Textbook - Polynomials 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) We have the quadratic polynomial $p(x) = x^{2}+7x+10$.
To find the zeroes,we set $p(x) = 0$:
$x^{2}+7x+10 = 0$
$x^{2}+5x+2x+10 = 0$
$x(x+5)+2(x+5) = 0$
$(x+2)(x+5) = 0$
Thus,the zeroes are $x = -2$ and $x = -5$.
Verification:
Sum of zeroes $= (-2) + (-5) = -7 = \frac{-7}{1} = \frac{-(\text{Coefficient of } x)}{\text{Coefficient of } x^{2}}$.
Product of zeroes $= (-2) \times (-5) = 10 = \frac{10}{1} = \frac{\text{Constant term}}{\text{Coefficient of } x^{2}}$.
Hence,the relationship is verified.

(A) Let the quadratic polynomial be $p(x) = ax^2 + bx + c,$ and its zeroes be $\alpha$ and $\beta$.
The relationship between the zeroes and coefficients of a quadratic polynomial is given by:
Sum of zeroes: $\alpha + \beta = -\frac{b}{a} = -3$
Product of zeroes: $\alpha \beta = \frac{c}{a} = 2$
If we assume $a = 1$,then:
$-b = -3 \implies b = 3$
$c = 2$
Substituting these values into the standard form $ax^2 + bx + c$,we get the polynomial $x^2 + 3x + 2$.
Thus,the required quadratic polynomial is $x^2 + 3x + 2$.

(A) Comparing the given polynomial $p(x) = 3x^3 - 5x^2 - 11x - 3$ with the standard form $ax^3 + bx^2 + cx + d$,we get $a = 3, b = -5, c = -11, d = -3$.
First,we verify the zeroes:
$p(3) = 3(3)^3 - 5(3)^2 - 11(3) - 3 = 81 - 45 - 33 - 3 = 0$
$p(-1) = 3(-1)^3 - 5(-1)^2 - 11(-1) - 3 = -3 - 5 + 11 - 3 = 0$
$p(-\frac{1}{3}) = 3(-\frac{1}{3})^3 - 5(-\frac{1}{3})^2 - 11(-\frac{1}{3}) - 3 = -\frac{1}{9} - \frac{5}{9} + \frac{11}{3} - 3 = -\frac{6}{9} + \frac{33}{9} - \frac{27}{9} = 0$
Thus,$3, -1, -\frac{1}{3}$ are the zeroes.
Let $\alpha = 3, \beta = -1, \gamma = -\frac{1}{3}$.
$1$. Sum of zeroes: $\alpha + \beta + \gamma = 3 - 1 - \frac{1}{3} = 2 - \frac{1}{3} = \frac{5}{3} = -\frac{b}{a}$.
$2$. Sum of product of zeroes taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = (3)(-1) + (-1)(-\frac{1}{3}) + (-\frac{1}{3})(3) = -3 + \frac{1}{3} - 1 = -4 + \frac{1}{3} = -\frac{11}{3} = \frac{c}{a}$.
$3$. Product of zeroes: $\alpha\beta\gamma = (3)(-1)(-\frac{1}{3}) = 1 = -\frac{d}{a} = -(\frac{-3}{3}) = 1$.
All relationships are verified.

(N/A) Given polynomial: $p(x) = x^{2}-2x-8$.
To find the zeroes,we set $p(x) = 0$:
$x^{2}-2x-8 = 0$
$x^{2}-4x+2x-8 = 0$
$x(x-4)+2(x-4) = 0$
$(x-4)(x+2) = 0$
Thus,the zeroes are $x = 4$ and $x = -2$.
Verification:
Comparing $x^{2}-2x-8$ with $ax^{2}+bx+c$,we get $a=1, b=-2, c=-8$.
Sum of zeroes $= 4 + (-2) = 2$.
Formula: $-b/a = -(-2)/1 = 2$.
Since $2 = 2$,the sum of zeroes is verified.
Product of zeroes $= 4 \times (-2) = -8$.
Formula: $c/a = -8/1 = -8$.
Since $-8 = -8$,the product of zeroes is verified.

