Relationship between Zeroes and Coefficients of a Polynomial Questions in English

(B) Let the two polynomials be $P(x)$ and $Q(x)$.
Given that $P(x) \cdot Q(x) = x^{2}+8x+15$.
We are given one polynomial $P(x) = x+3$.
To find the other polynomial $Q(x)$,we divide the product by the given polynomial:
$Q(x) = \frac{x^{2}+8x+15}{x+3}$.
We can factorize the quadratic polynomial $x^{2}+8x+15$ by splitting the middle term:
$x^{2}+5x+3x+15 = x(x+5)+3(x+5) = (x+3)(x+5)$.
Substituting this back into the division:
$Q(x) = \frac{(x+3)(x+5)}{x+3} = x+5$.
Therefore,the other polynomial is $x+5$.

(C) Given that $1$ is a zero of the polynomial $p(x) = ax^2 - 11x + 3$.
Therefore,$p(1) = 0$.
Substituting $x = 1$ into the polynomial:
$a(1)^2 - 11(1) + 3 = 0$
$a - 11 + 3 = 0$
$a - 8 = 0$
$a = 8$.

(C) Given that $3$ is a zero of the polynomial $p(x) = 3x^{3} - x^{2} - ax - 45$,it implies that $p(3) = 0$.
Substituting $x = 3$ into the polynomial:
$3(3)^{3} - (3)^{2} - a(3) - 45 = 0$
$3(27) - 9 - 3a - 45 = 0$
$81 - 9 - 3a - 45 = 0$
$27 - 3a = 0$
$3a = 27$
$a = 9$

(A) To find the quotient when $2x^3 - x^2 - 2x - 8$ is divided by $x - 2$,we can use synthetic division:
$\begin{array}{c|rrrr} 2 & 2 & -1 & -2 & -8 \\ & 0 & 4 & 6 & 8 \\ \hline & 2 & 3 & 4 & 0 \end{array}$
The coefficients of the quotient polynomial are $2, 3, 4$.
Therefore,the quotient polynomial is $2x^2 + 3x + 4$ and the remainder is $0$.

(B) polynomial is defined as an expression consisting of variables and coefficients,where the exponents of the variables must be non-negative integers.
$A$ cubic polynomial is a polynomial of degree $3$,meaning the highest power of the variable $x$ is $3$.
In option $A$,the term $\sqrt{x}$ is $x^{1/2}$,which is not an integer.
In option $B$,$p(x) = 2 - 3x - x^3$. The highest power of $x$ is $3$,and all exponents are non-negative integers. Thus,it is a cubic polynomial.
In option $C$,$\sqrt{x^3} = x^{3/2}$,which is not an integer.
In option $D$,the exponent $1/3$ is not an integer.
Therefore,the correct option is $B$.

(A) According to the Factor Theorem,if $(x-a)$ is a factor of a polynomial $p(x)$,then $p(a) = 0$.
Here,the given factor is $(x-1)$,so we must have $p(1) = 0$.
Let us check each option:
$A) p(1) = (1)^{2} + 2(1) - 3 = 1 + 2 - 3 = 0$. Since $p(1) = 0$,$(x-1)$ is a factor.
$B) p(1) = (1)^{2} + 4(1) + 3 = 1 + 4 + 3 = 8 \neq 0$.
$C) p(1) = (1)^{2} + 5(1) + 6 = 1 + 5 + 6 = 12 \neq 0$.
$D) p(1) = (1)^{2} + (1) - 6 = 1 + 1 - 6 = -4 \neq 0$.
Therefore,the correct option is $A$.

(A) To find the zeros of the polynomial $p(x) = 4x^3 - x$,we set $p(x) = 0$.
$4x^3 - x = 0$
Factor out $x$ from the expression:
$x(4x^2 - 1) = 0$
Using the difference of squares formula $a^2 - b^2 = (a - b)(a + b)$,we can factor $4x^2 - 1$ as $(2x - 1)(2x + 1)$:
$x(2x - 1)(2x + 1) = 0$
Setting each factor to zero,we get:
$x = 0$,$2x - 1 = 0 \implies x = 1/2$,and $2x + 1 = 0 \implies x = -1/2$.
Thus,the zeros are $0, 1/2, -1/2$.
There are $3$ zeros in total.

