Textbook - Polynomials Questions in English

(A) To evaluate $(102)^{3}$,we use the algebraic identity $(a+b)^{3} = a^{3} + b^{3} + 3ab(a+b)$.
Here,we can write $102$ as $(100 + 2)$.
Substituting $a = 100$ and $b = 2$ into the identity:
$(100 + 2)^{3} = (100)^{3} + (2)^{3} + 3(100)(2)(100 + 2)$
Calculating each term:
$(100)^{3} = 1000000$
$(2)^{3} = 8$
$3(100)(2) = 600$
Now,substitute these values back:
$1000000 + 8 + 600(102)$
$= 1000000 + 8 + 61200$
$= 1061208$

(B) We can write $998$ as $(1000 - 2)$.
Using the algebraic identity $(a - b)^{3} = a^{3} - b^{3} - 3ab(a - b)$,where $a = 1000$ and $b = 2$:
$(998)^{3} = (1000 - 2)^{3}$
$= (1000)^{3} - (2)^{3} - 3(1000)(2)(1000 - 2)$
$= 1,000,000,000 - 8 - 6000(998)$
$= 1,000,000,000 - 8 - 5,988,000$
$= 994,011,992$

(A) The given expression is $8 a^{3}+b^{3}+12 a^{2} b+6 a b^{2}$.
We can rewrite the expression as $(2 a)^{3}+(b)^{3}+3(2 a)^{2}(b)+3(2 a)(b)^{2}$.
This is in the form of the algebraic identity $(x+y)^{3} = x^{3}+y^{3}+3 x^{2} y+3 x y^{2}$,where $x = 2a$ and $y = b$.
Therefore,the expression simplifies to $(2 a+b)^{3}$.
Thus,the factorised form is $(2 a+b)(2 a+b)(2 a+b)$.

(A) The given expression is $8a^3 - b^3 - 12a^2b + 6ab^2$.
We can rewrite this as $(2a)^3 - (b)^3 - 3(2a)^2(b) + 3(2a)(b)^2$.
This expression is in the form of the algebraic identity $(x - y)^3 = x^3 - y^3 - 3x^2y + 3xy^2$,where $x = 2a$ and $y = b$.
Therefore,$8a^3 - b^3 - 12a^2b + 6ab^2 = (2a - b)^3$.
This can be written as $(2a - b)(2a - b)(2a - b)$.

(A) The given expression is $27-125 a^{3}-135 a+225 a^{2}$.
We can rewrite this as $(3)^{3}-(5 a)^{3}-3(3)^{2}(5 a)+3(3)(5 a)^{2}$.
This expression follows the algebraic identity $(x-y)^{3} = x^{3}-y^{3}-3x^{2}y+3xy^{2}$.
Here,$x = 3$ and $y = 5a$.
Therefore,$27-125 a^{3}-135 a+225 a^{2} = (3-5a)^{3}$.
This can be written as $(3-5a)(3-5a)(3-5a)$.

(A) The given expression is $64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}$.
This can be rewritten as $(4 a)^{3}-(3 b)^{3}-3(4 a)^{2}(3 b)+3(4 a)(3 b)^{2}$.
This expression follows the algebraic identity $(x-y)^{3} = x^{3}-y^{3}-3x^{2}y+3xy^{2}$,where $x = 4a$ and $y = 3b$.
Therefore,the expression simplifies to $(4 a-3 b)^{3}$.
Thus,the factors are $(4 a-3 b)(4 a-3 b)(4 a-3 b)$.

(A) The given expression is $27 p^{3} - \frac{1}{216} - \frac{9}{2} p^{2} + \frac{1}{4} p$.
We can rewrite this expression as:
$(3p)^{3} - (\frac{1}{6})^{3} - 3(3p)(\frac{1}{6})(3p - \frac{1}{6})$.
This is in the form of the algebraic identity $x^{3} - y^{3} - 3xy(x - y) = (x - y)^{3}$,where $x = 3p$ and $y = \frac{1}{6}$.
Therefore,the expression simplifies to $(3p - \frac{1}{6})^{3}$.
This can be written as $(3p - \frac{1}{6})(3p - \frac{1}{6})(3p - \frac{1}{6})$.

