Demo Questions in English

(A) Step $1$: Find the $HCF$ of $8$ and $6$.
Prime factorization of $8 = 2^3$.
Prime factorization of $6 = 2 \times 3$.
The common factor is $2$,so $HCF$ $(8, 6) = 2$.
Step $2$: Find the $LCM$ of $26$ and $91$.
Prime factorization of $26 = 2 \times 13$.
Prime factorization of $91 = 7 \times 13$.
$LCM$ $(26, 91) = 2 \times 7 \times 13 = 182$.
Therefore,$Q.1$ matches with $C$ and $Q.2$ matches with $B$.

(B) Let us assume,to the contrary,that $7\sqrt{5}$ is a rational number.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $7\sqrt{5} = \frac{a}{b}$.
Rearranging the equation,we get $\sqrt{5} = \frac{a}{7b}$.
Since $a$ and $b$ are integers,$\frac{a}{7b}$ is a rational number.
This implies that $\sqrt{5}$ is a rational number.
However,this contradicts the fact that $\sqrt{5}$ is an irrational number.
Therefore,our assumption is incorrect,and $7\sqrt{5}$ must be an irrational number.

(B) Let us assume,to the contrary,that $\sqrt{2}$ is rational.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $\sqrt{2} = \frac{a}{b}$.
Squaring both sides,we get $2 = \frac{a^2}{b^2}$,which implies $a^2 = 2b^2$.
This shows that $a^2$ is even,so $a$ must also be even.
Let $a = 2k$ for some integer $k$.
Substituting this into the equation,we get $(2k)^2 = 2b^2$,which simplifies to $4k^2 = 2b^2$,or $b^2 = 2k^2$.
This shows that $b^2$ is even,so $b$ must also be even.
Since both $a$ and $b$ are even,they have a common factor $2$,which contradicts our assumption that $a$ and $b$ are coprime.
Therefore,our assumption is false,and $\sqrt{2}$ is irrational.

(B) Let us assume,to the contrary,that $3 + 2\sqrt{5}$ is rational.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $3 + 2\sqrt{5} = \frac{a}{b}$.
Subtracting $3$ from both sides,we get $2\sqrt{5} = \frac{a}{b} - 3$.
$2\sqrt{5} = \frac{a - 3b}{b}$.
Dividing by $2$,we get $\sqrt{5} = \frac{a - 3b}{2b}$.
Since $a$ and $b$ are integers,$\frac{a - 3b}{2b}$ is a rational number.
This implies that $\sqrt{5}$ is a rational number.
However,this contradicts the fact that $\sqrt{5}$ is irrational.
Therefore,our assumption is incorrect,and $3 + 2\sqrt{5}$ is irrational.

(B) Assume,to the contrary,that $\sqrt{2}$ is a rational number.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $\sqrt{2} = \frac{a}{b}$.
Squaring both sides,we get $2 = \frac{a^2}{b^2}$,which implies $a^2 = 2b^2$.
This shows that $a^2$ is even,and therefore $a$ must also be even.
Let $a = 2k$ for some integer $k$.
Substituting this into the equation,we get $(2k)^2 = 2b^2$,so $4k^2 = 2b^2$,which simplifies to $b^2 = 2k^2$.
This shows that $b^2$ is even,and therefore $b$ must also be even.
Since both $a$ and $b$ are even,they have a common factor of $2$,which contradicts our assumption that $a$ and $b$ are coprime.
Thus,our assumption is false,and $\sqrt{2}$ is irrational.

(A) To find the time when Sonia and Ravi meet again at the starting point,we need to find the Least Common Multiple $(LCM)$ of the time taken by them to complete one round.
Time taken by Sonia = $18$ minutes.
Time taken by Ravi = $12$ minutes.
Prime factorization of $18 = 2 \times 3^2$.
Prime factorization of $12 = 2^2 \times 3$.
$LCM(18, 12) = 2^2 \times 3^2 = 4 \times 9 = 36$.
Therefore,they will meet again at the starting point after $36$ minutes.

(C) composite number is a positive integer that has at least one divisor other than $1$ and itself.
For the first expression: $7 \times 11 \times 13 + 13 = 13 \times (7 \times 11 + 1) = 13 \times (77 + 1) = 13 \times 78 = 13 \times 13 \times 6$.
Since this number can be expressed as a product of factors other than $1$ and itself,it is a composite number.
For the second expression: $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5 \times (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1) = 5 \times (1008 + 1) = 5 \times 1009$.
Since $1009$ is a prime number,the expression is $5 \times 1009$. This number has factors $1, 5, 1009$,and the number itself. Since it has more than two factors,it is a composite number.

