Geometrical Meaning of the Zeroes of a Polynomial Questions in English

(N/A) The number of zeroes of a polynomial $p(x)$ is equal to the number of points where the graph of $y = p(x)$ intersects the $x$-axis.
In the given graph,the curve intersects the $x$-axis at exactly one point.
Therefore,the number of zeroes of $p(x)$ is $1$.

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

(N/A) The number of zeroes of a polynomial $p(x)$ is equal to the number of points where the graph of $y=p(x)$ intersects the $X$-axis.
In the given graph,the curve intersects the $X$-axis at $3$ distinct points.
Therefore,the number of zeroes of the polynomial $p(x)$ is $3$.

(1) The number of zeroes of a polynomial $p(x)$ is equal to the number of points where the graph of $y = p(x)$ intersects the $X$-axis.
In the given graph,the line intersects the $X$-axis at exactly $1$ point.
Therefore,the number of zeroes of $p(x)$ is $1$.

(1) The number of zeroes of a polynomial $p(x)$ is equal to the number of points where the graph of $y=p(x)$ intersects the $X$-axis.
In the given graph,the parabola touches the $X$-axis at exactly one point.
Therefore,the number of zeroes of the polynomial $p(x)$ is $1$.

(N/A) The number of zeroes of a polynomial $p(x)$ is equal to the number of points where the graph of $y=p(x)$ intersects the $x$-axis.
By observing the given graph,we can see that the curve intersects the $x$-axis at $4$ distinct points.
Therefore,the number of zeroes of the polynomial $p(x)$ is $4$.

(0) The number of zeroes of a polynomial $p(x)$ is equal to the number of points where the graph of $y=p(x)$ intersects the $x$-axis.
In the given graph,the line is parallel to the $x$-axis and does not intersect it at any point.
Therefore,the number of zeroes of $p(x)$ is $0$.

(A) The number of zeroes of a polynomial $p(x)$ is equal to the number of points where the graph of $y=p(x)$ intersects the $x$-axis. In the given graph,the curve intersects the $x$-axis at exactly $1$ point. Therefore,the number of zeroes of the polynomial $p(x)$ is $1$.

(N/A) The number of zeroes of the polynomial $p(x)$ is equal to the number of points where the graph of $y=p(x)$ intersects the $x$-axis.
In the given graph,the curve intersects the $x$-axis at $3$ distinct points.
Therefore,the number of zeroes of the polynomial $p(x)$ is $3$.

(2) The number of zeroes of a polynomial $p(x)$ is equal to the number of points where the graph of $y=p(x)$ intersects the $x$-axis.
In the given graph,the curve intersects the $x$-axis at $2$ distinct points.
Therefore,the number of zeroes of the polynomial $p(x)$ is $2$.

(4) The number of zeroes of a polynomial $p(x)$ is equal to the number of points where the graph of $y=p(x)$ intersects the $x$-axis.
By observing the given graph,we can see that the curve intersects the $x$-axis at $4$ distinct points.
Therefore,the number of zeroes of the polynomial $p(x)$ is $4$.

(N/A) The number of zeroes of a polynomial $p(x)$ is equal to the number of points where the graph of $y=p(x)$ intersects the $x$-axis.
In the given graph,the curve intersects the $x$-axis at $3$ distinct points.
Therefore,the number of zeroes of the polynomial $p(x)$ is $3$.

(N/A) To divide $2x^2 + 3x + 1$ by $x + 2$, we perform polynomial long division:
$1$. Divide the first term of the dividend $(2x^2)$ by the first term of the divisor $(x)$ to get $2x$. This is the first term of the quotient.
$2$. Multiply the divisor $(x + 2)$ by $2x$ to get $2x^2 + 4x$. Subtract this from the dividend: $(2x^2 + 3x + 1) - (2x^2 + 4x) = -x + 1$.
$3$. Divide the first term of the new expression $(-x)$ by the first term of the divisor $(x)$ to get $-1$. This is the second term of the quotient.
$4$. Multiply the divisor $(x + 2)$ by $-1$ to get $-x - 2$. Subtract this from the current expression: $(-x + 1) - (-x - 2) = 3$.
Since the degree of the remainder ($3$, which is $0$) is less than the degree of the divisor ($x + 2$, which is $1$), we stop.
Thus, the quotient is $2x - 1$ and the remainder is $3$.
Verification:
$(2x - 1)(x + 2) + 3 = (2x^2 + 4x - x - 2) + 3 = 2x^2 + 3x + 1$.
Therefore, $\text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder}$.

