Mix Examples - Polynomials Questions in English

(A) The given expression is $64 x^{3} + 125 y^{3} + 240 x^{2} y + 300 x y^{2}$.
We can rewrite this expression as $(4x)^{3} + (5y)^{3} + 3(4x)^{2}(5y) + 3(4x)(5y)^{2}$.
This is in the form of the algebraic identity $a^{3} + b^{3} + 3a^{2}b + 3ab^{2} = (a + b)^{3}$,where $a = 4x$ and $b = 5y$.
Substituting these values,we get $(4x + 5y)^{3}$.
Therefore,the factorised form is $(4x + 5y)(4x + 5y)(4x + 5y)$.

(A) The given expression is $27 x^{3}-8 y^{3}-54 x^{2} y+36 x y^{2}$.
This can be rewritten as $(3x)^3 - (2y)^3 - 3(3x)^2(2y) + 3(3x)(2y)^2$.
Recall the algebraic identity: $(a - b)^3 = a^3 - b^3 - 3a^2b + 3ab^2$.
Here,$a = 3x$ and $b = 2y$.
Substituting these values into the identity,we get $(3x - 2y)^3$.
Thus,the factorised form is $(3x - 2y)(3x - 2y)(3x - 2y)$.

(A) To evaluate $(995)^{3}$,we can express $995$ as $(1000 - 5)$.
Using the algebraic identity $(a - b)^{3} = a^{3} - b^{3} - 3ab(a - b)$:
Here,$a = 1000$ and $b = 5$.
$(1000 - 5)^{3} = (1000)^{3} - (5)^{3} - 3(1000)(5)(1000 - 5)$
$= 1,000,000,000 - 125 - 15000(995)$
$= 1,000,000,000 - 125 - 14,925,000$
$= 1,000,000,000 - 14,925,125$
$= 985,074,875$.

(B) To evaluate $(105)^{3}$,we can express $105$ as $(100 + 5)$.
Using the algebraic identity $(a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)$:
Here,$a = 100$ and $b = 5$.
$(100 + 5)^{3} = (100)^{3} + (5)^{3} + 3(100)(5)(100 + 5)$
$= 1000000 + 125 + 1500(105)$
$= 1000000 + 125 + 157500$
$= 1157625$

(N/A) The given expression is of the form $a^{3}+b^{3}+c^{3}-3 a b c$,where $a = 2x$,$b = 3y$,and $c = 5z$.
Using the algebraic identity $a^{3}+b^{3}+c^{3}-3 a b c = (a+b+c)(a^{2}+b^{2}+c^{2}-a b-b c-c a)$:
Substitute $a = 2x$,$b = 3y$,and $c = 5z$ into the identity:
$8 x^{3}+27 y^{3}+125 z^{3}-3(2x)(3y)(5z) = (2x+3y+5z)((2x)^{2}+(3y)^{2}+(5z)^{2}-(2x)(3y)-(3y)(5z)-(5z)(2x))$
$= (2x+3y+5z)(4x^{2}+9y^{2}+25z^{2}-6xy-15yz-10zx)$.

(A) We use the algebraic identity: $a^{3}+b^{3}+c^{3}-3 a b c = (a+b+c)(a^{2}+b^{2}+c^{2}-a b-b c-c a)$.
Given expression: $27 x^{3}-y^{3}+64 z^{3}+36 x y z = (3 x)^{3}+(-y)^{3}+(4 z)^{3}-3(3 x)(-y)(4 z)$.
Here,$a = 3 x$,$b = -y$,and $c = 4 z$.
Substituting these values into the identity:
$= (3 x-y+4 z)((3 x)^{2}+(-y)^{2}+(4 z)^{2}-(3 x)(-y)-(-y)(4 z)-(4 z)(3 x))$.
$= (3 x-y+4 z)(9 x^{2}+y^{2}+16 z^{2}+3 x y+4 y z-12 z x)$.

