Geometrical meaning of the zeroes of a polynomial.

The axis of symmetry is at the average of zeroes, (\frac{(a-2)+(a+4)}{2}=a+1). Tip: take the average even for symbolic zeroes.

If a polynomial has a zero a, the graph meets the x-axis at (a,0) because the polynomial's value is zero at x = a. For zeros −2, 3 and 7 the x-intercepts are (−2,0), (3,0) and (7,0), which is option B. Distractor A incorrectly treats zeros as y-intercepts (0,a); C and D alter the order or signs of coordinates, so they are wrong. Exam tip: always write a zero a as the point (a,0) on the x-axis.

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

In the quadratic, the sum of zeroes is (5), so the other zero is (3). Tip: immediately convert a zero to ((x,0)).

The zeroes are (-5) and (5), which are opposite numbers. Tip: such zeroes are equally distant from the (y)-axis.

x-intercepts occur where \(p(x)=0\). From \(4x^2-25=0\) we get \(4x^2=25\), so \(x^2=\tfrac{25}{4}\) and hence \(x=\pm\tfrac{5}{2}\). Therefore the intercepts are \(\left(\tfrac{5}{2},0\right)\) and \(\left(-\tfrac{5}{2},0\right)\). Alternatively factor \(4x^2-25=(2x-5)(2x+5)\) to read off the roots. The closest distractor (B) reflects a common slip of using 5 instead of \(\tfrac{5}{2}\). Exam tip: treat \(4x^2\) as \((2x)^2\) or take square roots carefully to avoid fraction errors.

The zeroes are (-2) and (5), so the polynomial is ((x+2)(x-5)=x^2-3x-10). Tip: form factors from graph intersections.

Real zeroes are counted from (x)-axis intersections, not from the (y)-axis intercept. Tip: ((0,12)) is not a zero.

Direct answer: Option A, (-1,0) and (1,0). An x-axis intersection has y=0, so we solve p(x)=0: x^4-1=0. Factor it as (x^2-1)(x^2+1)=(x-1)(x+1)(x^2+1). The real solutions are x=1 and x=-1. The factor x^2+1=0 would give x^2=-1, which has no real solution. Therefore the graph meets the x-axis at (-1,0) and (1,0). Option A is correct. Option B wrongly includes (0,0), but p(0)=-1, not 0. Option C uses 4 as though it were a root, but p(4) is not zero. Option D is wrong because two real roots exist. Remember: factor first, then keep only real solutions and write each as (x,0).

For three distinct real zeroes, the degree must be at least (3). Tip: the number of distinct zeroes cannot exceed the degree.

(x^2-2x+5=(x-1)^2+4), so it cannot be zero for real (x). Tip: an always positive form gives no real intersection.

(x^3-4x^2-5x=x(x-5)(x+1)), so the zeroes are (0), (5), (-1). Tip: first take (x) as the common factor.

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.

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.