(N/A) Since two zeroes are $\sqrt{2}$ and $-\sqrt{2}$,$(x-\sqrt{2})(x+\sqrt{2}) = x^{2}-2$ is a factor of the given polynomial.
Now,we divide the given polynomial by $x^{2}-2$ to find the other factors.
Performing the division:
$2x^{4}-3x^{3}-3x^{2}+6x-2 = (x^{2}-2)(2x^{2}-3x+1)$
Now,we factorise the quadratic polynomial $2x^{2}-3x+1$ by splitting the middle term:
$2x^{2}-2x-x+1 = 2x(x-1)-1(x-1) = (2x-1)(x-1)$
Setting these factors to zero,we get:
$2x-1 = 0 \implies x = \frac{1}{2}$
$x-1 = 0 \implies x = 1$
Therefore,all the zeroes of the given polynomial are $\sqrt{2}, -\sqrt{2}, \frac{1}{2},$ and $1$.

(N/A) To divide $p(x) = x^{3} - 3x^{2} + 5x - 3$ by $g(x) = x^{2} - 2$:
$1$. Divide the first term of the dividend $(x^{3})$ by the first term of the divisor $(x^{2})$ to get $x$. This is the first term of the quotient.
$2$. Multiply $x$ by $(x^{2} - 2)$ to get $x^{3} - 2x$. Subtract this from $p(x)$ to get $-3x^{2} + 7x - 3$.
$3$. Divide the first term of the new dividend $(-3x^{2})$ by the first term of the divisor $(x^{2})$ to get $-3$. This is the second term of the quotient.
$4$. Multiply $-3$ by $(x^{2} - 2)$ to get $-3x^{2} + 6$. Subtract this from the current remainder to get $7x - 9$.
Thus,the quotient is $x - 3$ and the remainder is $7x - 9$.

(N/A) To divide $p(x)$ by $g(x)$,we first write them in standard form (descending order of powers):
$p(x) = x^{4} + 0x^{3} - 3x^{2} + 4x + 5$
$g(x) = x^{2} - x + 1$
Performing polynomial long division:
$1$. Divide the first term of $p(x)$ by the first term of $g(x)$: $x^{4} / x^{2} = x^{2}$. This is the first term of the quotient.
$2$. Multiply $x^{2}$ by $(x^{2} - x + 1) = x^{4} - x^{3} + x^{2}$. Subtract this from $p(x)$ to get $x^{3} - 4x^{2} + 4x + 5$.
$3$. Divide the first term of the new polynomial by the first term of $g(x)$: $x^{3} / x^{2} = x$. This is the second term of the quotient.
$4$. Multiply $x$ by $(x^{2} - x + 1) = x^{3} - x^{2} + x$. Subtract this to get $-3x^{2} + 3x + 5$.
$5$. Divide the first term of the new polynomial by the first term of $g(x)$: $-3x^{2} / x^{2} = -3$. This is the third term of the quotient.
$6$. Multiply $-3$ by $(x^{2} - x + 1) = -3x^{2} + 3x - 3$. Subtract this to get $8$.
Thus,the quotient is $x^{2} + x - 3$ and the remainder is $8$.

(A) To check if $t^{2}-3$ is a factor of $2t^{4}+3t^{3}-2t^{2}-9t-12$,we perform polynomial long division.
We write the divisor as $t^{2}+0t-3$.
Dividing $2t^{4}+3t^{3}-2t^{2}-9t-12$ by $t^{2}+0t-3$:
$1$. Divide $2t^{4}$ by $t^{2}$ to get $2t^{2}$. Multiply $2t^{2}(t^{2}+0t-3) = 2t^{4}+0t^{3}-6t^{2}$. Subtracting this from the dividend gives $3t^{3}+4t^{2}-9t-12$.
$2$. Divide $3t^{3}$ by $t^{2}$ to get $3t$. Multiply $3t(t^{2}+0t-3) = 3t^{3}+0t^{2}-9t$. Subtracting this gives $4t^{2}+0t-12$.
$3$. Divide $4t^{2}$ by $t^{2}$ to get $4$. Multiply $4(t^{2}+0t-3) = 4t^{2}+0t-12$. Subtracting this gives a remainder of $0$.
Since the remainder is $0$,$t^{2}-3$ is a factor of $2t^{4}+3t^{3}-2t^{2}-9t-12$.