(B) The given expression is $x^{3}-8 y^{3}-27-18 x y$.
This can be rewritten as $x^{3} + (-2y)^{3} + (-3)^{3} - 3(x)(-2y)(-3)$.
We use the algebraic identity $a^{3} + b^{3} + c^{3} - 3abc = (a+b+c)(a^{2} + b^{2} + c^{2} - ab - bc - ca)$.
Here,$a = x$,$b = -2y$,and $c = -3$.
Substituting these values into the identity:
$= (x - 2y - 3)(x^{2} + (-2y)^{2} + (-3)^{2} - (x)(-2y) - (-2y)(-3) - (-3)(x))$
$= (x - 2y - 3)(x^{2} + 4y^{2} + 9 + 2xy - 6y + 3x)$.

(A) The given expression is of the form $a^{3} + b^{3} + c^{3} - 3abc$,where $a = 2x$,$b = 5y$,and $c = 7$.
We know the identity: $a^{3} + b^{3} + c^{3} - 3abc = (a + b + c)(a^{2} + b^{2} + c^{2} - ab - bc - ca)$.
Here,$3abc = 3(2x)(5y)(7) = 210xy$.
Substituting the values into the identity:
$8x^{3} + 125y^{3} + 343 - 210xy = (2x + 5y + 7)((2x)^{2} + (5y)^{2} + (7)^{2} - (2x)(5y) - (5y)(7) - (7)(2x))$.
Simplifying the terms:
$= (2x + 5y + 7)(4x^{2} + 25y^{2} + 49 - 10xy - 35y - 14x)$.

(A) We use the algebraic identity: If $a+b+c=0$,then $a^3+b^3+c^3=3abc$.
Let $a = (x-2y)$,$b = (2y-3z)$,and $c = (3z-x)$.
Now,calculate the sum $a+b+c$:
$a+b+c = (x-2y) + (2y-3z) + (3z-x)$
$a+b+c = x - x - 2y + 2y - 3z + 3z = 0$.
Since the sum is $0$,the expression $(x-2y)^3 + (2y-3z)^3 + (3z-x)^3$ simplifies to $3abc$.
Therefore,the factorised form is $3(x-2y)(2y-3z)(3z-x)$.

(D) We use the algebraic identity: If $x + y + z = 0$,then $x^3 + y^3 + z^3 = 3xyz$.
Given expression: $(31)^3 + (-16)^3 + (-15)^3$.
Let $x = 31$,$y = -16$,and $z = -15$.
Check the sum: $x + y + z = 31 + (-16) + (-15) = 31 - 31 = 0$.
Since the sum is $0$,the value of the expression is $3xyz$.
Value $= 3 \times (31) \times (-16) \times (-15)$.
Value $= 3 \times 31 \times 240$.
Value $= 93 \times 240 = 22320$.

(A) We know the algebraic identity: If $x + y + z = 0$,then $x^{3} + y^{3} + z^{3} = 3xyz$.
Given expression: $(14)^{3} + (27)^{3} - (41)^{3}$.
This can be rewritten as: $(14)^{3} + (27)^{3} + (-41)^{3}$.
Let $x = 14$,$y = 27$,and $z = -41$.
Check the sum: $x + y + z = 14 + 27 + (-41) = 41 - 41 = 0$.
Since the sum is $0$,the expression equals $3xyz$.
Value $= 3 \times (14) \times (27) \times (-41)$.
Value $= 3 \times 378 \times (-41) = 1134 \times (-41) = -46494$.