(B) For any number to end with the digit $0$,its prime factorization must contain the factors $2$ and $5$.
The prime factorization of $6^n$ is $(2 \times 3)^n = 2^n \times 3^n$.
Since the prime factorization of $6^n$ only contains $2$ and $3$,it does not contain the factor $5$.
Therefore,by the Fundamental Theorem of Arithmetic,$6^n$ cannot end with the digit $0$ for any natural number $n$.

(A) We know that for any two positive integers $a$ and $b$,the relationship between their $HCF$ and $LCM$ is given by:
$HCF(a, b) \times LCM(a, b) = a \times b$
Given $a = 306$,$b = 657$,and $HCF(306, 657) = 9$.
Substituting these values into the formula:
$9 \times LCM(306, 657) = 306 \times 657$
$LCM(306, 657) = \frac{306 \times 657}{9}$
$LCM(306, 657) = 34 \times 657$
$LCM(306, 657) = 22338$
Therefore,the $LCM$ is $22338$.

(A) To find the $HCF$ and $LCM$ of $8$,$9$,and $25$ using the prime factorization method:
Step $1$: Find the prime factors of each number.
$8 = 2 \times 2 \times 2 = 2^3$
$9 = 3 \times 3 = 3^2$
$25 = 5 \times 5 = 5^2$
Step $2$: The $HCF$ is the product of the smallest power of each common prime factor. Since there are no common prime factors other than $1$,the $HCF$ is $1$.
Step $3$: The $LCM$ is the product of the greatest power of each prime factor involved in the numbers.
$LCM = 2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25 = 72 \times 25 = 1800$.
Thus,$HCF$ = $1$ and $LCM$ = $1800$.

(A) The prime factorization of the given numbers is as follows:
$17 = 17 \times 1$
$23 = 23 \times 1$
$29 = 29 \times 1$
Since $17, 23,$ and $29$ are all prime numbers,they have no common factor other than $1$.
Therefore,the Highest Common Factor $(HCF)$ is $1$.
The Least Common Multiple $(LCM)$ is the product of the highest powers of all prime factors involved:
$LCM = 17 \times 23 \times 29 = 11339$.
Thus,$HCF$ = $1$ and $LCM$ = $11339$.

(A) To find the $HCF$ and $LCM$ of $12$,$15$,and $21$ using prime factorization:
Step $1$: Write the prime factorization of each number.
$12 = 2^2 \times 3^1$
$15 = 3^1 \times 5^1$
$21 = 3^1 \times 7^1$
Step $2$: Find the $HCF$ (Highest Common Factor).
The $HCF$ is the product of the smallest power of each common prime factor.
The only common prime factor is $3$,and its smallest power is $3^1$.
Therefore,$HCF = 3$.
Step $3$: Find the $LCM$ (Least Common Multiple).
The $LCM$ is the product of the greatest power of each prime factor involved.
The prime factors involved are $2, 3, 5,$ and $7$.
The greatest powers are $2^2, 3^1, 5^1,$ and $7^1$.
$LCM = 2^2 \times 3^1 \times 5^1 \times 7^1 = 4 \times 3 \times 5 \times 7 = 420$.
Thus,$HCF = 3$ and $LCM = 420$.

(A) Step $1$: Prime factorization of $336$ and $54$.
$336 = 2^4 \times 3^1 \times 7^1$
$54 = 2^1 \times 3^3$
Step $2$: Find $HCF$.
$HCF = 2^{\min(4,1)} \times 3^{\min(1,3)} = 2^1 \times 3^1 = 6$.
Step $3$: Find $LCM$.
$LCM = 2^{\max(4,1)} \times 3^{\max(1,3)} \times 7^1 = 2^4 \times 3^3 \times 7 = 16 \times 27 \times 7 = 3024$.
Step $4$: Verify $HCF \times LCM = \text{Product of integers}$.
$HCF \times LCM = 6 \times 3024 = 18144$.
$\text{Product of integers} = 336 \times 54 = 18144$.
Since $18144 = 18144$, the relationship is verified.

(A) Step $1$: Find the prime factorization of $510$ and $92$.
$510 = 2 \times 3 \times 5 \times 17$
$92 = 2^2 \times 23$
Step $2$: Find $HCF$ and $LCM$.
$HCF = 2^1 = 2$
$LCM = 2^2 \times 3 \times 5 \times 17 \times 23 = 23460$
Step $3$: Verify the relationship $HCF \times LCM = \text{product of the two numbers}$.
$HCF \times LCM = 2 \times 23460 = 46920$
$\text{Product of numbers} = 510 \times 92 = 46920$
Since $46920 = 46920$,the relationship is verified.