(C) To find the zeros of the polynomial $p(x) = x^{2} - 2x - 15$,we set $p(x) = 0$.
$x^{2} - 2x - 15 = 0$
We need to factorize the quadratic equation by splitting the middle term. We look for two numbers whose product is $-15$ and whose sum is $-2$.
These numbers are $-5$ and $3$.
$x^{2} - 5x + 3x - 15 = 0$
$x(x - 5) + 3(x - 5) = 0$
$(x - 5)(x + 3) = 0$
Setting each factor to zero:
$x - 5 = 0 \implies x = 5$
$x + 3 = 0 \implies x = -3$
Therefore,the zeros of the polynomial are $5$ and $-3$.

(A) quadratic polynomial with zeros $\alpha$ and $\beta$ is given by the formula: $p(x) = k[x^{2} - (\alpha + \beta)x + (\alpha \cdot \beta)]$,where $k$ is a constant.
Given zeros are $\alpha = -3$ and $\beta = 4$.
Sum of zeros: $\alpha + \beta = -3 + 4 = 1$.
Product of zeros: $\alpha \cdot \beta = (-3) \times 4 = -12$.
Substituting these values into the formula (taking $k=1$): $p(x) = x^{2} - (1)x + (-12) = x^{2} - x - 12$.
Therefore,the correct polynomial is $p(x) = x^{2} - x - 12$.

(B) Given the quadratic polynomial $p(x) = x^{2} - 3x + 2$.
Comparing this with the standard form $ax^{2} + bx + c$,we get $a = 1$,$b = -3$,and $c = 2$.
For a quadratic polynomial,the sum of zeros $\alpha + \beta = -\frac{b}{a} = -(\frac{-3}{1}) = 3$.
The product of zeros $\alpha \beta = \frac{c}{a} = \frac{2}{1} = 2$.
We need to find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$.
Taking the common denominator,we get $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta}$.
Substituting the values,$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{3}{2}$.

(B) For a cubic polynomial $p(x) = ax^3 + bx^2 + cx + d,$ the relationships between zeros and coefficients are:
$1$. Sum of zeros: $\alpha + \beta + \gamma = -\frac{b}{a}$
$2$. Sum of product of zeros taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$
$3$. Product of zeros: $\alpha\beta\gamma = -\frac{d}{a}$
Now,we need to find the value of $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}.$
Taking the least common multiple $(LCM)$ of the denominators:
$\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\beta\gamma + \alpha\gamma + \alpha\beta}{\alpha\beta\gamma}$
Substituting the values from the relationships:
$= \frac{c/a}{-d/a} = \frac{c}{a} \times \left(-\frac{a}{d}\right) = -\frac{c}{d}.$

(B) For a cubic polynomial $p(x) = ax^3 + bx^2 + cx + d$,where $a, b, c, d$ are real numbers and $a \neq 0$.
Let the zeros of the polynomial be $\alpha, \beta,$ and $\gamma$.
According to the relationship between zeros and coefficients of a cubic polynomial:
$1$. The sum of zeros: $\alpha + \beta + \gamma = -\frac{b}{a}$.
$2$. The sum of the product of zeros taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$.
$3$. The product of zeros: $\alpha\beta\gamma = -\frac{d}{a}$.
Here,$b$ is the coefficient of $x^2$ and $a$ is the coefficient of $x^3$.
Therefore,$\alpha + \beta + \gamma = -\frac{\text{coefficient of } x^2}{\text{coefficient of } x^3}$.

(C) For a cubic polynomial $p(x) = ax^3 + bx^2 + cx + d$,the relationship between the zeros $\alpha, \beta, \gamma$ and the coefficients is given by Vieta's formulas:
$1$. Sum of zeros: $\alpha + \beta + \gamma = -\frac{b}{a} = -\frac{\text{coefficient of } x^2}{\text{coefficient of } x^3}$
$2$. Sum of the product of zeros taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{\text{coefficient of } x}{\text{coefficient of } x^3}$
$3$. Product of zeros: $\alpha\beta\gamma = -\frac{d}{a} = -\frac{\text{constant term}}{\text{coefficient of } x^3}$
Therefore,$\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{\text{coefficient of } x}{\text{coefficient of } x^3}$.