(B) Let $x = \frac{1}{2}$,$y = \frac{1}{3}$,and $z = -\frac{5}{6}$.
First,check the sum of $x, y,$ and $z$:
$x + y + z = \frac{1}{2} + \frac{1}{3} - \frac{5}{6} = \frac{3+2-5}{6} = \frac{0}{6} = 0$.
We know the algebraic identity: If $x + y + z = 0$,then $x^3 + y^3 + z^3 = 3xyz$.
Substituting the values:
$x^3 + y^3 + z^3 = 3 \times \left(\frac{1}{2}\right) \times \left(\frac{1}{3}\right) \times \left(-\frac{5}{6}\right)$.
$= 3 \times \left(\frac{1}{6}\right) \times \left(-\frac{5}{6}\right) = 1 \times \left(-\frac{5}{6}\right) = -\frac{5}{6}$.
Wait,re-calculating: $3 \times \frac{1}{2} \times \frac{1}{3} \times (-\frac{5}{6}) = 1 \times (-\frac{5}{6}) = -\frac{5}{6}$.
Correction: The expression is $3 \times \frac{1}{2} \times \frac{1}{3} \times (-\frac{5}{6}) = \frac{3}{6} \times (-\frac{5}{6}) = \frac{1}{2} \times (-\frac{5}{6}) = -\frac{5}{12}$.

(C) Let $a = 0.2$,$b = -0.3$,and $c = 0.1$.
We observe that $a + b + c = 0.2 + (-0.3) + 0.1 = 0.3 - 0.3 = 0$.
We know the algebraic identity: If $a + b + c = 0$,then $a^{3} + b^{3} + c^{3} = 3abc$.
Substituting the values:
$(0.2)^{3} + (-0.3)^{3} + (0.1)^{3} = 3(0.2)(-0.3)(0.1)$.
Calculating the product: $3 \times 0.2 = 0.6$; $0.6 \times (-0.3) = -0.18$; $-0.18 \times 0.1 = -0.018$.
Thus,$(0.2)^{3} - (0.3)^{3} + (0.1)^{3} = -0.018$.

(A) The area of a square is given by the formula: $\text{Area} = (\text{side})^2$.
Given, $\text{Area} = 9x^2 + 30x + 25$.
We can factorize this quadratic expression:
$9x^2 + 30x + 25 = (3x)^2 + 2(3x)(5) + (5)^2$.
Using the algebraic identity $(a + b)^2 = a^2 + 2ab + b^2$, where $a = 3x$ and $b = 5$, we get:
$9x^2 + 30x + 25 = (3x + 5)^2$.
Since $\text{Area} = (\text{side})^2$, the length of the side is $(3x + 5)$ units.

(A) The area of a rectangle is given by the product of its length and breadth: $\text{Area} = \text{Length} \times \text{Breadth}$.
Given area = $20 x^{2} + 22 x + 6$.
To find the length and breadth,we factorize the quadratic expression:
$20 x^{2} + 22 x + 6 = 20 x^{2} + 10 x + 12 x + 6$.
Grouping the terms: $(20 x^{2} + 10 x) + (12 x + 6)$.
Factor out common terms: $10 x(2 x + 1) + 6(2 x + 1)$.
Thus,the factors are $(10 x + 6)(2 x + 1)$ or $(5 x + 3)(4 x + 2)$.
Therefore,the possible expressions for length and breadth are $(5 x + 3)$ units and $(4 x + 2)$ units.

(A) The volume of a cuboid is given by the product of its dimensions: $V = \text{length} \times \text{breadth} \times \text{height}$.
Given $V = 2x^3 + 15x^2 + 33x + 20$.
To find the dimensions, we factorize the polynomial.
Let $p(x) = 2x^3 + 15x^2 + 33x + 20$.
By testing values, if $x = -1$, $p(-1) = 2(-1)^3 + 15(-1)^2 + 33(-1) + 20 = -2 + 15 - 33 + 20 = 0$.
Thus, $(x + 1)$ is a factor.
Dividing $2x^3 + 15x^2 + 33x + 20$ by $(x + 1)$ using synthetic division or long division gives $2x^2 + 13x + 20$.
Now, factorize the quadratic $2x^2 + 13x + 20$:
$2x^2 + 8x + 5x + 20 = 2x(x + 4) + 5(x + 4) = (2x + 5)(x + 4)$.
Therefore, the dimensions are $(2x + 5), (x + 4), (x + 1)$.