(A) Step $1$: Prime factorization of $26$ and $91$.
$26 = 2 \times 13$
$91 = 7 \times 13$
Step $2$: Find $HCF$ and $LCM$.
$HCF = \text{Product of the smallest power of each common prime factor} = 13$.
$LCM = \text{Product of the greatest power of each prime factor involved} = 2 \times 7 \times 13 = 182$.
Step $3$: Verify the relationship $HCF \times LCM = \text{Product of the two numbers}$.
$HCF \times LCM = 13 \times 182 = 2366$.
$\text{Product of the two numbers} = 26 \times 91 = 2366$.
Since $2366 = 2366$, the relationship is verified.

(A) To find the prime factors of $7429$,we test divisibility by prime numbers sequentially.
$7429$ is not divisible by $2, 3, 5, 7, 11,$ or $13$.
Testing $17$: $7429 \div 17 = 437$.
Now,we find the prime factors of $437$.
Testing $19$: $437 \div 19 = 23$.
Since $23$ is a prime number,the prime factorization is $17 \times 19 \times 23$.

(A) To find the prime factors of $5005$,we perform successive division by prime numbers:
$1$. Since the last digit is $5$,it is divisible by $5$: $5005 \div 5 = 1001$.
$2$. Now,divide $1001$ by the next prime number,$7$: $1001 \div 7 = 143$.
$3$. Next,divide $143$ by the next prime number,$11$: $143 \div 11 = 13$.
$4$. Finally,$13$ is a prime number,so $13 \div 13 = 1$.
Therefore,the prime factorization of $5005$ is $5 \times 7 \times 11 \times 13$.

(B) To find the prime factors of $3825$,we perform successive division by prime numbers:
$1$. Since the last digit is $5$,it is divisible by $5$: $3825 \div 5 = 765$.
$2$. $765$ is also divisible by $5$: $765 \div 5 = 153$.
$3$. Now,check for $3$: The sum of digits of $153$ is $1+5+3 = 9$,which is divisible by $3$. So,$153 \div 3 = 51$.
$4$. $51$ is divisible by $3$: $51 \div 3 = 17$.
$5$. $17$ is a prime number,so $17 \div 17 = 1$.
Thus,the prime factorization is $3825 = 3 \times 3 \times 5 \times 5 \times 17 = 3^2 \times 5^2 \times 17$.

(A) To express $156$ as a product of its prime factors,we perform prime factorization:
$156 = 2 \times 78$
$78 = 2 \times 39$
$39 = 3 \times 13$
Combining these,we get $156 = 2 \times 2 \times 3 \times 13$.
In exponential form,this is $2^2 \times 3 \times 13$.

(B) To express $140$ as a product of its prime factors,we perform prime factorization:
$1$. Divide $140$ by the smallest prime number,$2$: $140 \div 2 = 70$.
$2$. Divide $70$ by $2$: $70 \div 2 = 35$.
$3$. Divide $35$ by the next smallest prime number,$5$: $35 \div 5 = 7$.
$4$. Divide $7$ by the prime number $7$: $7 \div 7 = 1$.
Thus,$140 = 2 \times 2 \times 5 \times 7$.
In exponential form,this is $2^2 \times 5 \times 7$.

(A) To find the $HCF$ of $867$ and $255$ using Euclid's division algorithm,we follow these steps:
Step $1$: Since $867 > 255$,we apply Euclid's division lemma to $867$ and $255$:
$867 = 255 \times 3 + 102$
Step $2$: Since the remainder $102 \neq 0$,we apply the lemma to $255$ and $102$:
$255 = 102 \times 2 + 51$
Step $3$: Since the remainder $51 \neq 0$,we apply the lemma to $102$ and $51$:
$102 = 51 \times 2 + 0$
Since the remainder is now $0$,the divisor at this stage is the $HCF$.
Therefore,the $HCF$ of $867$ and $255$ is $51$.

(A) To find the $HCF$ of $135$ and $225$ using Euclid's division algorithm,we follow these steps:
Step $1$: Since $225 > 135$,we apply Euclid's division lemma to $225$ and $135$:
$225 = 135 \times 1 + 90$
Step $2$: Since the remainder $90 \neq 0$,we apply the lemma to $135$ and $90$:
$135 = 90 \times 1 + 45$
Step $3$: Since the remainder $45 \neq 0$,we apply the lemma to $90$ and $45$:
$90 = 45 \times 2 + 0$
Since the remainder is $0$,the divisor at this stage is the $HCF$.
Therefore,the $HCF$ of $135$ and $225$ is $45$.