(D) For a cubic polynomial $p(x) = ax^3 + bx^2 + cx + d$,where $a \neq 0$,let the zeros be $\alpha, \beta,$ and $\gamma$.
According to the relationship between zeros and coefficients:
$1$. Sum of zeros: $\alpha + \beta + \gamma = -\frac{b}{a} = -\frac{\text{coefficient of } x^2}{\text{coefficient of } x^3}$
$2$. Sum of product of zeros taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{\text{coefficient of } x}{\text{coefficient of } x^3}$
$3$. Product of zeros: $\alpha\beta\gamma = -\frac{d}{a} = -\frac{\text{constant term}}{\text{coefficient of } x^3}$
Since the given options do not include $-\frac{d}{a}$,the correct choice is $D$.

(C) For a cubic polynomial $p(x) = ax^3 + bx^2 + cx + d$,the relationship between the zeros and the coefficients is given by Vieta's formulas.
Let the zeros be $\alpha, \beta$,and $\gamma$.
The sum of the zeros is given by the formula:
$\alpha + \beta + \gamma = -\frac{\text{coefficient of } x^2}{\text{coefficient of } x^3}$
Substituting the coefficients from the given polynomial:
$\alpha + \beta + \gamma = -\frac{b}{a}$
Therefore,the correct option is $C$.

(A) For a cubic polynomial $p(x) = ax^3 + bx^2 + cx + d$,where $a, b, c, d$ are real numbers and $a \neq 0$,let the zeros be $\alpha, \beta$,and $\gamma$.
According to the relationship between the coefficients and the zeros of a polynomial:
$1$. The sum of the zeros: $\alpha + \beta + \gamma = -\frac{b}{a}$
$2$. The sum of the product of the zeros taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$
$3$. The product of the zeros: $\alpha\beta\gamma = -\frac{d}{a}$
Therefore,the value of $\alpha\beta + \beta\gamma + \gamma\alpha$ is $\frac{c}{a}$.

(B) For a cubic polynomial $p(x) = ax^3 + bx^2 + cx + d$ with zeros $\alpha, \beta,$ and $\gamma$:
$1$. The sum of zeros is $\alpha + \beta + \gamma = -\frac{b}{a}$.
$2$. The sum of the product of zeros taken two at a time is $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$.
$3$. The product of the zeros is $\alpha\beta\gamma = -\frac{d}{a}$.
Thus,the correct option is $B$.

(B) For a cubic polynomial of the form $p(x) = ax^{3} + bx^{2} + cx + d$,the sum of the zeros $\alpha + \beta + \gamma$ is given by the formula $-\frac{b}{a}$.
Given the polynomial $p(x) = x^{3} - 6x^{2} - 7x$,we identify the coefficients as $a = 1$,$b = -6$,$c = -7$,and $d = 0$.
Substituting these values into the formula:
$\alpha + \beta + \gamma = -\frac{-6}{1} = 6$.

(A) For a cubic polynomial of the form $p(x) = ax^{3} + bx^{2} + cx + d$,the relationship between the coefficients and the zeros $\alpha, \beta, \gamma$ is given by Vieta's formulas.
Specifically,the sum of the product of zeros taken two at a time is given by $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$.
Comparing the given polynomial $p(x) = x^{3} + x^{2} - 17x + 15$ with the standard form,we have $a = 1$,$b = 1$,$c = -17$,and $d = 15$.
Substituting these values into the formula:
$\alpha\beta + \beta\gamma + \gamma\alpha = \frac{-17}{1} = -17$.

(B) According to the Fundamental Theorem of Algebra,a polynomial of degree $n$ has at most $n$ real zeros.
Given that the degree of the polynomial is $k+1$,the maximum number of zeros it can have is equal to its degree.
Therefore,the maximum number of zeros is $k+1$.

(C) Given the cubic polynomial $p(x) = ax^3 + bx^2 + cx + d$,the relationships between the zeros $\alpha, \beta, \gamma$ and the coefficients are:
$\alpha + \beta + \gamma = -\frac{b}{a}$
$\alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a}$
$\alpha \beta \gamma = -\frac{d}{a}$
We need to evaluate the expression: $\alpha^2 \beta \gamma + \alpha \beta^2 \gamma + \alpha \beta \gamma^2$
Factor out the common term $\alpha \beta \gamma$:
$= \alpha \beta \gamma (\alpha + \beta + \gamma)$
Substitute the known values:
$= (-\frac{d}{a}) \times (-\frac{b}{a})$
$= \frac{bd}{a^2}$