(B) The statement is 'False'.
$A$ quadratic polynomial $p(x) = ax^2 + bx + c$ can intersect or touch the $x$-axis at only one point if the discriminant $D = b^2 - 4ac$ is equal to $0$.
In this case,the quadratic polynomial has two equal real roots,and the graph touches the $x$-axis at exactly one point (the vertex).
Therefore,a quadratic polynomial can indeed intersect the $x$-axis at only one point.

(A) The statement is 'True'.
If the graph of a polynomial intersects the $x$-axis at exactly two points,it does not necessarily have to be a quadratic polynomial.
$A$ polynomial of degree $n > 2$ can also intersect the $x$-axis at exactly two points if it has two real roots and the remaining $(n-2)$ roots are imaginary (non-real complex roots).
For example,a polynomial of degree $4$ can have two real roots and two imaginary roots,resulting in a graph that crosses the $x$-axis at only two points.

(N/A) Here,$p(x)=x^{2}-3x-4=(x-4)(x+1)$.
To find the zeros of $p(x)$,consider $p(x)=0$.
$\therefore (x-4)(x+1)=0$.
$\therefore x=4$ or $x=-1$.
$\therefore 4$ and $-1$ are the zeros of $p(x)$.
To draw the graph of this polynomial,we take some different values of $x$ and prepare the following table:

$x$	$-2$	$-1$	$0$	$3$	$4$	$5$
$p(x)=x^2-3x-4$	$6$	$0$	$-4$	$-4$	$0$	$6$

Plot these points on a graph paper. Joining all these points $(-2, 6), (-1, 0), (0, -4), (3, -4), (4, 0)$ and $(5, 6)$,we get the shape of the graph as a parabola opening upwards. We can see that this graph intersects the $X$-axis at two points $(-1, 0)$ and $(4, 0)$. Their $X$-coordinates are the zeros of this polynomial. Thus,$-1$ and $4$ are the zeros of $p(x)$.

(N/A) Here,$p(x)=x^{3}-4 x$
$=x(x^{2}-4)$
$=x(x-2)(x+2)$
To find the zeros of $p(x)$,consider $p(x)=0$.
$\therefore x(x-2)(x+2)=0$
$\therefore x=0, x=2$ or $x=-2$.
$\therefore 0, 2$ and $-2$ are the zeros of $p(x)$.
To draw the graph of this polynomial,we take some different values of $x$ and prepare the following table:

$x$	$-2$	$-1$	$0$	$1$	$2$
$p(x)=x^3-4x$	$0$	$3$	$0$	$-3$	$0$

Plot these points as shown in the graph. We can see that this graph intersects the $X$-axis at three distinct points $(-2, 0)$,$(0, 0)$ and $(2, 0)$. Their $X$-coordinates are the zeros of this polynomial. So,the zeros of $p(x)$ are $-2, 0$ and $2$.

(0) The number of 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 graph of $y=p(x)$ is a line parallel to the $X-$axis,which does not intersect the $X-$axis at any point.
Therefore,the number of real zeros of $p(x)$ is $0$.

(0) The number of 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 graph of $y=p(x)$ 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 $0$.

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

(A) The number of 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 graph of $y=p(x)$ intersects the $X-$axis at exactly $1$ point.
Therefore,the number of real zeros of $p(x)$ is $1$.

(4) The number of zeros of a polynomial $p(x)$ is equal to the number of points where the graph of $y=p(x)$ intersects the $X-$axis.
By observing the given graph,we can see that the curve intersects the $X-$axis at $4$ distinct points.
Therefore,the number of real zeros of $p(x)$ is $4$.

(C) To find the zeros of the polynomial $p(x) = 5x - 4$,we set $p(x) = 0$.
Thus,$5x - 4 = 0$.
Adding $4$ to both sides,we get $5x = 4$.
Dividing by $5$,we get $x = \frac{4}{5}$.
Since this is a linear polynomial of the form $ax + b$ (where $a \neq 0$),it has exactly one real zero.
Therefore,the number of real zeros is $1$.

(A) To find the zero of the polynomial $p(x) = 2x - 5$,we set $p(x) = 0$.
$2x - 5 = 0$
$2x = 5$
$x = 5/2$
Thus,the zero of the polynomial is $5/2$ or $2.5$.
To draw the graph,we find two points on the line:
$1$. If $x = 0$,$p(0) = 2(0) - 5 = -5$. Point: $(0, -5)$.
$2$. If $x = 2.5$,$p(2.5) = 2(2.5) - 5 = 0$. Point: $(2.5, 0)$.
Plotting these points and drawing a straight line through them gives the graph of the linear polynomial.