(A) To determine if the decimal expansion of a rational number $\frac{p}{q}$ is terminating,we examine the prime factorization of the denominator $q$.
If $q$ is of the form $2^n \times 5^m$,where $n$ and $m$ are non-negative integers,then the decimal expansion is terminating.
Given the rational number $\frac{13}{3125}$,the denominator is $q = 3125$.
Prime factorization of $3125 = 5 \times 5 \times 5 \times 5 \times 5 = 5^5$.
This can be written as $2^0 \times 5^5$.
Since the denominator is in the form $2^n \times 5^m$ (where $n=0$ and $m=5$),the decimal expansion of $\frac{13}{3125}$ is terminating.

(N/A) Let us assume,to the contrary,that $\sqrt{2}$ is rational.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $\sqrt{2} = \frac{a}{b}$.
Squaring both sides,we get $2 = \frac{a^2}{b^2}$,which implies $a^2 = 2b^2$.
This shows that $a^2$ is divisible by $2$,and consequently,$a$ is also divisible by $2$ (by the Fundamental Theorem of Arithmetic).
So,we can write $a = 2k$ for some integer $k$.
Substituting this into the equation $a^2 = 2b^2$,we get $(2k)^2 = 2b^2$,which simplifies to $4k^2 = 2b^2$,or $b^2 = 2k^2$.
This shows that $b^2$ is divisible by $2$,and consequently,$b$ is also divisible by $2$.
Thus,both $a$ and $b$ have at least $2$ as a common factor,which contradicts our assumption that $a$ and $b$ are coprime.
Therefore,our assumption that $\sqrt{2}$ is rational is false,and $\sqrt{2}$ must be irrational.

(A) Let us assume,to the contrary,that $\sqrt{2}$ is rational.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $\sqrt{2} = \frac{a}{b}$.
Squaring both sides,we get $2 = \frac{a^2}{b^2}$,which implies $a^2 = 2b^2$.
This means $a^2$ is divisible by $2$,and consequently,$a$ is also divisible by $2$ (by the Fundamental Theorem of Arithmetic).
So,we can write $a = 2k$ for some integer $k$.
Substituting this into $a^2 = 2b^2$,we get $(2k)^2 = 2b^2$,which simplifies to $4k^2 = 2b^2$,or $b^2 = 2k^2$.
This implies $b^2$ is divisible by $2$,and consequently,$b$ is also divisible by $2$.
Thus,both $a$ and $b$ have at least $2$ as a common factor,which contradicts our assumption that $a$ and $b$ are coprime.
Therefore,our assumption is false,and $\sqrt{2}$ is irrational.

(N/A) Let us assume,to the contrary,that $3\sqrt{2}$ is a rational number.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $3\sqrt{2} = \frac{a}{b}$.
This can be rearranged as $\sqrt{2} = \frac{a}{3b}$.
Since $a$ and $b$ are integers,$\frac{a}{3b}$ is a rational number.
This implies that $\sqrt{2}$ is a rational number.
However,this contradicts the fact that $\sqrt{2}$ is an irrational number.
Therefore,our assumption that $3\sqrt{2}$ is a rational number is incorrect.
Hence,$3\sqrt{2}$ is an irrational number.

(B) Let us assume,to the contrary,that $5 - \sqrt{3}$ is a rational number.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $5 - \sqrt{3} = \frac{a}{b}$.
Rearranging the equation,we get $5 - \frac{a}{b} = \sqrt{3}$,which simplifies to $\frac{5b - a}{b} = \sqrt{3}$.
Since $a$ and $b$ are integers,$\frac{5b - a}{b}$ is a rational number.
This implies that $\sqrt{3}$ is a rational number.
However,this contradicts the fact that $\sqrt{3}$ is an irrational number.
Therefore,our assumption is incorrect,and $5 - \sqrt{3}$ must be an irrational number.

(N/A) Let us assume,to the contrary,that $\sqrt{2}$ is a rational number.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $\sqrt{2} = \frac{a}{b}$.
Squaring both sides,we get $2 = \frac{a^2}{b^2}$,which implies $a^2 = 2b^2$.
This means $a^2$ is divisible by $2$,so $a$ must also be divisible by $2$ (by the Fundamental Theorem of Arithmetic).
Let $a = 2k$ for some integer $k$.
Substituting this into $a^2 = 2b^2$,we get $(2k)^2 = 2b^2$,which simplifies to $4k^2 = 2b^2$,or $b^2 = 2k^2$.
This means $b^2$ is divisible by $2$,so $b$ must also be divisible by $2$.
Since both $a$ and $b$ are divisible by $2$,they have a common factor of $2$,which contradicts our assumption that $a$ and $b$ are coprime.
Therefore,our assumption is false,and $\sqrt{2}$ is an irrational number.

