Geometrical meaning of the zeroes of a polynomial.

(x^2+6x+18=(x+3)^2+9), so there is no real zero. Tip: an always positive form gives no intersection.

(x^3-10x^2+24x=x(x-4)(x-6)), so the zeroes are (0), (4), (6). Tip: first take (x) as the common factor.

(14=\frac{8+20}{2}), so it is the midpoint of the two zeroes. Tip: the midpoint value need not be a zero.

The graph behavior follows from the multiplicity of the real zero. Factor the polynomial: 5x^2 - 50x + 125 = 5(x^2 - 10x + 25) = 5(x - 5)^2. Thus the only zero is x = 5, repeated twice. A parabola with a repeated real zero reaches the x-axis at that point and turns back, so it touches rather than crosses the axis. Therefore option A is correct. Option B uses the wrong sign for the root. Option C would be appropriate if the quadratic had two distinct real roots, and option D would be appropriate only if the quadratic had no real roots. The non-zero factor 5 changes vertical scaling but does not change the zero.

For (x<-11), both factors are negative and the outside negative makes the value negative. Tip: check factor signs first.

The x-values where the graph touches or crosses the x-axis are the zeroes. Here the zeroes are 9 and −5, so their product is 9×(−5)=−45. Note: touching usually means an even multiplicity and crossing means an odd multiplicity, but the numerical root values remain 9 and −5; unless the question asks for product counting multiplicities explicitly, use these root values. Option B (45) is the closest distractor but has the wrong sign. Exam tip: remember touch → even multiplicity, cross → odd multiplicity; for product-of-roots questions, multiply the root values given by the graph.

The x-coordinates of the x-intercepts are the polynomial's zeros: 0, d and −d. Their product is 0\times d\times(−d)=0 because any product containing 0 equals 0. Note that \(d\neq0\) ensures the other two zeros are nonzero, but the presence of the zero root makes the whole product zero. The closest distractor, \(-d^2\), would be the product of d and −d alone and is wrong because it ignores the zero root. Exam tip: always check intercepts for a root equal to 0 first — it instantly gives the product as 0.

Answer: A, 1 unit. The graph's zeroes are the x-values for which p(x)=0. Factor the quadratic by finding numbers with product 72 and sum 17: 8 and 9. Thus x²−17x+72=(x−8)(x−9), so the zeroes are 8 and 9. The graph meets the x-axis at (8,0) and (9,0). The distance between them is |9−8|=1 unit. Option A is correct. Option B is one of the zeroes, not the distance. Option C is their sum, 8+9=17. Option D is their product, 8×9=72. These values are useful for identifying the roots but must not be confused with the distance between them. Memory cue: after factorising, distance between real x-axis zeroes is the absolute difference of the roots.

For a downward-opening parabola, values between the two zeroes are positive. Tip: (x=4) lies between the zeroes.

For real (x), (x^{10}\geq0), so (x^{10}+1>0). Tip: an always positive polynomial does not cut the (x)-axis.

A value x is a zero of the polynomial exactly when \(p(x)=0\). Here \(p(-10)=0\), \(p(3)=0\) and \(p(12)=0\), so there are three zeroes. Since \(p(-1)=4\) is not zero, it is not counted. Exam tip: verify the function value equals 0 exactly rather than assuming from sign or proximity.

Factorize: \(x^2-ex=x(x-e)\). x-axis intercepts occur at x-values where the polynomial equals zero. Thus x=0 and x=e give intercepts \((0,0)\) and \((e,0)\). The closest distractor (option B) is wrong because it uses \(-e\) instead of \(+e\); sign matters when solving \(x-e=0\). Exam tip: always factor out the common factor first and set each factor to zero to find x-intercepts.

For equal distance from the (y)-axis, zeroes should be opposites, so (8) is needed with (-8). Tip: symmetric zeroes are (a) and (-a).

The zeroes are (-4) and (12), and ((x+4)^2) causes touching at (-4). Tip: the outside (9) does not change the zeroes.

In this interval the signs are (+), (-), (-), so the product is positive. Tip: the product of two negative factors is positive.

The discriminant is (f^2-4f^2=-3f^2<0), so there are no real zeroes. Tip: a negative discriminant means no (x)-axis intersection.

The vertex lies on the (x)-axis, so the parabola touches at ((12,0)). Tip: if the vertex has (y=0), there is one distinct zero.

The discriminant is (256-336=-80), so there are no real zeroes. Tip: with negative discriminant a parabola does not meet the (x)-axis.

It is ((x-c)^2-25), so (x-c=\pm5) and the zeroes are (c-5), (c+5). Tip: use difference of squares.

The mean is the average of the x-coordinates of the x-intercepts:
\(\frac{-3+5+13}{3}=\frac{15}{3}=5\). So 5 is correct. Closest distractor D (15) is the sum of the zeros, not the mean; C (13) is just one root; B (−5) reflects a sign error. Exam tip: read the x-values of intercepts first, sum them, then divide by the number of zeros to get the mean.