(N/A) Given polynomial: $p(s) = 4s^{2}-4s+1$.
To find the zeroes,set $p(s) = 0$:
$4s^{2}-4s+1 = 0$
$(2s-1)^{2} = 0$
$2s-1 = 0 \implies s = \frac{1}{2}$.
Thus,the zeroes are $\frac{1}{2}$ and $\frac{1}{2}$.
Verification:
Sum of zeroes $= \frac{1}{2} + \frac{1}{2} = 1 = \frac{-(-4)}{4} = \frac{-(\text{Coefficient of } s)}{\text{Coefficient of } s^{2}}$.
Product of zeroes $= \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \frac{\text{Constant term}}{\text{Coefficient of } s^{2}}$.
Hence,the relationship is verified.

(N/A) First,rewrite the polynomial in standard form: $6x^{2}-7x-3$.
To find the zeroes,factorize the quadratic polynomial:
$6x^{2}-7x-3 = 6x^{2}-9x+2x-3 = 3x(2x-3)+1(2x-3) = (3x+1)(2x-3)$.
The value of $6x^{2}-7x-3$ is zero when $3x+1=0$ or $2x-3=0$,which gives $x = -1/3$ or $x = 3/2$.
Therefore,the zeroes are $\alpha = -1/3$ and $\beta = 3/2$.
Verification of the relationship between zeroes and coefficients:
Sum of zeroes: $\alpha + \beta = -1/3 + 3/2 = (-2+9)/6 = 7/6 = -(-7)/6 = -(\text{Coefficient of } x) / (\text{Coefficient of } x^{2})$.
Product of zeroes: $\alpha \times \beta = (-1/3) \times (3/2) = -3/6 = -1/2 = -3/6 = (\text{Constant term}) / (\text{Coefficient of } x^{2})$.

(N/A) Given polynomial: $p(u) = 4u^{2} + 8u$.
To find the zeroes,set $p(u) = 0$:
$4u^{2} + 8u = 0$
$4u(u + 2) = 0$
This implies $4u = 0$ or $u + 2 = 0$.
So,$u = 0$ or $u = -2$.
The zeroes are $0$ and $-2$.
Verification:
Comparing $4u^{2} + 8u$ with $au^{2} + bu + c$,we get $a = 4, b = 8, c = 0$.
Sum of zeroes $= 0 + (-2) = -2$.
Relationship: $\frac{-b}{a} = \frac{-8}{4} = -2$.
Since Sum of zeroes $= \frac{-b}{a}$,the relationship is verified.
Product of zeroes $= 0 \times (-2) = 0$.
Relationship: $\frac{c}{a} = \frac{0}{4} = 0$.
Since Product of zeroes $= \frac{c}{a}$,the relationship is verified.

(N/A) Given polynomial: $p(t) = t^{2}-15$.
To find the zeroes,set $p(t) = 0$:
$t^{2}-15 = 0$
$t^{2} = 15$
$t = \pm \sqrt{15}$
So,the zeroes are $\alpha = \sqrt{15}$ and $\beta = -\sqrt{15}$.
Comparing $t^{2}-15$ with the standard form $at^{2}+bt+c$,we get $a=1, b=0, c=-15$.
Verification:
Sum of zeroes: $\alpha + \beta = \sqrt{15} + (-\sqrt{15}) = 0$.
From coefficients: $\frac{-b}{a} = \frac{-0}{1} = 0$.
Thus,$\alpha + \beta = \frac{-b}{a}$ is verified.
Product of zeroes: $\alpha \cdot \beta = (\sqrt{15})(-\sqrt{15}) = -15$.
From coefficients: $\frac{c}{a} = \frac{-15}{1} = -15$.
Thus,$\alpha \cdot \beta = \frac{c}{a}$ is verified.

