Geometrical Meaning of the Zeroes of a Polynomial Questions in English

(D) The number of real zeros of a polynomial $y=p(x)$ is equal to the number of points where the graph of the polynomial intersects the $X$-axis.
In the given figure,the parabola $y=p(x)$ lies entirely above the $X$-axis and does not intersect or touch the $X$-axis at any point.
Therefore,the number of real zeros of the polynomial $p(x)$ is $0$.

(A) The number of real zeros of a polynomial $y=p(x)$ is equal to the number of points where the graph of the polynomial intersects the $X$-axis.
In the given figure,the parabola $y=p(x)$ touches the $X$-axis at only one point,which is the origin $(0, 0)$.
Since the graph intersects/touches the $X$-axis at exactly one point,the number of real zeros is $1$.

(B) The number of real zeros of a polynomial $y=p(x)$ is equal to the number of points where the graph of the polynomial intersects the $x$-axis.
In the given figure,the curve $y=p(x)$ intersects the $x$-axis at $3$ distinct points.
Therefore,the number of real zeros of $p(x)$ is $3$.

(B) The number of real zeros of a polynomial $y = p(x)$ is equal to the number of points where the graph of the polynomial intersects the $x$-axis.
In the given figure,the parabola $y = p(x)$ intersects the $x$-axis at exactly $2$ distinct points.
Therefore,the number of real zeros of the polynomial is $2$.

(A) The number of real zeros of a polynomial $y=p(x)$ is equal to the number of points where the graph of the polynomial intersects the $x$-axis.
In the given figure,the curve $y=p(x)$ intersects the $x$-axis at $4$ distinct points.
Therefore,the number of real zeros of $p(x)$ is $4$.

(B) The given expression $p(x) = ax^4 + bx^3 + cx^2 + dx + e$ is a polynomial of degree $4$ because the highest power of the variable $x$ is $4$ and $a \neq 0$.
According to the Fundamental Theorem of Algebra,a polynomial of degree $n$ has at most $n$ real zeros.
Since the degree of the given polynomial is $4$,the maximum number of zeros it can have is $4$.

(C) The number of real zeros of a polynomial $p(x)$ is equal to the number of points where its graph $y=p(x)$ intersects the $X$-axis.
In the given figure,the curve $y=p(x)$ intersects the $X$-axis at two distinct points.
Therefore,the polynomial $p(x)$ has $2$ real zeros.

(A) To determine the number of points where the graph of a polynomial $p(x)$ intersects the $X$-axis,we find the number of real roots of the equation $p(x) = 0$.
$1.$ $p(x) = x^{2}-5x+6 = (x-2)(x-3)$. The roots are $x=2, 3$. It intersects the $X$-axis at two points. So,$1-c$.
$2.$ $p(x) = x^{2}-10x+25 = (x-5)^{2}$. The root is $x=5$ (repeated). It intersects the $X$-axis at one point. So,$2-a$.
$3.$ $p(x) = x^{3}-4x = x(x^{2}-4) = x(x-2)(x+2)$. The roots are $x=0, 2, -2$. It intersects the $X$-axis at three points. So,$3-d$.
$4.$ $p(x) = x^{2}+4x+5$. The discriminant $D = b^{2}-4ac = 16 - 4(1)(5) = 16 - 20 = -4$. Since $D < 0$,there are no real roots. It intersects the $X$-axis at zero points. So,$4-b$.
Thus,the correct matching is $(1-c), (2-a), (3-d), (4-b)$.

(A) linear polynomial is of the form $p(x) = ax + b$,where $a \neq 0$. The graph of a linear polynomial always represents a straight line in the Cartesian plane. Since $p(x) = 5x + 3$ is a linear polynomial,its graph is a straight line.

(C) To find the point where the graph of the polynomial $p(x) = ax - b$ intersects the $X$-axis,we set $p(x) = 0$.
Setting $ax - b = 0$ gives $ax = b$.
Solving for $x$,we get $x = \frac{b}{a}$.
Since the point lies on the $X$-axis,its y-coordinate is $0$.
Therefore,the graph intersects the $X$-axis at the point $(\frac{b}{a}, 0)$.

(A) To find the zeros of the polynomial $p(x) = x^3 - x$,we set $p(x) = 0$.
$x^3 - x = 0$
Factor out $x$:
$x(x^2 - 1) = 0$
Using the difference of squares formula $a^2 - b^2 = (a - b)(a + b)$:
$x(x - 1)(x + 1) = 0$
Setting each factor to zero,we get:
$x = 0$,$x - 1 = 0 \implies x = 1$,and $x + 1 = 0 \implies x = -1$.
Thus,the zeros are $0, 1, -1$.
There are $3$ real zeros.