(N/A) To verify the identity,we expand the right-hand side ($R$.$H$.$S$.).
$R.H.S. = (x+y)(x^{2}-xy+y^{2})$
$= x(x^{2}-xy+y^{2}) + y(x^{2}-xy+y^{2})$
$= (x^{3} - x^{2}y + xy^{2}) + (x^{2}y - xy^{2} + y^{3})$
By grouping the like terms,we get:
$= x^{3} + (-x^{2}y + x^{2}y) + (xy^{2} - xy^{2}) + y^{3}$
$= x^{3} + 0 + 0 + y^{3}$
$= x^{3} + y^{3} = L.H.S.$
Hence,the identity is verified.

(N/A) To verify the identity,we start with the Right Hand Side ($R$.$H$.$S$.):
$R.H.S. = (x-y)(x^{2}+xy+y^{2})$
Distribute the terms:
$= x(x^{2}+xy+y^{2}) - y(x^{2}+xy+y^{2})$
$= (x^{3} + x^{2}y + xy^{2}) - (x^{2}y + xy^{2} + y^{3})$
$= x^{3} + x^{2}y + xy^{2} - x^{2}y - xy^{2} - y^{3}$
Cancel the like terms with opposite signs ($x^{2}y - x^{2}y = 0$ and $xy^{2} - xy^{2} = 0$):
$= x^{3} - y^{3}$
$= L.H.S.$
Since $L.H.S. = R.H.S.$,the identity is verified.

(A) Using the algebraic identity $a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})$,we can factorise the expression.
Here,$27 y^{3}+125 z^{3} = (3 y)^{3}+(5 z)^{3}$.
Comparing this with $a^{3}+b^{3}$,we get $a = 3y$ and $b = 5z$.
Substituting these values into the identity:
$(3 y)^{3}+(5 z)^{3} = (3 y+5 z)((3 y)^{2}-(3 y)(5 z)+(5 z)^{2})$
$= (3 y+5 z)(9 y^{2}-15 y z+25 z^{2})$.

(A) Using the algebraic identity $x^{3} - y^{3} = (x - y)(x^{2} + xy + y^{2})$,we can factorise the expression.
Given expression: $64m^{3} - 343n^{3}$
This can be written as: $(4m)^{3} - (7n)^{3}$
Applying the identity where $x = 4m$ and $y = 7n$:
$(4m - 7n)((4m)^{2} + (4m)(7n) + (7n)^{2})$
$= (4m - 7n)(16m^{2} + 28mn + 49n^{2})$

(A) The algebraic identity is: $x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$.
We can rewrite the given expression as:
$27x^3 + y^3 + z^3 - 9xyz = (3x)^3 + (y)^3 + (z)^3 - 3(3x)(y)(z)$.
Comparing this with the identity,where $x$ is replaced by $3x$,$y$ by $y$,and $z$ by $z$,we get:
$= (3x + y + z)((3x)^2 + y^2 + z^2 - (3x)(y) - (y)(z) - (z)(3x))$.
Simplifying the terms inside the second bracket:
$= (3x + y + z)(9x^2 + y^2 + z^2 - 3xy - yz - 3zx)$.

(N/A) Start with the Right Hand Side ($R$.$H$.$S$.):
$\text{R.H.S.} = \frac{1}{2}(x+y+z)[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}]$
Expand the squares inside the bracket:
$= \frac{1}{2}(x+y+z)[(x^{2}-2xy+y^{2})+(y^{2}-2yz+z^{2})+(z^{2}-2zx+x^{2})]$
Combine like terms:
$= \frac{1}{2}(x+y+z)[2x^{2}+2y^{2}+2z^{2}-2xy-2yz-2zx]$
Factor out $2$ from the expression inside the square bracket:
$= \frac{1}{2}(x+y+z) \cdot 2[x^{2}+y^{2}+z^{2}-xy-yz-zx]$
$= (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$
Using the algebraic identity $x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$,we get:
$= x^{3}+y^{3}+z^{3}-3xyz = \text{L.H.S.}$

(N/A) Given that $x+y+z=0.$
Therefore,$x+y=-z.$
Cubing both sides,we get $(x+y)^{3}=(-z)^{3}.$
Using the identity $(x+y)^{3} = x^{3}+y^{3}+3xy(x+y),$ we have:
$x^{3}+y^{3}+3xy(x+y) = -z^{3}.$
Since $x+y=-z,$ substitute this into the equation:
$x^{3}+y^{3}+3xy(-z) = -z^{3}.$
$x^{3}+y^{3}-3xyz = -z^{3}.$
Rearranging the terms,we get:
$x^{3}+y^{3}+z^{3} = 3xyz.$
Hence,if $x+y+z=0,$ then $x^{3}+y^{3}+z^{3}=3xyz.$