(N/A) Given polynomial: $p(x) = 3x^{2}-x-4$.
To find the zeroes,we set $p(x) = 0$:
$3x^{2}-x-4 = 0$
$3x^{2}-4x+3x-4 = 0$
$x(3x-4)+1(3x-4) = 0$
$(3x-4)(x+1) = 0$
Thus,the zeroes are $x = \frac{4}{3}$ and $x = -1$.
Verification:
Sum of zeroes $= \frac{4}{3} + (-1) = \frac{4-3}{3} = \frac{1}{3}$.
From the polynomial,$-\frac{\text{coefficient of } x}{\text{coefficient of } x^{2}} = -\frac{-1}{3} = \frac{1}{3}$.
Since $\frac{1}{3} = \frac{1}{3}$,the sum is verified.
Product of zeroes $= \frac{4}{3} \times (-1) = -\frac{4}{3}$.
From the polynomial,$\frac{\text{constant term}}{\text{coefficient of } x^{2}} = \frac{-4}{3}$.
Since $-\frac{4}{3} = -\frac{4}{3}$,the product is verified.

(A) Let the quadratic polynomial be $ax^2 + bx + c$ and its zeroes be $\alpha$ and $\beta$.
Given that the sum of zeroes $\alpha + \beta = \frac{1}{4}$ and the product of zeroes $\alpha \beta = -1$.
We know that a quadratic polynomial can be expressed as $k[x^2 - (\alpha + \beta)x + \alpha \beta]$,where $k$ is a non-zero constant.
Substituting the given values,we get $k[x^2 - (\frac{1}{4})x + (-1)]$.
To simplify,let $k = 4$.
Then the polynomial becomes $4[x^2 - \frac{1}{4}x - 1] = 4x^2 - x - 4$.
Thus,the required quadratic polynomial is $4x^2 - x - 4$.

(D) Let the quadratic polynomial be $p(x) = ax^2 + bx + c$,and its zeroes be $\alpha$ and $\beta$.
The sum of the zeroes is given by $\alpha + \beta = \sqrt{2} = -\frac{b}{a}$.
The product of the zeroes is given by $\alpha \beta = \frac{1}{3} = \frac{c}{a}$.
To express these with a common denominator $a$,we write $\alpha + \beta = \frac{3\sqrt{2}}{3} = -\frac{b}{a}$.
Comparing the coefficients,we get $a = 3$,$b = -3\sqrt{2}$,and $c = 1$.
Substituting these values into the general form $ax^2 + bx + c$,we get the quadratic polynomial $3x^2 - 3\sqrt{2}x + 1$.

(D) Let the quadratic polynomial be $p(x) = ax^2 + bx + c$,and its zeroes be $\alpha$ and $\beta$.
The sum of the zeroes is given by $\alpha + \beta = 0 = \frac{0}{1} = -\frac{b}{a}$.
The product of the zeroes is given by $\alpha \times \beta = \sqrt{5} = \frac{\sqrt{5}}{1} = \frac{c}{a}$.
Comparing the coefficients,if we take $a = 1$,then $b = 0$ and $c = \sqrt{5}$.
Substituting these values into the general form $ax^2 + bx + c$,we get the quadratic polynomial $x^2 + 0x + \sqrt{5}$,which simplifies to $x^2 + \sqrt{5}$.

(A) Let the quadratic polynomial be $p(x) = ax^2 + bx + c$,and its zeroes be $\alpha$ and $\beta$.
The sum of zeroes is given by $\alpha + \beta = 1 = \frac{-b}{a}$.
The product of zeroes is given by $\alpha \times \beta = 1 = \frac{c}{a}$.
If we assume $a = 1$,then:
$-b = 1 \implies b = -1$
$c = 1$
Substituting these values into the general form $ax^2 + bx + c$,we get the polynomial $x^2 - x + 1$.

(A) Let the quadratic polynomial be $p(x) = ax^2 + bx + c$,and its zeroes be $\alpha$ and $\beta$.
The sum of zeroes is given by $\alpha + \beta = -\frac{1}{4} = -\frac{b}{a}$.
The product of zeroes is given by $\alpha \times \beta = \frac{1}{4} = \frac{c}{a}$.
Comparing the ratios,if we take $a = 4$,then $b = 1$ and $c = 1$.
Substituting these values into the general form $ax^2 + bx + c$,we get the polynomial $4x^2 + x + 1$.

