes touching at (x=2). Tip: the outside (3) does not change the zeroes.

The average of the two zeroes is (1), so the other zero is (7). Tip: the axis of symmetry passes through the midpoint of zeroes.

The even-power factor ((x-3)^2) gives touching and the single factor (x+4) gives crossing. Tip: identify behavior from factor power.

(x=-3) lies between the two zeroes, and an upward parabola is below the axis there. Tip: check the sign between zeroes.

In this interval the first factor is positive and the second is negative, and the outside negative makes the value positive. Tip: check each factor's sign separately.

The graph touches the x-axis when the polynomial has a repeated zero. Here, the expression can be rewritten as a perfect square: \(p(x)=x^2-4kx+4k^2=(x-2k)^2\). A square is zero only when its inside expression is zero, so \(x-2k=0\), giving \(x=2k\). Thus the graph has one repeated intercept and touches the x-axis at the point whose x-coordinate is \(2k\), not at \(k\), \(-2k\), or \(4k\). Recognising the perfect-square form is the quickest method.

Equivalently, the quadratic formula gives a discriminant of \((-4k)^2-4(1)(4k^2)=0\), confirming that both zeroes coincide. Their common value is \(\frac{4k}{2}=2k\). Therefore option B is correct. The graph does not cross the axis at this repeated zero; it merely touches it, because \((x-2k)^2\) is never negative.

Factor the polynomial: \(p(x)=x^2-(m+n)x+mn=(x-m)(x-n)\). Hence the roots are \(x=m\) and \(x=n\), so the x-intercepts are (m,0) and (n,0). Closest distractor (\(-m,0\),\(-n,0\)) is incorrect because the signs are negated; (0,m),(0,n) are y-intercepts; (m,n) are arbitrary coordinate points and do not represent roots. Exam tip: set \(p(x)=0\) or factorize to find roots, then report each root as (root,0).

The average of the two zeroes is (2), so the other zero is (-5). Tip: set ( \frac{a+b}{2} ) equal to the axis of symmetry.

A zero of p(x) is an input x for which p(x)=0. From the information given, p(-6)=0 and p(4)=0, so the stated zeroes are -6 and 4. The values p(-1)>0 and p(7)<0 are sign information, not additional zeroes, because neither value equals zero. Their product is (-6)(4)=-24. Therefore Option A is correct. Option B loses the negative sign, Option C incorrectly multiplies or combines the two x-values as a difference, and Option D would be possible only if one of the zeroes were 0. The inequalities help describe signs near the roots but do not change the requested product.

The governing concept is that distinct real zeroes correspond to distinct points where the graph meets the x-axis. First factor out the common factor x: x^3-16x = x(x^2-16). Then use the difference of squares: x(x^2-16)=x(x-4)(x+4). Setting each factor equal to zero gives x=0, x=4 and x=-4. These are three different real x-values, so the graph has three distinct x-axis intersection points. Option C is correct. Option A omits two roots, Option B omits one root, and Option D incorrectly counts beyond the three factors or treats multiplicity as an additional distinct point.

Both factors have even powers, so the graph touches at both places. Tip: an even-power zero usually gives touching, not crossing.