(C) Let $x = -12, y = 7,$ and $z = 5$.
Then,$x + y + z = -12 + 7 + 5 = 0$.
We know that if $x + y + z = 0$,then $x^{3} + y^{3} + z^{3} = 3xyz$.
Therefore,$(-12)^{3} + (7)^{3} + (5)^{3} = 3[(-12)(7)(5)]$,since $(-12) + 7 + 5 = 0$.
Calculating the product: $3[-420] = -1260$.
Thus,$(-12)^{3} + (7)^{3} + (5)^{3} = -1260$.

(D) Let $x = 28$,$y = -15$,and $z = -13$.
First,calculate the sum of $x, y,$ and $z$:
$x + y + z = 28 + (-15) + (-13) = 28 - 28 = 0$.
We know the algebraic identity: If $x + y + z = 0$,then $x^{3} + y^{3} + z^{3} = 3xyz$.
Substituting the values into the identity:
$(28)^{3} + (-15)^{3} + (-13)^{3} = 3(28)(-15)(-13)$.
Calculating the product:
$3 \times 28 = 84$
$(-15) \times (-13) = 195$
$84 \times 195 = 16380$.
Therefore,the value is $16380$.

(A) The area of a rectangle is given by the formula: $\text{Area} = \text{Length} \times \text{Breadth}$.
Given the area: $25a^2 - 35a + 12$.
To find the length and breadth,we need to factorise the quadratic polynomial $25a^2 - 35a + 12$ by splitting the middle term.
We look for two numbers whose product is $25 \times 12 = 300$ and whose sum is $-35$.
These two numbers are $-20$ and $-15$,since $(-20) \times (-15) = 300$ and $(-20) + (-15) = -35$.
Now,rewrite the expression:
$25a^2 - 20a - 15a + 12$
Factor by grouping:
$= 5a(5a - 4) - 3(5a - 4)$
$= (5a - 4)(5a - 3)$
Thus,the possible expressions for the length and breadth are $(5a - 3)$ and $(5a - 4)$.

(A) The area of a rectangle is given by the formula: $\text{Area} = \text{Length} \times \text{Breadth}$.
Given the area is $35y^2 + 13y - 12$,we need to factorise this quadratic polynomial.
To factorise $35y^2 + 13y - 12$,we split the middle term $13y$ into two parts such that their sum is $13y$ and their product is equal to the product of the coefficient of $y^2$ and the constant term $(35 \times -12 = -420)$.
We find that $28y$ and $-15y$ satisfy these conditions because $28y - 15y = 13y$ and $28y \times (-15y) = -420y^2$.
Now,rewrite the expression:
$35y^2 + 28y - 15y - 12$
Group the terms:
$(35y^2 + 28y) - (15y + 12)$
Factor out the common terms:
$7y(5y + 4) - 3(5y + 4)$
$(7y - 3)(5y + 4)$
Thus,the possible expressions for the length and breadth are $(7y - 3)$ and $(5y + 4)$.

(N/A) The volume of a cuboid is given by the formula: $\text{Volume} = \text{Length} \times \text{Breadth} \times \text{Height}$.
Given volume $= 3x^2 - 12x$.
To find the possible dimensions,we factorize the expression:
$3x^2 - 12x = 3x(x - 4)$.
Thus,the expression can be written as the product of three factors: $3$,$x$,and $(x - 4)$.
Therefore,the possible dimensions of the cuboid are $3$,$x$,and $(x - 4)$ units.

(A) The volume of a cuboid is given by the formula: $\text{Volume} = \text{Length} \times \text{Breadth} \times \text{Height}$.
Given volume $= 12ky^{2} + 8ky - 20k$.
First,factor out the common term $4k$:
$12ky^{2} + 8ky - 20k = 4k(3y^{2} + 2y - 5)$.
Next,factor the quadratic expression $(3y^{2} + 2y - 5)$ by splitting the middle term:
$3y^{2} + 2y - 5 = 3y^{2} + 5y - 3y - 5$
$= y(3y + 5) - 1(3y + 5)$
$= (3y + 5)(y - 1)$.
Combining these,the volume is $4k \times (3y + 5) \times (y - 1)$.
Therefore,the possible dimensions of the cuboid are $4k$,$(3y + 5)$,and $(y - 1)$ units.