(A) Let the quadratic polynomial be $p(x) = ax^2 + bx + c$,where $\alpha$ and $\beta$ are its zeroes.
The sum of the zeroes is given by $\alpha + \beta = 4 = \frac{4}{1} = \frac{-b}{a}$.
The product of the zeroes is given by $\alpha \times \beta = 1 = \frac{1}{1} = \frac{c}{a}$.
By comparing the ratios,if we assume $a = 1$,then $b = -4$ and $c = 1$.
Substituting these values into the general form $ax^2 + bx + c$,we get the quadratic polynomial $x^2 - 4x + 1$.

(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) We first arrange the terms of the dividend and the divisor in the decreasing order of their degrees. This is called writing the polynomials in standard form. The dividend is already in standard form,and the divisor,in standard form,is $x^{2}+2x+1$.
Step $1$: To obtain the first term of the quotient,divide the highest degree term of the dividend $(3x^{3})$ by the highest degree term of the divisor $(x^{2})$. This gives $3x$. Multiplying the divisor by $3x$,we get $3x(x^{2}+2x+1) = 3x^{3}+6x^{2}+3x$. Subtracting this from the dividend,we get $(3x^{3}+x^{2}+2x+5) - (3x^{3}+6x^{2}+3x) = -5x^{2}-x+5$.
Step $2$: To obtain the second term of the quotient,divide the highest degree term of the new dividend $(-5x^{2})$ by the highest degree term of the divisor $(x^{2})$. This gives $-5$. Multiplying the divisor by $-5$,we get $-5(x^{2}+2x+1) = -5x^{2}-10x-5$. Subtracting this from the current dividend,we get $(-5x^{2}-x+5) - (-5x^{2}-10x-5) = 9x+10$.
Step $3$: The degree of the remainder $9x+10$ is $1$,which is less than the degree of the divisor $(2)$. Thus,the division process stops.
Therefore,the quotient is $3x-5$ and the remainder is $9x+10$.

(N/A) Note that the given polynomials are not in standard form. To carry out division,we first write both the dividend and divisor in decreasing order of their degrees.
So,dividend $= -x^{3}+3x^{2}-3x+5$ and divisor $= -x^{2}+x-1$.
Performing the division:
$(-x^{2}+x-1) \overline{) -x^{3}+3x^{2}-3x+5}$
$1$. 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 the divisor $(-x^{2}+x-1)$ by $x$ to get $-x^{3}+x^{2}-x$. Subtract this from the dividend to get $2x^{2}-2x+5$.
$3$. Divide the first term of the new dividend $2x^{2}$ by the first term of the divisor $-x^{2}$ to get $-2$. This is the second term of the quotient.
$4$. Multiply the divisor $(-x^{2}+x-1)$ by $-2$ to get $2x^{2}-2x+2$. Subtract this from the current dividend to get $3$.
We stop here since the degree of the remainder $(3)$ is $0$,which is less than the degree of the divisor $(-x^{2}+x-1)$,which is $2$.
So,quotient $= x-2$,remainder $= 3$.
Verification:
Divisor $\times$ Quotient $+$ Remainder
$= (-x^{2}+x-1)(x-2)+3$
$= -x^{3}+2x^{2}+x^{2}-2x-x+2+3$
$= -x^{3}+3x^{2}-3x+5$
$= \text{Dividend}$.
Thus,the division algorithm is verified.

(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 $8$.

(N/A) To divide $p(x) = x^{4} - 5x + 6$ by $g(x) = -x^{2} + 2$,we arrange the terms in descending order of their degrees:
$p(x) = x^{4} + 0x^{3} + 0x^{2} - 5x + 6$
$g(x) = -x^{2} + 2$
Performing 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 $(-x^{2} + 2)$ to get $x^{4} - 2x^{2}$. Subtract this from $p(x)$ to get $2x^{2} - 5x + 6$.
$3$. Divide the first term of the new polynomial $(2x^{2})$ by the first term of $g(x)$ $(-x^{2})$: $2x^{2} / (-x^{2}) = -2$. This is the second term of the quotient.
$4$. Multiply $-2$ by $(-x^{2} + 2)$ to get $2x^{2} - 4$. Subtract this from $2x^{2} - 5x + 6$ to get $-5x + 10$.
Thus,the quotient is $-x^{2} - 2$ and the remainder is $-5x + 10$.