(B) To find the zeros of the polynomial $p(x) = 3 - 2x - x^2$,we set $p(x) = 0$.
$3 - 2x - x^2 = 0$
Multiply by $-1$ to simplify: $x^2 + 2x - 3 = 0$.
Factor the quadratic equation: $x^2 + 3x - x - 3 = 0$.
$x(x + 3) - 1(x + 3) = 0$.
$(x - 1)(x + 3) = 0$.
Setting each factor to zero,we get $x - 1 = 0$ or $x + 3 = 0$.
Thus,the zeros are $x = 1$ and $x = -3$.

(C) To find the zeros of the polynomial $p(x) = x^{2} + x - 12$,we set $p(x) = 0$.
Thus,$x^{2} + x - 12 = 0$.
Factoring the quadratic equation,we look for two numbers that multiply to $-12$ and add to $1$.
These numbers are $4$ and $-3$.
So,$x^{2} + 4x - 3x - 12 = 0$.
$x(x + 4) - 3(x + 4) = 0$.
$(x - 3)(x + 4) = 0$.
Setting each factor to zero,we get $x - 3 = 0$ or $x + 4 = 0$.
Therefore,$x = 3$ or $x = -4$.
The zeros of the polynomial are $-4$ and $3$.

(D) To find the zeros of the polynomial $p(x) = x^3 - 2x^2$,we set $p(x) = 0$.
$x^3 - 2x^2 = 0$
Factor out $x^2$ from the expression:
$x^2(x - 2) = 0$
This gives us two cases:
$x^2 = 0 \implies x = 0$
$x - 2 = 0 \implies x = 2$
Thus,the zeros of the polynomial are $0$ and $2$.

(A) To find the zeros of the polynomial $p(x) = x^3$,we set $p(x) = 0$.
This gives the equation $x^3 = 0$.
Taking the cube root on both sides,we get $x = 0$.
Graphically,the curve $y = x^3$ passes through the origin $(0, 0)$,which is the only point where it intersects the $x$-axis.
Therefore,the zero of the polynomial $p(x) = x^3$ is $0$.

(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 $y = p(x)$ touches the $x$-axis at exactly one point.
Therefore,the number of real zeros of the polynomial is $1$.

(C) 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 line representing $y=p(x)$ intersects the $x$-axis at exactly one point.
Therefore,the number of real zeros of $p(x)$ is $1$.
Thus,the correct option is $C$.

(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 curve $y=p(x)$ intersects the $x$-axis at $3$ distinct points.
Therefore,the number of real zeros of the polynomial is $3$.

(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 curve $y=p(x)$ intersects the $x$-axis at $3$ distinct points.
Therefore,the number of real zeros of $p(x)$ is $3$.

(C) To find the points where the graph of $p(x) = x^2 - 9$ intersects the $X$-axis,we set $p(x) = 0$.
$x^2 - 9 = 0$
$(x - 3)(x + 3) = 0$
This gives $x = 3$ and $x = -3$.
Since there are two distinct real roots,the graph intersects the $X$-axis at two distinct points: $(3, 0)$ and $(-3, 0)$.
Therefore,the correct option is $C$.

(C) linear polynomial is of the form $p(x) = ax + b$,where $a \neq 0$ and $a, b$ are real numbers.
The graph of the equation $y = ax + b$ represents a straight line in the Cartesian plane.
Therefore,the graph of a linear polynomial is always a straight line.

(A) linear polynomial is defined as $p(x) = ax + b$,where $a \neq 0$.
To find the intersection with the $X$-axis,we set $p(x) = 0$.
$ax + b = 0$
$ax = -b$
$x = -b/a$
Since $a \neq 0$,there is exactly one unique value of $x$ for which $p(x) = 0$.
Therefore,the graph of a linear polynomial intersects the $X$-axis at exactly one point,which is $(-b/a, 0)$.

(D) To find the point where the graph of the polynomial $p(x) = 3x - 6$ intersects the $X$-axis,we set $p(x) = 0$.
Setting the polynomial to zero gives: $3x - 6 = 0$.
Adding $6$ to both sides: $3x = 6$.
Dividing by $3$: $x = 2$.
Since the point lies on the $X$-axis,its $y$-coordinate is $0$.
Therefore,the graph intersects the $X$-axis at the point $(2, 0)$.