(C) Let the polynomial be $p(x) = x^{2} + 9x + 20$.
According to the Remainder Theorem,if $p(x)$ is divided by $(x + 2)$,the remainder is $p(-2)$.
Substitute $x = -2$ into the polynomial:
$p(-2) = (-2)^{2} + 9(-2) + 20$
$p(-2) = 4 - 18 + 20$
$p(-2) = 6$.
For the polynomial to be divisible by $(x + 2)$,the remainder must be $0$.
Therefore,we must subtract the remainder $6$ from the polynomial $p(x)$ so that the new polynomial becomes divisible by $(x + 2)$.

(D) Let the polynomial be $p(x) = x^{2} - 8x + 10$. Let the value to be added be $k$. The new polynomial is $g(x) = x^{2} - 8x + 10 + k$. For $g(x)$ to be divisible by $x - 3$,the remainder when $g(x)$ is divided by $x - 3$ must be $0$. According to the Remainder Theorem,the remainder is $g(3)$. Thus,$g(3) = 0$. Substituting $x = 3$ into $g(x)$: $(3)^{2} - 8(3) + 10 + k = 0$. $9 - 24 + 10 + k = 0$. $-15 + 10 + k = 0$. $-5 + k = 0$. Therefore,$k = 5$.

(A) Given the equation $x+y = -4$.
Cubing both sides,we get $(x+y)^3 = (-4)^3$.
Using the identity $(x+y)^3 = x^3 + y^3 + 3xy(x+y)$,we have $x^3 + y^3 + 3xy(x+y) = -64$.
Substitute the value of $(x+y) = -4$ into the equation:
$x^3 + y^3 + 3xy(-4) = -64$.
$x^3 + y^3 - 12xy = -64$.
Adding $64$ to both sides,we get $x^3 + y^3 - 12xy + 64 = -64 + 64$.
Therefore,$x^3 + y^3 - 12xy + 64 = 0$.

(B) Given the equation $x = 2y + 6$,we can rewrite it as $x - 2y = 6$.
To find the value of $x^3 - 8y^3 - 36xy - 216$,we can express it as $x^3 + (-2y)^3 + (-6)^3 - 3(x)(-2y)(-6)$.
Recall the algebraic identity: $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$.
Alternatively,consider the identity $(a + b + c)^3 = a^3 + b^3 + c^3 + 3(a + b)(b + c)(c + a)$.
Let $a = x$,$b = -2y$,and $c = -6$.
Then $a + b + c = x - 2y - 6$. Since $x - 2y = 6$,we have $a + b + c = 6 - 6 = 0$.
If $a + b + c = 0$,then $a^3 + b^3 + c^3 = 3abc$.
Substituting the values: $x^3 + (-2y)^3 + (-6)^3 = 3(x)(-2y)(-6)$.
$x^3 - 8y^3 - 216 = 36xy$.
Rearranging the terms: $x^3 - 8y^3 - 36xy - 216 = 0$.

(C) Let $p(x) = ax^3 + 4x^2 + 3x - 4$ and $q(x) = x^3 - 4x + a$.
According to the Remainder Theorem,when a polynomial $f(x)$ is divided by $x - c$,the remainder is $f(c)$.
Here,$c = 3$. Therefore,the remainder for $p(x)$ is $p(3)$ and for $q(x)$ is $q(3)$.
$p(3) = a(3)^3 + 4(3)^2 + 3(3) - 4 = 27a + 36 + 9 - 4 = 27a + 41$.
$q(3) = (3)^3 - 4(3) + a = 27 - 12 + a = 15 + a$.
Since the remainders are equal,$p(3) = q(3)$.
$27a + 41 = 15 + a$.
$27a - a = 15 - 41$.
$26a = -26$.
$a = -1$.