(A) To find the $HCF$ and $LCM$ of $6$,$72$,and $120$ using prime factorization:
Step $1$: Find the prime factors of each number.
$6 = 2^1 \times 3^1$
$72 = 2^3 \times 3^2$
$120 = 2^3 \times 3^1 \times 5^1$
Step $2$: The $HCF$ is the product of the smallest power of each common prime factor.
$HCF = 2^1 \times 3^1 = 6$
Step $3$: The $LCM$ is the product of the greatest power of each prime factor involved.
$LCM = 2^3 \times 3^2 \times 5^1 = 8 \times 9 \times 5 = 360$
Thus,the $HCF$ is $6$ and the $LCM$ is $360$.

(A) Step $1$: Find the prime factorization of $96$ and $404$.
$96 = 2^5 \times 3^1$
$404 = 2^2 \times 101^1$
Step $2$: Find the $HCF$ by taking the smallest power of each common prime factor.
$HCF(96, 404) = 2^2 = 4$.
Step $3$: Use the relationship $LCM(a, b) \times HCF(a, b) = a \times b$.
$LCM(96, 404) = \frac{96 \times 404}{HCF(96, 404)}$
$LCM(96, 404) = \frac{96 \times 404}{4} = 96 \times 101 = 9696$.

(A) Step $1$: Find the prime factorization of the given numbers.
$6 = 2 \times 3$
$20 = 2^2 \times 5$
Step $2$: To find the $HCF$,take the product of the smallest power of each common prime factor.
Common prime factor is $2$. The smallest power is $2^1$.
Therefore,$HCF$ = $2$.
Step $3$: To find the $LCM$,take the product of the highest power of each prime factor involved.
Prime factors involved are $2, 3,$ and $5$.
Highest powers are $2^2, 3^1,$ and $5^1$.
$LCM$ = $2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60$.
Thus,$HCF$ = $2$ and $LCM$ = $60$.

(B) For a number to end with the digit $0$,it must be divisible by $10$.
This means its prime factorization must contain both $2$ and $5$ as factors.
The prime factorization of $4^n$ is $(2^2)^n = 2^{2n}$.
Since the only prime factor of $4^n$ is $2$,it does not contain $5$ as a factor.
Therefore,$4^n$ can never end with the digit $0$ for any positive integer $n$.

(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 $75$.

(D) To find the value of $p(-2)$,substitute $x = -2$ into the given polynomial $p(x) = 2x^4 - 3x^3 + 7x + 5$.
$p(-2) = 2(-2)^4 - 3(-2)^3 + 7(-2) + 5$
Calculate the powers:
$(-2)^4 = 16$
$(-2)^3 = -8$
Substitute these values back into the expression:
$p(-2) = 2(16) - 3(-8) + 7(-2) + 5$
$p(-2) = 32 + 24 - 14 + 5$
$p(-2) = 56 - 14 + 5$
$p(-2) = 42 + 5$
$p(-2) = 47$

(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) 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.

(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 $D$.

(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.

(C) To find the degree of the polynomial $(x+1)(x^2 - x - x^4 + 1)$,we identify the highest power of $x$ in each factor.
The first factor is $(x+1)$,which has a degree of $1$.
The second factor is $(x^2 - x - x^4 + 1)$,which has a degree of $4$ (since the highest power of $x$ is $x^4$).
When multiplying two polynomials,the degree of the resulting polynomial is the sum of the degrees of the individual factors.
Therefore,the degree of the product is $1 + 4 = 5$.

(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 $-\frac{d}{a}$.

(A) Given the equation: $\frac{x^{3}-1}{p(x)} = \frac{x^{2}+x+1}{x-1}$.
We know the algebraic identity for the difference of cubes: $x^{3}-1 = (x-1)(x^{2}+x+1)$.
Substituting this into the equation: $\frac{(x-1)(x^{2}+x+1)}{p(x)} = \frac{x^{2}+x+1}{x-1}$.
By canceling $(x^{2}+x+1)$ from both sides (assuming $x^{2}+x+1 \neq 0$): $\frac{x-1}{p(x)} = \frac{1}{x-1}$.
Cross-multiplying gives: $p(x) = (x-1)(x-1) = (x-1)^{2}$.

(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 $12$.