(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 $2t^{4}+3t^{3}-2t^{2}-9t-12$.

(A) To check if the first polynomial is a factor of the second,we perform long division of $3x^{4}+5x^{3}-7x^{2}+2x+2$ by $x^{2}+3x+1$.
Step $1$: Divide $3x^{4}$ by $x^{2}$ to get $3x^{2}$. Multiply $3x^{2}(x^{2}+3x+1) = 3x^{4}+9x^{3}+3x^{2}$. Subtracting this from the dividend gives $-4x^{3}-10x^{2}+2x+2$.
Step $2$: Divide $-4x^{3}$ by $x^{2}$ to get $-4x$. Multiply $-4x(x^{2}+3x+1) = -4x^{3}-12x^{2}-4x$. Subtracting this gives $2x^{2}+6x+2$.
Step $3$: Divide $2x^{2}$ by $x^{2}$ to get $2$. Multiply $2(x^{2}+3x+1) = 2x^{2}+6x+2$. Subtracting this gives a remainder of $0$.
Since the remainder is $0$,the first polynomial is a factor of the second polynomial.

(N/A) To check if the first polynomial is a factor of the second,we perform polynomial long division.
Dividing $x^{5}-4x^{3}+x^{2}+3x+1$ by $x^{3}-3x+1$:
$1$. Divide the first term of the dividend $(x^{5})$ by the first term of the divisor $(x^{3})$ to get $x^{2}$.
$2$. Multiply $x^{2}$ by $(x^{3}-3x+1)$ to get $x^{5}-3x^{3}+x^{2}$.
$3$. Subtract this from the dividend: $(x^{5}-4x^{3}+x^{2}+3x+1) - (x^{5}-3x^{3}+x^{2}) = -x^{3}+3x+1$.
$4$. Divide the first term of the new polynomial $(-x^{3})$ by the first term of the divisor $(x^{3})$ to get $-1$.
$5$. Multiply $-1$ by $(x^{3}-3x+1)$ to get $-x^{3}+3x-1$.
$6$. Subtract this from the current remainder: $(-x^{3}+3x+1) - (-x^{3}+3x-1) = 2$.
Since the remainder is $2$ (which is $\neq 0$),the first polynomial is not a factor of the second polynomial.

(N/A) Let $p(x) = 3x^{4} + 6x^{3} - 2x^{2} - 10x - 5$.
Since $\sqrt{\frac{5}{3}}$ and $-\sqrt{\frac{5}{3}}$ are two zeroes of $p(x)$,then $(x - \sqrt{\frac{5}{3}})(x + \sqrt{\frac{5}{3}}) = (x^{2} - \frac{5}{3})$ is a factor of $p(x)$.
To find the other zeroes,we divide $p(x)$ by $(x^{2} - \frac{5}{3})$:
$3x^{4} + 6x^{3} - 2x^{2} - 10x - 5 = (x^{2} - \frac{5}{3})(3x^{2} + 6x + 3)$
$= 3(x^{2} - \frac{5}{3})(x^{2} + 2x + 1)$
$= 3(x^{2} - \frac{5}{3})(x + 1)^{2}$
Now,to find the remaining zeroes,we set $(x + 1)^{2} = 0$,which gives $x + 1 = 0$,so $x = -1$.
Since the factor is $(x + 1)^{2}$,the zero $x = -1$ is repeated twice.
Thus,the other two zeroes of the polynomial are $-1$ and $-1$.