(A) According to the Remainder Theorem,if $p(x)$ is divided by $(x + 1)$,the remainder is $p(-1)$.
Given $p(-1) = 19$.
$p(-1) = (-1)^{4} - 2(-1)^{3} + 3(-1)^{2} - a(-1) + 3a - 7 = 19$
$1 + 2 + 3 + a + 3a - 7 = 19$
$4a - 1 = 19$
$4a = 20 \implies a = 5$.
Now,$p(x) = x^{4} - 2x^{3} + 3x^{2} - 5x + 3(5) - 7 = x^{4} - 2x^{3} + 3x^{2} - 5x + 8$.
To find the remainder when $p(x)$ is divided by $(x + 2)$,we calculate $p(-2)$.
$p(-2) = (-2)^{4} - 2(-2)^{3} + 3(-2)^{2} - 5(-2) + 8$
$p(-2) = 16 - 2(-8) + 3(4) + 10 + 8$
$p(-2) = 16 + 16 + 12 + 10 + 8 = 62$.
Thus,$a = 5$ and the remainder is $62$.

(B) The given polynomial is $p(x) = x^{2}-5x+4$.
$A$ polynomial is called a linear polynomial if its degree (the highest power of the variable) is $1$.
In the given expression $x^{2}-5x+4$,the highest power of the variable $x$ is $2$.
Since the degree of the polynomial is $2$,it is a quadratic polynomial,not a linear polynomial.
Therefore,the statement is False.

(B) The degree of a polynomial is defined as the highest power of the variable present in the expression.
In the polynomial $5x^{2} - 7x + 2$,the powers of the variable $x$ are $2$,$1$,and $0$ (since $2 = 2x^{0}$).
The highest power among these is $2$.
Therefore,the degree of the polynomial $5x^{2} - 7x + 2$ is $2$,not $5$.
Hence,the given statement is False.

(A) To verify the statement,we expand the right-hand side:
$(x-6)(x-2) = x(x-2) - 6(x-2)$
$= x^{2} - 2x - 6x + 12$
$= x^{2} - 8x + 12$
Since the expanded form matches the left-hand side,the statement is True.

(B) To find the zero of the polynomial $p(x) = 2x + 3$,we set $p(x) = 0$.
$2x + 3 = 0$
$2x = -3$
$x = -\frac{3}{2}$
Since the calculated zero is $-\frac{3}{2}$ and not $\frac{3}{2}$,the given statement is False.

(B) The given statement is False.
In the polynomial $5x^{3} - 3x^{2} + 11x - 14$,the term containing $x^{3}$ is $5x^{3}$.
The coefficient of $x^{3}$ is the numerical factor multiplying $x^{3}$,which is $5$,not $3$.

(B) To find $p(-2)$,substitute $x = -2$ into the polynomial $p(x) = x^{3} + 9x^{2} + 26x + 24$.
$p(-2) = (-2)^{3} + 9(-2)^{2} + 26(-2) + 24$
$p(-2) = -8 + 9(4) - 52 + 24$
$p(-2) = -8 + 36 - 52 + 24$
$p(-2) = (-8 - 52) + (36 + 24)$
$p(-2) = -60 + 60$
$p(-2) = 0$

(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 $p(x).$

(D) To find the zeros of the polynomial $p(x) = x^{3} + 7x^{2} + 11x + 5$,we test the given options by substituting them into the polynomial.
If $x = -1$:
$p(-1) = (-1)^{3} + 7(-1)^{2} + 11(-1) + 5$
$p(-1) = -1 + 7(1) - 11 + 5$
$p(-1) = -1 + 7 - 11 + 5$
$p(-1) = 12 - 12 = 0$
Since $p(-1) = 0$,$x = -1$ is a zero of the polynomial $p(x)$.

(A) To find the remainder when $p(x) = x^{3} + 125$ is divided by $(x - 5)$,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,$a = 5$.
So,the remainder is $p(5) = (5)^{3} + 125$.
$p(5) = 125 + 125 = 250$.
Therefore,the remainder is $250$.

(B) To calculate $85 \times 75$,we can use the algebraic identity $(a - b)(a + b) = a^2 - b^2$.
We can write $85$ as $(80 + 5)$ and $75$ as $(80 - 5)$.
Thus,$85 \times 75 = (80 + 5)(80 - 5)$.
Using the identity $a^2 - b^2$,where $a = 80$ and $b = 5$:
$= 80^2 - 5^2$
$= 6400 - 25$
$= 6375$.