(A) The given expression is $y^{2} + \sqrt{2}$.
This can be written as $y^{2} + \sqrt{2}y^{0}$.
$A$ polynomial in one variable is an algebraic expression where the exponent of the variable is a non-negative integer (a whole number).
In the expression $y^{2} + \sqrt{2}$,the exponent of the variable $y$ is $2$,which is a whole number.
Therefore,$y^{2} + \sqrt{2}$ is a polynomial in one variable.

(D) The given expression is $3 \sqrt{t} + t \sqrt{2}$.
This can be rewritten as $3 t^{1/2} + \sqrt{2} \cdot t$.
$A$ polynomial in one variable is an algebraic expression where the exponent of the variable must be a non-negative integer (a whole number).
In the term $3 t^{1/2}$,the exponent of the variable $t$ is $\frac{1}{2}$,which is not a whole number.
Therefore,$3 \sqrt{t} + t \sqrt{2}$ is not a polynomial.

(D) The given expression is $y + \frac{2}{y}$.
This can be rewritten as $y + 2 \cdot y^{-1}$.
$A$ polynomial in one variable is an algebraic expression where the exponent of the variable must be a non-negative integer (a whole number).
In the term $2 \cdot y^{-1}$,the exponent of the variable $y$ is $-1$.
Since $-1$ is not a whole number,the expression $y + \frac{2}{y}$ is not a polynomial.

(D) The given expression is $x^{10} + y^{3} + t^{50}$.
$1$. $A$ polynomial in one variable is an algebraic expression where the exponent of the variable is a non-negative integer and there is only one variable present.
$2$. In the expression $x^{10} + y^{3} + t^{50}$,there are three distinct variables: $x$,$y$,and $t$.
$3$. Although the exponent of each variable $(10, 3, 50)$ is a whole number,the expression contains three variables.
$4$. Therefore,$x^{10} + y^{3} + t^{50}$ is a polynomial in three variables,not in one variable.

(A) $(i)$ In the expression $\frac{\pi}{2} x^2 + x$,the term containing $x^2$ is $\frac{\pi}{2} x^2$. Therefore,the coefficient of $x^2$ is $\frac{\pi}{2}$.
$(ii)$ In the expression $\sqrt{2} x - 1$,there is no $x^2$ term. We can write this expression as $\sqrt{2} x - 1 + 0 \cdot x^2$. Therefore,the coefficient of $x^2$ is $0$.

(B) The degree of a polynomial is defined as the highest power of the variable present in the polynomial.
$(i)$ $5t - \sqrt{7}$
This is a polynomial in the variable $t$. The highest power of the variable $t$ is $1$. Therefore,the degree of this polynomial is $1$.
$(ii)$ $3$
This is a constant polynomial. The degree of any non-zero constant polynomial is always $0$,because $3$ can be written as $3 \times t^0$.

(N/A) $(i)$ $1+x$
Since the degree of $1+x$ is $1$,it is a linear polynomial.
$(ii)$ $3t$
Since the degree of $3t$ is $1$,it is a linear polynomial.
$(iii)$ $r^{2}$
Since the degree of $r^{2}$ is $2$,it is a quadratic polynomial.
$(iv)$ $7x^{3}$
Since the degree of $7x^{3}$ is $3$,it is a cubic polynomial.

$(16+\sqrt{11})$ Given the polynomial $q(y) = 3y^3 - 4y + \sqrt{11}$.
To find the value of the polynomial at $y = 2$,we substitute $2$ for $y$ in the expression:
$q(2) = 3(2)^3 - 4(2) + \sqrt{11}$
Calculate the power: $2^3 = 8$.
Substitute back: $q(2) = 3(8) - 8 + \sqrt{11}$.
Perform multiplication: $q(2) = 24 - 8 + \sqrt{11}$.
Perform subtraction: $q(2) = 16 + \sqrt{11}$.

(N/A) Given the polynomial $p(t) = 4t^4 + 5t^3 - t^2 + 6$.
To find the value of the polynomial at $t = a$,we substitute $a$ for $t$ in the expression.
$p(a) = 4(a)^4 + 5(a)^3 - (a)^2 + 6$.
Therefore,the value of the polynomial at $t = a$ is $4a^4 + 5a^3 - a^2 + 6$.