(N/A) Given:
Dividend $p(x) = x^{3}-3 x^{2}+x+2$
Quotient $q(x) = x-2$
Remainder $r(x) = -2 x+4$
Divisor $g(x) = ?$
Using the division algorithm for polynomials:
$p(x) = g(x) \cdot q(x) + r(x)$
$x^{3}-3 x^{2}+x+2 = g(x) \cdot (x-2) + (-2 x+4)$
Rearranging the terms to isolate $g(x) \cdot (x-2)$:
$g(x) \cdot (x-2) = (x^{3}-3 x^{2}+x+2) - (-2 x+4)$
$g(x) \cdot (x-2) = x^{3}-3 x^{2}+x+2+2 x-4$
$g(x) \cdot (x-2) = x^{3}-3 x^{2}+3 x-2$
Now,divide $(x^{3}-3 x^{2}+3 x-2)$ by $(x-2)$ to find $g(x)$:
By performing polynomial long division:
$(x^{3}-3 x^{2}+3 x-2) \div (x-2) = x^{2}-x+1$
Therefore,$g(x) = x^{2}-x+1$.

(N/A) According to the division algorithm,if $p(x)$ and $g(x)$ are two polynomials with $g(x) \neq 0,$ then we can find polynomials $q(x)$ and $r(x)$ such that $p(x) = g(x) \cdot q(x) + r(x),$ where $r(x) = 0$ or $\operatorname{deg} r(x) < \operatorname{deg} g(x).$
The degree of a polynomial is the highest power of the variable in the polynomial.
To satisfy $\operatorname{deg} p(x) = \operatorname{deg} q(x),$ the degree of the quotient must be equal to the degree of the dividend. This occurs when the divisor $g(x)$ is a constant.
Let us assume the division of $p(x) = 6x^2 + 2x + 2$ by $g(x) = 2.$
Here,$p(x) = 6x^2 + 2x + 2,$
$g(x) = 2,$
$q(x) = 3x^2 + x + 1,$
$r(x) = 0.$
The degree of $p(x)$ is $2$ and the degree of $q(x)$ is $2,$ so $\operatorname{deg} p(x) = \operatorname{deg} q(x).$
Checking the division algorithm:
$p(x) = g(x) \cdot q(x) + r(x)$
$6x^2 + 2x + 2 = 2(3x^2 + x + 1) + 0$
$6x^2 + 2x + 2 = 6x^2 + 2x + 2.$
Thus,the division algorithm is satisfied.

(N/A) According to the division algorithm,if $p(x)$ and $g(x)$ are two polynomials with $g(x) \neq 0$,then we can find polynomials $q(x)$ and $r(x)$ such that $p(x) = g(x) \cdot q(x) + r(x)$,where $r(x) = 0$ or $\operatorname{deg} r(x) < \operatorname{deg} g(x)$.
We need to find examples such that $\operatorname{deg} q(x) = \operatorname{deg} r(x)$.
Let us consider $p(x) = x^3 + x$ and $g(x) = x^2$.
Performing the division:
$x^3 + x = (x^2) \cdot x + x$.
Here,$q(x) = x$ and $r(x) = x$.
Checking the conditions:
$1$. $\operatorname{deg} q(x) = \operatorname{deg}(x) = 1$.
$2$. $\operatorname{deg} r(x) = \operatorname{deg}(x) = 1$.
Since $1 = 1$,the condition $\operatorname{deg} q(x) = \operatorname{deg} r(x)$ is satisfied.
$3$. $\operatorname{deg} r(x) < \operatorname{deg} g(x)$ is $1 < 2$,which is true.
Thus,the division algorithm is satisfied.

(A) According to the division algorithm,if $p(x)$ and $g(x)$ are two polynomials with $g(x) \neq 0$,then we can find polynomials $q(x)$ and $r(x)$ such that $p(x) = g(x) \times q(x) + r(x)$,where $r(x) = 0$ or $\operatorname{deg} r(x) < \operatorname{deg} g(x)$.
The degree of a polynomial is the highest power of the variable in the polynomial.
For $\operatorname{deg} r(x) = 0$,the remainder must be a non-zero constant.
Let us consider the division of $x^3 + 1$ by $x^2$.
Here,$p(x) = x^3 + 1$,$g(x) = x^2$.
Performing the division: $(x^3 + 1) \div x^2$ gives quotient $q(x) = x$ and remainder $r(x) = 1$.
Clearly,the degree of $r(x) = 1$ is $0$.
Checking the division algorithm:
$p(x) = g(x) \times q(x) + r(x)$
$x^3 + 1 = (x^2) \times x + 1$
$x^3 + 1 = x^3 + 1$
Thus,the division algorithm is satisfied.