(C) To factorize the quadratic expression $4x^{2} - 20x + 25$,we observe that it follows the algebraic identity $(a - b)^{2} = a^{2} - 2ab + b^{2}$.
Here,$a^{2} = 4x^{2} \implies a = 2x$.
Also,$b^{2} = 25 \implies b = 5$.
Checking the middle term: $-2ab = -2(2x)(5) = -20x$,which matches the given expression.
Therefore,$4x^{2} - 20x + 25 = (2x - 5)^{2}$.

(D) Given the quadratic expression $x^{2}-10x+21 = (x+m)(x+n)$.
Expanding the right side,we get $(x+m)(x+n) = x^{2} + nx + mx + mn = x^{2} + (m+n)x + mn$.
Comparing the coefficients of $x$ and the constant term with the original expression $x^{2}-10x+21$,we get:
$m+n = -10$
$mn = 21$
Therefore,the value of $m+n$ is $-10$.

(A) Let $p(x) = x^{3}-3x^{2}+ax+24$.
Since $(x-2)$ is a factor of $p(x)$,by the Factor Theorem,$p(2) = 0$.
Substituting $x = 2$ into the polynomial:
$p(2) = (2)^{3} - 3(2)^{2} + a(2) + 24 = 0$.
$8 - 3(4) + 2a + 24 = 0$.
$8 - 12 + 2a + 24 = 0$.
$20 + 2a = 0$.
$2a = -20$.
$a = -10$.

(B) polynomial is classified based on its degree (the highest power of the variable).
For the expression $4 x^{2}+11 x-3$,the highest power of the variable $x$ is $2$.
$A$ polynomial of degree $2$ is called a quadratic polynomial.
Therefore,$4 x^{2}+11 x-3$ is a quadratic polynomial.

(C) To factorize the quadratic polynomial $x^{2}-23x+120$,we need to find two numbers whose sum is $-23$ and whose product is $120$.
Let the two numbers be $a$ and $b$.
We have $a+b = -23$ and $ab = 120$.
By testing factors of $120$,we find that $-15$ and $-8$ satisfy these conditions because $(-15) + (-8) = -23$ and $(-15) \times (-8) = 120$.
Now,rewrite the middle term $-23x$ as $-15x - 8x$:
$x^{2}-15x-8x+120$
Group the terms:
$(x^{2}-15x) - (8x-120)$
Factor out the common terms:
$x(x-15) - 8(x-15)$
$(x-15)(x-8)$
Thus,the factors are $(x-15)(x-8)$.

(D) To find the factor of the polynomial $p(x) = x^3 - 3x^2 + 7x - 5$,we use the Factor Theorem.
According to the Factor Theorem,$(x-a)$ is a factor of $p(x)$ if $p(a) = 0$.
Let us test the given options:
For option $D$,$x-1 = 0$ implies $x = 1$.
$p(1) = (1)^3 - 3(1)^2 + 7(1) - 5$
$p(1) = 1 - 3 + 7 - 5$
$p(1) = 8 - 8 = 0$.
Since $p(1) = 0$,$(x-1)$ is a factor of the polynomial $p(x)$.

(A) According to the Remainder Theorem,if a polynomial $p(x)$ is divided by $(x - a)$,the remainder is $p(a)$.
Here,we are dividing $p(x) = x^{2} + 12x + 36$ by $(x + 5)$,which is equivalent to $(x - (-5))$.
Therefore,the remainder is $p(-5)$.
Substitute $x = -5$ into the polynomial:
$p(-5) = (-5)^{2} + 12(-5) + 36$
$p(-5) = 25 - 60 + 36$
$p(-5) = 61 - 60$
$p(-5) = 1$
Thus,the remainder is $1$.

(B) Given the quadratic expression $x^{2}-8x-20 = (x+a)(x+b)$.
Expanding the right side,we get $(x+a)(x+b) = x^{2} + (a+b)x + ab$.
Comparing this with the given expression $x^{2}-8x-20$,we equate the constant terms.
The constant term on the left side is $-20$.
The constant term on the right side is $ab$.
Therefore,$ab = -20$.