(B) Let the polynomial be $p(x) = 5x - 4x^2 + 3$.
To find the value at $x = -1$,substitute $-1$ for $x$ in the polynomial:
$p(-1) = 5(-1) - 4(-1)^2 + 3$
$p(-1) = -5 - 4(1) + 3$
$p(-1) = -5 - 4 + 3$
$p(-1) = -9 + 3$
$p(-1) = -6$

(C) Given polynomial is $p(x) = 5x - 4x^2 + 3$.
To find the value of the polynomial at $x = 2$,substitute $2$ for $x$ in the expression:
$p(2) = 5(2) - 4(2)^2 + 3$
$p(2) = 10 - 4(4) + 3$
$p(2) = 10 - 16 + 3$
$p(2) = -6 + 3$
$p(2) = -3$
Therefore,the value of the polynomial at $x = 2$ is $-3$.

(B) According to the Factor Theorem,$(x + 1)$ is a factor of $p(x)$ if and only if $p(-1) = 0$.
Given $p(x) = x^{4} + x^{3} + x^{2} + x + 1$.
Substituting $x = -1$ into the polynomial:
$p(-1) = (-1)^{4} + (-1)^{3} + (-1)^{2} + (-1) + 1$
Calculating each term:
$p(-1) = 1 - 1 + 1 - 1 + 1$
Simplifying the expression:
$p(-1) = 1$
Since $p(-1) \neq 0$,by the Factor Theorem,$(x + 1)$ is not a factor of the given polynomial.

(NONE) According to the Factor Theorem,$(x + 1)$ is a factor of a polynomial $p(x)$ if and only if $p(-1) = 0$.
Let $p(x) = x^{4} + 3x^{3} + 3x^{2} + x + 1$.
Substitute $x = -1$ into the polynomial:
$p(-1) = (-1)^{4} + 3(-1)^{3} + 3(-1)^{2} + (-1) + 1$
$p(-1) = 1 + 3(-1) + 3(1) - 1 + 1$
$p(-1) = 1 - 3 + 3 - 1 + 1$
$p(-1) = 1$
Since $p(-1) = 1 \neq 0$,by the Factor Theorem,$(x + 1)$ is not a factor of the given polynomial.

(NONE) According to the Factor Theorem, $(x + 1)$ is a factor of a polynomial $p(x)$ if $p(-1) = 0$.
Let $p(x) = x^{3} - x^{2} - (2 + \sqrt{2})x + \sqrt{2}$.
Substitute $x = -1$ into the polynomial:
$p(-1) = (-1)^{3} - (-1)^{2} - (2 + \sqrt{2})(-1) + \sqrt{2}$
Calculate each term:
$p(-1) = -1 - 1 + (2 + \sqrt{2}) + \sqrt{2}$
$p(-1) = -2 + 2 + \sqrt{2} + \sqrt{2}$
$p(-1) = 2\sqrt{2}$
Since $p(-1) = 2\sqrt{2} \neq 0$, $(x + 1)$ is not a factor of the given polynomial.

(D) According to the Factor Theorem,$g(x) = x - a$ is a factor of $p(x)$ if and only if $p(a) = 0$.
Here,$g(x) = x + 2$,so we set $x + 2 = 0$,which gives $x = -2$.
Now,we evaluate $p(-2)$:
$p(-2) = (-2)^3 + 3(-2)^2 + 3(-2) + 1$
$p(-2) = -8 + 3(4) - 6 + 1$
$p(-2) = -8 + 12 - 6 + 1$
$p(-2) = 4 - 6 + 1 = -1$
Since $p(-2) \neq 0$,by the Factor Theorem,$g(x)$ is not a factor of $p(x)$.

(A) According to the Factor Theorem,$g(x) = x - a$ is a factor of $p(x)$ if $p(a) = 0$.
Here,$g(x) = x - 3$,so $a = 3$.
We calculate $p(3)$ by substituting $x = 3$ into $p(x) = x^3 - 4x^2 + x + 6$:
$p(3) = (3)^3 - 4(3)^2 + (3) + 6$
$p(3) = 27 - 4(9) + 3 + 6$
$p(3) = 27 - 36 + 3 + 6$
$p(3) = 36 - 36 = 0$
Since $p(3) = 0$,by the Factor Theorem,$g(x) = x - 3$ is a factor of $p(x) = x^3 - 4x^2 + x + 6$.