(A) Let $p(x) = 2x^3 + x^2 - 5x + 2$.
The given values are $\alpha = \frac{1}{2}, \beta = 1, \gamma = -2$.
First,we verify that these are zeroes:
$p(\frac{1}{2}) = 2(\frac{1}{8}) + \frac{1}{4} - \frac{5}{2} + 2 = \frac{1}{4} + \frac{1}{4} - \frac{5}{2} + 2 = \frac{1}{2} - \frac{5}{2} + 2 = -2 + 2 = 0$.
$p(1) = 2(1)^3 + (1)^2 - 5(1) + 2 = 2 + 1 - 5 + 2 = 0$.
$p(-2) = 2(-8) + 4 - 5(-2) + 2 = -16 + 4 + 10 + 2 = 0$.
Since $p(\frac{1}{2}) = 0, p(1) = 0,$ and $p(-2) = 0$,these are indeed the zeroes.
Comparing $p(x)$ with $ax^3 + bx^2 + cx + d$,we get $a = 2, b = 1, c = -5, d = 2$.
Verification of relationships:
$1$. Sum of zeroes: $\alpha + \beta + \gamma = \frac{1}{2} + 1 - 2 = -\frac{1}{2} = \frac{-b}{a}$.
$2$. Sum of product of zeroes taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = (\frac{1}{2})(1) + (1)(-2) + (-2)(\frac{1}{2}) = \frac{1}{2} - 2 - 1 = \frac{1-6}{2} = -\frac{5}{2} = \frac{c}{a}$.
$3$. Product of zeroes: $\alpha\beta\gamma = (\frac{1}{2})(1)(-2) = -1 = \frac{-d}{a} = \frac{-2}{2} = -1$.
Thus,the relationship is verified.

(A) Let the polynomial be $p(x) = x^{3} - 4x^{2} + 5x - 2$.
The given numbers are $2, 1, 1$.
For $x = 2$: $p(2) = (2)^{3} - 4(2)^{2} + 5(2) - 2 = 8 - 16 + 10 - 2 = 0$.
For $x = 1$: $p(1) = (1)^{3} - 4(1)^{2} + 5(1) - 2 = 1 - 4 + 5 - 2 = 0$.
Since $p(2) = 0$ and $p(1) = 0$,the numbers $2, 1, 1$ are indeed the zeroes of the polynomial.
Comparing $p(x)$ with the standard form $ax^{3} + bx^{2} + cx + d$,we get $a = 1, b = -4, c = 5, d = -2$.
Verification of relationships:
$1$. Sum of zeroes: $2 + 1 + 1 = 4 = -(-4)/1 = -b/a$.
$2$. Sum of product of zeroes taken two at a time: $(2)(1) + (1)(1) + (2)(1) = 2 + 1 + 2 = 5 = 5/1 = c/a$.
$3$. Product of zeroes: $2 \times 1 \times 1 = 2 = -(-2)/1 = -d/a$.
Thus,the relationship between the zeroes and the coefficients is verified.

(A) Let the cubic polynomial be $p(x) = ax^3 + bx^2 + cx + d$ and its zeroes be $\alpha, \beta,$ and $\gamma$.
The relationships between the coefficients and the zeroes are given by:
$1.$ Sum of zeroes: $\alpha + \beta + \gamma = -b/a = 2/1$
$2.$ Sum of the product of zeroes taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = c/a = -7/1$
$3.$ Product of zeroes: $\alpha\beta\gamma = -d/a = -14/1$
Comparing the values,we get $a = 1, b = -2, c = -7,$ and $d = 14$.
Substituting these values into the general form $ax^3 + bx^2 + cx + d$,we get the polynomial:
$p(x) = x^3 - 2x^2 - 7x + 14$.