(C) To find the zeros of the polynomial $p(x) = x^{3}-3x^{2}-5x+15$,we set $p(x) = 0$.
Factor by grouping:
$x^{2}(x-3) - 5(x-3) = 0$
$(x^{2}-5)(x-3) = 0$
The zeros are found by setting each factor to zero:
$x-3 = 0 \implies x = 3$
$x^{2}-5 = 0 \implies x^{2} = 5 \implies x = \pm\sqrt{5}$
Comparing these values with the given options,$3$ is one of the zeros.

(D) Let $p(x) = x^{3} + 12x^{2} + ax + 60$.
Since $(x+3)$ is a factor of $p(x)$,by the Factor Theorem,$p(-3) = 0$.
Substituting $x = -3$ into the polynomial:
$(-3)^{3} + 12(-3)^{2} + a(-3) + 60 = 0$
$-27 + 12(9) - 3a + 60 = 0$
$-27 + 108 - 3a + 60 = 0$
$141 - 3a = 0$
$3a = 141$
$a = 47$.

(A) The degree of a polynomial is defined as the highest power of the variable present in the expression.
In the given polynomial $5x^{2} - 7x - 11$,the powers of the variable $x$ are $2$,$1$,and $0$ (since $11 = 11x^{0}$).
The highest power among these is $2$.
Therefore,the degree of the polynomial is $2$.

(B) To find the zero of the polynomial $p(x) = 5x - 10$,we set $p(x) = 0$.
$5x - 10 = 0$
$5x = 10$
$x = \frac{10}{5}$
$x = 2$
Therefore,the zero of the polynomial is $2$.

(B) According to the Factor Theorem,if $p(a) = 0$,then $(x - a)$ is a factor of the polynomial $p(x)$.
Given that $p(-3) = 0$,we substitute $a = -3$ into the expression $(x - a)$.
This gives us $(x - (-3)) = x + 3$.
Therefore,$(x + 3)$ is a factor of $p(x)$.

(B) The given expression is in the form of the algebraic identity $(a+b)(a-b) = a^{2}-b^{2}$.
Here,$a = 5x$ and $b = 3$.
Applying the identity:
$(5x+3)(5x-3) = (5x)^{2} - (3)^{2}$
$= 25x^{2} - 9$.

(A) The given polynomial is $p(x) = 5x^{3} - 3x^{2} + 11x - 4$.
To find the coefficient of $x^{2}$,we look at the term containing $x^{2}$,which is $-3x^{2}$.
The numerical factor multiplying $x^{2}$ is $-3$.
Therefore,the coefficient of $x^{2}$ is $-3$.

(A) Given the polynomial $p(x) = x^{3} - 3x^{2} + 8x + 12$.
To find $p(-1)$,substitute $x = -1$ into the polynomial:
$p(-1) = (-1)^{3} - 3(-1)^{2} + 8(-1) + 12$
$p(-1) = -1 - 3(1) - 8 + 12$
$p(-1) = -1 - 3 - 8 + 12$
$p(-1) = -12 + 12$
$p(-1) = 0$
Thus,the correct option is $A$.

(C) To find the degree of the polynomial,we first simplify the expression:
$7x^5 - 4x^4 + 2(x^3)^2 - x^2 + 35$
$= 7x^5 - 4x^4 + 2x^6 - x^2 + 35$
The degree of a polynomial is the highest power of the variable $x$ present in the expression.
In the simplified expression $2x^6 + 7x^5 - 4x^4 - x^2 + 35$,the highest power of $x$ is $6$.
Therefore,the degree of the polynomial is $6$.

(A) To find the zero of the polynomial $p(x) = bx + m$,we set $p(x) = 0$.
$bx + m = 0$
$bx = -m$
$x = -\frac{m}{b}$
Therefore,the zero of the polynomial is $-\frac{m}{b}$.