(C) Given the polynomial $p(x) = 3x^2 - x - 4$,comparing it with $ax^2 + bx + c$,we have $a = 3$,$b = -1$,and $c = -4$.
The sum of the zeros is $\alpha + \beta = -\frac{b}{a} = -\frac{-1}{3} = \frac{1}{3}$.
The product of the zeros is $\alpha \beta = \frac{c}{a} = \frac{-4}{3} = -\frac{4}{3}$.
We need to find the value of $\alpha^2 \beta + \alpha \beta^2$.
Factoring out $\alpha \beta$,we get $\alpha^2 \beta + \alpha \beta^2 = \alpha \beta(\alpha + \beta)$.
Substituting the values,we get $\left(-\frac{4}{3}\right) \left(\frac{1}{3}\right) = -\frac{4}{9}$.

(A) To find the remainder when the polynomial $p(x) = x^{3}+2x^{2}+3x+5$ is divided by $x+1$,we use the Remainder Theorem.
According to the Remainder Theorem,if a polynomial $p(x)$ is divided by $(x-a)$,the remainder is $p(a)$.
Here,the divisor is $x+1$,which can be written as $x-(-1)$. Thus,$a = -1$.
Now,we calculate $p(-1)$:
$p(-1) = (-1)^{3} + 2(-1)^{2} + 3(-1) + 5$
$p(-1) = -1 + 2(1) - 3 + 5$
$p(-1) = -1 + 2 - 3 + 5$
$p(-1) = 3$
Therefore,the remainder is $3$.

(C) The degree of a polynomial is the highest power of the variable present in the expression.
For option $A$: $x^{3} + x^{2}$,the degree is $3$.
For option $B$: $5x + x^{2}$,the degree is $2$.
For option $C$: $x^{3}(x^{2} + 1) = x^{5} + x^{3}$,the degree is $5$.
For option $D$: $x(x^{5} - 2) = x^{6} - 2x$,the degree is $6$.
Therefore,the polynomial with degree $5$ is $x^{3}(x^{2} + 1)$.

(B) According to the Remainder Theorem,if a polynomial $p(x)$ is divided by $(x - a)$,the remainder is $p(a)$.
Here,the divisor is $x + 2$,which can be written as $x - (-2)$.
Therefore,$a = -2$.
Now,substitute $x = -2$ into the polynomial $p(x) = 40x^2 + 11x - 63$:
$p(-2) = 40(-2)^2 + 11(-2) - 63$
$p(-2) = 40(4) - 22 - 63$
$p(-2) = 160 - 22 - 63$
$p(-2) = 160 - 85$
$p(-2) = 75$
Thus,the remainder is $75$.

(C) According to the Factor Theorem,if $P(a) = 0$ for a polynomial $P(x)$,then $(x - a)$ is a factor of $P(x)$.
Given that $P(-7) = 0$,we substitute $a = -7$ into the theorem.
Therefore,$(x - (-7))$ is a factor of $P(x)$.
This simplifies to $(x + 7)$ being a factor of $P(x)$.

(B) Let the zeros of the cubic polynomial $p(x) = ax^3 + bx^2 + cx + d$ be $\alpha, \beta,$ and $\gamma$.
Given: $\alpha = \sqrt{5}$,$\beta = -\sqrt{5}$,and we need to find $\gamma$.
According to the relationship between zeros and coefficients of a cubic polynomial,the sum of the zeros is given by:
$\alpha + \beta + \gamma = -\frac{b}{a}$
From the polynomial $p(x) = x^3 + x^2 - 5x - 5$,we have $a = 1$ and $b = 1$.
Substituting the values:
$\sqrt{5} + (-\sqrt{5}) + \gamma = -\frac{1}{1}$
$0 + \gamma = -1$
$\gamma = -1$
Thus,the third zero is $-1$.

(A) Given the polynomial $p(x) = ax^2 + bx + c$ with zeros $\alpha$ and $\beta$.
From the relationship between zeros and coefficients:
Sum of zeros: $\alpha + \beta = -\frac{b}{a}$
Product of zeros: $\alpha \cdot \beta = \frac{c}{a}$
We need to find $\frac{1}{\alpha} + \frac{1}{\beta}$.
Taking the common denominator: $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \cdot \beta}$.
Substituting the values: $\frac{-\frac{b}{a}}{\frac{c}{a}} = -\frac{b}{a} \cdot \frac{a}{c} = -\frac{b}{c}$.
Therefore,the correct option is $A$.