(A) Given polynomial is $p(x) = x^{3}-3x^{2}+x+1$.
The zeroes are given as $a-b, a, a+b$.
Comparing the polynomial with the standard form $Ax^{3}+Bx^{2}+Cx+D$,we get $A=1, B=-3, C=1, D=1$.
Using the relationship between zeroes and coefficients:
Sum of zeroes $= (a-b) + a + (a+b) = -B/A$.
$3a = -(-3)/1 = 3$.
$a = 1$.
Product of zeroes $= (a-b)(a)(a+b) = -D/A$.
$(1-b)(1)(1+b) = -1/1$.
$1-b^{2} = -1$.
$b^{2} = 2$.
$b = \pm\sqrt{2}$.
Thus,$a=1$ and $b=\pm\sqrt{2}$.

(7, -5) Given that $2+\sqrt{3}$ and $2-\sqrt{3}$ are zeroes of the given polynomial.
Therefore,$(x-(2+\sqrt{3}))(x-(2-\sqrt{3})) = ((x-2)-\sqrt{3})((x-2)+\sqrt{3}) = (x-2)^{2} - (\sqrt{3})^{2} = x^{2}-4x+4-3 = x^{2}-4x+1$ is a factor of the given polynomial.
To find the remaining zeroes,we divide the polynomial $x^{4}-6 x^{3}-26 x^{2}+138 x-35$ by $x^{2}-4 x+1$:
$x^{4}-6 x^{3}-26 x^{2}+138 x-35 = (x^{2}-4 x+1)(x^{2}-2 x-35)$
Now,we factorize the quadratic polynomial $x^{2}-2 x-35$:
$x^{2}-2 x-35 = x^{2}-7x+5x-35 = x(x-7)+5(x-7) = (x-7)(x+5)$
Setting these factors to zero,we get $x-7=0$ or $x+5=0$,which gives $x=7$ or $x=-5$.
Hence,the other two zeroes are $7$ and $-5$.

(A) By the division algorithm,we have:
Dividend $=$ Divisor $\times$ Quotient $+$ Remainder
Therefore,Dividend $-$ Remainder $=$ Divisor $\times$ Quotient.
Subtracting the remainder $(x+a)$ from the dividend:
$(x^{4}-6x^{3}+16x^{2}-25x+10) - (x+a) = x^{4}-6x^{3}+16x^{2}-26x+10-a$.
This resulting polynomial must be perfectly divisible by $x^{2}-2x+k$.
Performing long division of $(x^{4}-6x^{3}+16x^{2}-26x+10-a)$ by $(x^{2}-2x+k)$:
$1$. Dividing $x^{4}$ by $x^{2}$ gives $x^{2}$. Multiplying $(x^{2}-2x+k)$ by $x^{2}$ gives $x^{4}-2x^{3}+kx^{2}$. Subtracting this from the dividend leaves $-4x^{3}+(16-k)x^{2}-26x$.
$2$. Dividing $-4x^{3}$ by $x^{2}$ gives $-4x$. Multiplying $(x^{2}-2x+k)$ by $-4x$ gives $-4x^{3}+8x^{2}-4kx$. Subtracting this leaves $(8-k)x^{2}+(4k-26)x+(10-a)$.
$3$. Dividing $(8-k)x^{2}$ by $x^{2}$ gives $(8-k)$. Multiplying $(x^{2}-2x+k)$ by $(8-k)$ gives $(8-k)x^{2}-2(8-k)x+k(8-k)$.
Subtracting this from the previous remainder,we get the final remainder as $[(4k-26) + 2(8-k)]x + [10-a - k(8-k)] = 0$.
Simplifying the coefficient of $x$: $4k-26+16-2k = 2k-10$. Setting $2k-10=0$ gives $k=5$.
Simplifying the constant term: $10-a-8k+k^{2} = 0$. Substituting $k=5$: $10-a-8(5)+25 = 10-a-40+25 = -5-a = 0$,which gives $a=-5$.
Thus,$k=5$ and $a=-5$.