(D) The given polynomial is $p(x) = x^{2} + 5x + 6$.
This is a quadratic polynomial of the form $ax^{2} + bx + c$,where $a = 1$,$b = 5$,and $c = 6$.
The graph of any quadratic polynomial $p(x) = ax^{2} + bx + c$ (where $a \neq 0$) is always a curve known as a parabola.
Since the coefficient of $x^{2}$ is positive $(a = 1 > 0)$,the parabola opens upwards.
Therefore,the graph of $p(x) = x^{2} + 5x + 6$ is a parabola.

(A) The given polynomial is a quadratic polynomial of the form $p(x) = ax^{2} + bx + c$,where $a = 1$,$b = 6$,and $c = 8$.
Since the coefficient of $x^{2}$ is $a = 1$,which is greater than $0$ $(a > 0)$,the graph of the quadratic polynomial is a parabola that opens upwards.
Therefore,the correct option is $A$.

(B) To determine the number of points where the graph of the quadratic polynomial $p(x) = x^{2} - x - 2$ intersects the $X$-axis,we need to find the number of real roots of the equation $p(x) = 0$.
$1$. Set the polynomial equal to zero: $x^{2} - x - 2 = 0$.
$2$. Calculate the discriminant $D = b^{2} - 4ac$,where $a = 1$,$b = -1$,and $c = -2$.
$3$. $D = (-1)^{2} - 4(1)(-2) = 1 + 8 = 9$.
$4$. Since $D > 0$,the quadratic equation has two distinct real roots.
$5$. Therefore,the graph of the polynomial intersects the $X$-axis at two distinct points.

(D) The given polynomial is a quadratic polynomial of the form $p(x) = ax^2 + bx + c$,where $a = -1$,$b = 1$,and $c = 6$.
Since the degree of the polynomial is $2$,its graph is a parabola.
In a quadratic polynomial $ax^2 + bx + c$,if $a > 0$,the parabola opens upwards,and if $a < 0$,the parabola opens downwards.
Here,$a = -1$,which is less than $0$.
Therefore,the graph of $p(x) = -x^2 + x + 6$ is a parabola opening downwards.

(C) To determine how the graph of $p(x) = x^{2} + 6x + 9$ intersects the $X$-axis,we find the zeros of the polynomial by setting $p(x) = 0$.
$x^{2} + 6x + 9 = 0$
This is a perfect square trinomial,which can be factored as $(x + 3)^{2} = 0$.
Thus,$x = -3$ is the only root (a repeated root).
Since there is only one distinct real root,the parabola represented by the quadratic polynomial touches the $X$-axis at exactly one point,which is $(-3, 0)$.

(D) polynomial of degree $1$ is called a linear polynomial.
The graph of a linear polynomial $p(x) = ax + b$ (where $a \neq 0$) is always a straight line.
In the given options:
$A) p(x) = x^{2} + 8$ is a quadratic polynomial (degree $2$).
$B) p(x) = x^{3} - 2$ is a cubic polynomial (degree $3$).
$C) p(x) = x^{2} + 6x + 9$ is a quadratic polynomial (degree $2$).
$D) p(x) = 5x - 10$ is a linear polynomial (degree $1$).
Therefore,the graph of $p(x) = 5x - 10$ is a line.

(A) The graph of a quadratic polynomial $p(x) = ax^2 + bx + c$ is a parabola.
If $a > 0$,the parabola opens upwards.
If $a < 0$,the parabola opens downwards.
In option $A$,$p(x) = x^2 + 8x + 12$,here $a = 1$,which is greater than $0$.
Therefore,the graph of $p(x) = x^2 + 8x + 12$ is a curve open upwards.

(B) For a quadratic polynomial $p(x) = ax^{2} + bx + c$,the graph is a parabola.
If $a > 0$,the parabola opens upwards.
If $a < 0$,the parabola opens downwards.
Let us examine the coefficient of $x^{2}$ for each option:
$A) p(x) = x^{2} - 4x + 3$,here $a = 1 > 0$ (opens upwards).
$B) p(x) = 6x - x^{2} - 9$,which is $p(x) = -x^{2} + 6x - 9$. Here $a = -1 < 0$ (opens downwards).
$C) p(x) = 3 - 6x + 3x^{2}$,here $a = 3 > 0$ (opens upwards).
$D) p(x) = x^{2} + 6x + 5$,here $a = 1 > 0$ (opens upwards).
Therefore,the correct option is $B$.

(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 line representing $y = p(x)$ intersects the $x$-axis at exactly one point.
Therefore,the number of real zeros of $y = p(x)$ 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 representing $y=p(x)$ intersects the $x$-axis at exactly $2$ distinct points.
Therefore,the number of real zeros of $y=p(x)$ is $2$.