(C) To factorize the quadratic polynomial $p(x) = x^2 + 4x - 5$,we need to find two numbers whose product is $-5$ and whose sum is $4$.
These two numbers are $5$ and $-1$,because $5 \times (-1) = -5$ and $5 + (-1) = 4$.
Now,rewrite the middle term $4x$ as $5x - x$:
$p(x) = x^2 + 5x - x - 5$
Group the terms:
$p(x) = (x^2 + 5x) - (x + 5)$
Factor out the common terms:
$p(x) = x(x + 5) - 1(x + 5)$
Finally,factor out $(x + 5)$:
$p(x) = (x + 5)(x - 1)$.

(D) For a quadratic polynomial of the form $ax^2 + bx + c$,the sum of the zeros $\alpha + \beta$ is given by the formula $\alpha + \beta = -b/a$.
Comparing $p(x) = x^2 - 4x + 3$ with $ax^2 + bx + c$,we get $a = 1$,$b = -4$,and $c = 3$.
Substituting these values into the formula: $\alpha + \beta = -(-4)/1 = 4/1 = 4$.
Since $4$ is a positive integer,the correct option is $D$.

(A) For a quadratic polynomial $ax^2 + bx + c$,the sum of zeros $\alpha + \beta = -b/a$ and the product of zeros $\alpha\beta = c/a$.
Given the polynomial $P(x) = x^2 - 3x + 2k$,we have $a = 1$,$b = -3$,and $c = 2k$.
Thus,$\alpha + \beta = -(-3)/1 = 3$ and $\alpha\beta = 2k/1 = 2k$.
According to the problem,$\alpha + \beta = \alpha\beta$.
Substituting the values,we get $3 = 2k$.
Therefore,$k = 3/2$.

(B) To find the zero of the polynomial $p(x) = \sqrt{5}x - 5$,we set $p(x) = 0$.
So,$\sqrt{5}x - 5 = 0$.
Adding $5$ to both sides,we get $\sqrt{5}x = 5$.
Dividing both sides by $\sqrt{5}$,we get $x = \frac{5}{\sqrt{5}}$.
Rationalizing the denominator,$x = \frac{5 \cdot \sqrt{5}}{\sqrt{5} \cdot \sqrt{5}} = \frac{5\sqrt{5}}{5} = \sqrt{5}$.
Therefore,the zero of the polynomial is $\sqrt{5}$.

(D) For a quadratic polynomial of the form $ax^{2}+bx+c$,the product of the zeros is given by the formula $\frac{c}{a}$.
Comparing the given polynomial $x^{2}-4x+3$ with the standard form $ax^{2}+bx+c$,we have $a=1$,$b=-4$,and $c=3$.
Therefore,the product of the zeros = $\frac{c}{a} = \frac{3}{1} = 3$.
Thus,the correct option is $D$.

(A) For a cubic polynomial $p(x) = ax^3 + bx^2 + cx + d$,the relationship between the coefficients and the zeros $(\alpha, \beta, \gamma)$ are given by:
$1$. Sum of zeros: $\alpha + \beta + \gamma = -\frac{b}{a}$
$2$. Sum of product of zeros taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$
$3$. Product of zeros: $\alpha\beta\gamma = -\frac{d}{a}$
Therefore,the product of the zeros is $-\frac{d}{a}$.

(C) polynomial is an algebraic expression where the exponents of the variable must be non-negative integers.
In option $A$,$4x^2 + \sqrt{7}$,the exponent of $x$ is $2$,which is a non-negative integer.
In option $B$,$3x^2 + x - 1$,the exponents of $x$ are $2$ and $1$,which are non-negative integers.
In option $C$,$3x^2 - 5\sqrt{x} + 2$,the term $\sqrt{x}$ can be written as $x^{1/2}$. Since $1/2$ is not an integer,this expression is not a polynomial.
In option $D$,$3x - 1$,the exponent of $x$ is $1$,which is a non-negative integer.
Therefore,$3x^2 - 5\sqrt{x} + 2$ is not a polynomial.

(B) Let the polynomial be $p(x) = x^{2} + 7x + m$.
Since $(x+4)$ is a factor of $p(x)$,by the Factor Theorem,$p(-4) = 0$.
Substituting $x = -4$ into the polynomial:
$(-4)^{2} + 7(-4) + m = 0$
$16 - 28 + m = 0$
$-12 + m = 0$
$m = 12$.
Therefore,the value of $m$ is $12$.