Geometrical meaning of the zeroes of a polynomial.

A zero (root) of the polynomial is an x-value where \(p(x)=0\). On the graph the y-coordinate equals \(p(x)\), so when \(p(x)=0\) the point lies on the x-axis (where \(y=0\)). Why others are wrong: (B) the y-axis corresponds to \(x=0\), not \(p(x)=0\). (C) a highest point (maximum) need not have \(y=0\). (D) any point with \(p(x)\neq 0\) cannot be a zero. Exam tip: remember “zero of p(x)” means the graph crosses or touches the x-axis at that x-value (i.e. \(y=0\)).

The number of real zeroes equals the number of times the graph cuts the (x)-axis. Tip: a linear polynomial has at most one zero.

Two intersection points show two real zeroes. Tip: a parabola can cut the (x)-axis at most twice.

A zero of a polynomial is represented by a point where its graph meets the x-axis. When a parabola only touches the x-axis, it meets the axis at exactly one point, so it has one distinct real zero. Algebraically, this corresponds to a repeated real root and the quadratic has discriminant zero. Thus the root may occur twice in factor form, but the number of distinct real zeroes is one, not two.

There is no (x)-axis intersection so there is no real zero. Tip: count only meetings with the (x)-axis.

The point \((4,0)\) means the polynomial value is \(p(4)=0\), so 4 is a root (zero) of the polynomial. Option B is wrong because 0 would be a root only if \(x=0\) (i.e. point \((0,0)\) or \(p(0)=0\)); here x=4. Option C is incorrect because a y-intercept occurs at \(x=0\), not at \(x=4\). Option D is wrong since a single x-intercept does not imply the polynomial is constant. Exam tip: x-intercepts \((a,0)\) correspond to roots a; to find y-intercept set \(x=0\).

Passing through the origin (0,0) means p(0)=0, so x=0 is necessarily a root of the polynomial. Other choices (1, -1, 2) would be roots only if p(1)=0, p(-1)=0, or p(2)=0 respectively; the graph passing through the origin does not guarantee those. Exam tip: substitute x=0 into the polynomial to check quickly whether 0 is a root.

The zero of a polynomial is the x-value where its graph intersects the x-axis (y=0). Since the graph intersects the x-axis at -2, the zero is -2. Option B (2) is the sign-opposite and therefore incorrect; options C (0) and D (-1) do not match the given intercept. Exam tip: remember that x-axis intersections give x-values (y=0) and always double-check the sign.

The x-intercepts give the x-values where the function equals zero (y = 0). Since the graph meets the x-axis at x = -3 and x = 5, the real zeros are x = -3 and x = 5. A close distractor is option C (-3 and -5) which has the first value correct but the second with the wrong sign. Exam tip: always use the x-coordinate of x-axis intersections as zeros and double-check the signs.

A graph parallel to the x-axis is a horizontal line of the form \(y=c\). If it does not intersect the x-axis then \(c\neq 0\). Roots are solutions of \(p(x)=0\); here the polynomial equals the nonzero constant \(c\), so there is no solution and hence zero zeroes. The distractor "one" would apply if the graph merely touched the x-axis (a single root), while "infinite" applies when \(c=0\) (the graph lies on the x-axis and every x is a root). Exam tip: for horizontal lines check whether the constant equals zero to decide roots.

An x-intercept occurs where the polynomial equals zero, i.e. \(p(x)=0\). For \(p(x)=x-7\), solving \(x-7=0\) gives \(x=7\), so the point is (7, 0). Option B (−7, 0) is incorrect because that would correspond to \(x=-7\), not a root of this polynomial. Exam tip: set the polynomial equal to zero and solve for x to find x-axis intersections.

An x-axis intersection (x-intercept) occurs where the polynomial equals zero (i.e. \(p(x)=0\)) so the y-coordinate is 0. For \(p(x)=x+4\), setting \(x+4=0\) gives \(x=-4\), hence the intercept is \((-4,0)\). Option B (4,0) is incorrect due to the wrong sign; options C and D are points on the y-axis (x=0), not x-axis intercepts. Exam tip: to find x-intercepts, set the function equal to zero and solve for x.

Set the polynomial equal to zero: \(p(x)=0\) ⇒ \(2x-6=0\). Solving gives \(x=3\), so the root on the graph is at \(x=3\). Why other choices fail: for \(x=2\), \(p(2)=-2\) and for \(x=6\), \(p(6)=6\); neither is zero. Exam tip: always start with \(p(x)=0\) and solve the resulting linear equation directly.

Any point on the x‑axis has y = 0. Since the graph crosses the x‑axis at (6, 0), the function value p(6) is 0. Option B (6) is misleading because 6 is the x‑coordinate, not the y‑value. Exam tip: an intersection with the x‑axis always means the function value at that x is 0 (p(x)=0).

The statement \(p(-1)=0\) means at \(x=-1\) the value \(y=p(x)\) is zero, so the point \((-1,0)\) lies on the graph. Option B is incorrect because that would require \(p(0)=-1\), which is not given. Option C is wrong since intersection with the y-axis depends on \(p(0)\), not on \(p(-1)\). Option D is false because only degree‑1 polynomials are straight lines. Exam tip: x-intercepts come from solving \(p(x)=0\); the y-intercept is \(p(0)\).

A zero x = α of a polynomial satisfies p(α)=0. The point (0,5) means p(0)=5, which is not zero, so x=0 is not a zero. A common distractor says “x=0 so it is a zero”; that is wrong because being at x=0 does not force p(0)=0. Exam tip: always substitute the x-coordinate into p(x) and check whether the value equals 0 before calling it a zero.

A zero of a polynomial is an x-value for which the polynomial’s value is zero. Geometrically, it is the x-coordinate of a point where the graph meets the x-axis. The word “touches” does not change this definition: a graph may cross the axis or merely touch it, and either event gives a zero. At the point (2, 0), the x-coordinate is 2 and the y-coordinate is 0, so f(2) = 0. Therefore the zero is 2, and option A is correct. Option B confuses the y-coordinate with the required input; option C invents an additional zero not shown by the graph; and option D is wrong because touching the x-axis still counts as meeting it and establishes a zero.

The graph can stay above the (x)-axis so there is no real zero. Tip: if there is no intersection do not write a zero.

If a polynomial has a zero a, its graph meets the x-axis at (a,0) because the y-value is zero there. So zeros 1 and 4 give intersection points (1,0) and (4,0). Option B is wrong because those are points on the y-axis, C is wrong since y≠0 at those coordinates, and D is wrong because the signs of the zeros are incorrect. Exam tip: Always convert a zero a to the point (a,0) on the graph.

The x-coordinates where the graph meets the x-axis are the zeros. Here the zeros are -5 and 2, so their sum = -5 + 2 = -3. Option B (3) is wrong because it ignores the negative sign of -5; option C (7) would be the sum of absolute values (5 + 2) which is a sign error; option D (-7) assumes both were negative. Exam tip: always take the x-values of x-intercepts and keep their signs before adding.