Geometrical meaning of the zeroes of a polynomial.

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.

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

The midpoint is \(\left(\frac{-9+3}{2},0\right)=(-3,0)\). Tip: on the (x)-axis the midpoint has \(y=0\).

The direct answer is option A: the x-axis intersections are (7,0) and (-3,0). Factor the polynomial: x^2-4x-21=(x-7)(x+3), because the numbers -7 and 3 multiply to -21 and add to -4. Set p(x)=0: (x-7)(x+3)=0. Hence x=7 or x=-3. An x-axis point has y=0, so the points are (7,0) and (-3,0). Option A is correct. Option B has both signs reversed; substituting 7 and -3 into the factors shows the actual roots. Option C lists points on the y-axis because their x-coordinate is 0, not x-axis intersections. Option D incorrectly uses the constant and x-coefficient as roots; coefficients are not automatically zeroes. Memory cue: factor, set each factor to zero, then write (root,0).

A zero (root) is an x-value a for which the function value is zero, i.e. the graph passes through (a,0). If the graph meets the y-axis at (0,0), that point lies on the x-axis as well, so \(p(0)=0\) and x=0 is a root. The closest distractor (C) is wrong because although a generic y-intercept need not be a root, the specific y-intercept here is the origin, which is also an x-intercept. Exam tip: any point of the form (a,0) on the graph immediately tells you x=a is a zero of the polynomial.

(x^2-12x+36=(x-6)^2), so the touching point is ((6,0)). Tip: in a perfect square, change the sign to get the zero.

The zeroes are (a), (b), (c), so their product is (abc). Tip: even in symbolic points, the first coordinate is the zero.

((x+2)^2+4) is always positive, so (p(x)=0) will not occur. Tip: adding a positive number to a square can prevent intersection.

((x-3)^2) is an even-power factor, so the graph touches at (x=3). Tip: power (2) shows a repeated zero.

The axis of symmetry is at the average of the zeroes, (\frac{(b-5)+(b+1)}{2}=b-2). Tip: the average rule also works with symbols.

A zero r corresponds to the point (r,0) on the x-axis because at x=r the function value y=0. Therefore for zeros -4, 1 and 8 the x-axis intersections are (-4,0), (1,0) and (8,0), which is option B. Option A is a common error: it places the zeros as y-coordinates, giving points on the y-axis. Exam tip: always treat a zero as the x-coordinate (write it as (x,0)) and keep the given order unless asked otherwise.

In this interval the first two factors are positive and the third is negative, so the product is negative. Tip: check factor signs separately.

In the quadratic, the sum of zeroes is (7), so the other zero is (4). Tip: quickly convert a zero to ((x,0)).

Direct answer: Option B. The graph cuts the x-axis at x=-6 and x=6, so the zeroes are -6 and 6. They have equal magnitudes but opposite signs: 6=-(-6). Their sum is (-6)+6=0. Their product is (-6)(6)=-36, not 36. Option A, equal zeroes, is false because -6 and 6 are different. Option B is correct because the numbers are opposites and their sum is zero. Option C is false because one zero is negative and the other is positive. Option D is false because the product is negative 36. On a graph, opposite zeroes lie the same distance from the y-axis, one on each side. Exam cue: when roots are a and -a, their sum is 0 but their product is -a^2.

Set p(x)=0 to find x-intercepts: 9x^2-16=0 ⇒ 9x^2=16 ⇒ x^2=16/9 ⇒ x=±4/3. Alternatively factor as a difference of squares: 9x^2-16=(3x-4)(3x+4), giving x=4/3 and x=-4/3. Distractor C (±3/4) is the reciprocal mistake; B (±4) would be from x^2-16=0. Exam tip: either factor as (3x-4)(3x+4) or take square roots after isolating x^2 to avoid arithmetic slips.

Real zeroes are counted from (x)-axis intersections, not from the (y)-axis intercept. Tip: ((0,-15)) does not show a zero.

The direct answer is option A: the real x-axis intersections are (-2,0) and (2,0). Set x^4-16=0. Using difference of squares, x^4-16=(x^2-4)(x^2+4)=(x-2)(x+2)(x^2+4). Thus x=2 or x=-2 from the first factors. The equation x^2+4=0 gives x^2=-4, which has no real solution because a real square cannot be negative. Therefore the real points are (-2,0) and (2,0). Option A is correct. Option B confuses 4 with its square root and gives ±4. Option C wrongly includes 0; p(0)=-16, not 0. Option D is wrong because two real roots exist. The question asks for real intersections, so non-real complex roots are not plotted on the real coordinate plane. Memory cue: solve every factor, then discard non-real solutions when real intersections are requested.

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

(x^2+10x+29=(x+5)^2+4), so there is no real zero. Tip: an always positive form gives no intersection.

The direct answer is option A: the zero set is {0,1,5}. First take the common factor x: x^3-6x^2+5x=x(x^2-6x+5). The quadratic factors as (x-1)(x-5), since -1 and -5 multiply to 5 and add to -6. Therefore p(x)=x(x-1)(x-5). Set each factor equal to zero: x=0, x=1, or x=5. Hence the graph has x-intercepts (0,0), (1,0), and (5,0), and its zero set is 0,1,5. Option A is correct. Option B changes both positive roots to negative values. Option C includes 6, which is a coefficient in the expression, not a root, and omits 0. Option D is wrong because all three values make the polynomial zero. Memory cue: extract the common factor first, then factor the remaining quadratic.

(8=\frac{4+12}{2}), so it is the midpoint of the two zeroes. Tip: the midpoint need not be a zero.

For (x<-8), both factors are negative and the outside negative makes the value negative. Tip: check factor signs first.

The x-intercepts \\((5,0)\\) and \\((-3,0)\\) correspond to zeroes 5 and −3. Their sum is \\(5+(-3)=2\\). Note: a touch-point is still a zero (usually even multiplicity) and a cross-point is a zero with odd multiplicity, but multiplicity does not change the numerical values of the zeroes. Exam tip: use the x-coordinates of the given intercepts as the roots directly.