Mix Examples - Polynomials Questions in English

(A) To find the number of prizes left undistributed,we need to divide the total number of prizes $P(x) = 3x^3 + 10x^2 + 7x - 2$ by the number of students $(x + 2)$.
Using polynomial long division:
Step $1$: Divide $3x^3$ by $x$ to get $3x^2$.
Step $2$: Multiply $3x^2(x + 2) = 3x^3 + 6x^2$.
Step $3$: Subtract $(3x^3 + 10x^2 + 7x - 2) - (3x^3 + 6x^2) = 4x^2 + 7x - 2$.
Step $4$: Divide $4x^2$ by $x$ to get $4x$.
Step $5$: Multiply $4x(x + 2) = 4x^2 + 8x$.
Step $6$: Subtract $(4x^2 + 7x - 2) - (4x^2 + 8x) = -x - 2$.
Step $7$: Divide $-x$ by $x$ to get $-1$.
Step $8$: Multiply $-1(x + 2) = -x - 2$.
Step $9$: Subtract $(-x - 2) - (-x - 2) = 0$.
The remainder is $0$. Therefore,no prizes are left undistributed.

(A) According to the division algorithm for polynomials, $p(x) = s(x) \cdot q(x) + r(x)$.
Given $p(x) = x^{3} - 3x^{2} + x + 2$, $q(x) = x - 2$, and $r(x) = -2x + 4$.
Substituting these values into the formula: $x^{3} - 3x^{2} + x + 2 = s(x) \cdot (x - 2) + (-2x + 4)$.
Rearranging to solve for $s(x) \cdot (x - 2)$: $s(x) \cdot (x - 2) = (x^{3} - 3x^{2} + x + 2) - (-2x + 4)$.
$s(x) \cdot (x - 2) = x^{3} - 3x^{2} + x + 2 + 2x - 4$.
$s(x) \cdot (x - 2) = x^{3} - 3x^{2} + 3x - 2$.
Now, divide $x^{3} - 3x^{2} + 3x - 2$ by $(x - 2)$ to find $s(x)$.
Using polynomial division: $(x^{3} - 2x^{2}) - (x^{2} - 2x) + (x - 2) = x^{2}(x - 2) - x(x - 2) + 1(x - 2) = (x^{2} - x + 1)(x - 2)$.
Therefore, $s(x) = x^{2} - x + 1$.

(B) To find what must be added to $p(x)$ to make it divisible by $g(x) = 3x^2 - 2x + 4$,we perform polynomial long division of $p(x)$ by $g(x)$.
Dividing $6x^5 + 5x^4 + 11x^3 - 3x^2 + x + 1$ by $3x^2 - 2x + 4$:
$1$. Divide $6x^5$ by $3x^2$ to get $2x^3$. Multiply $2x^3(3x^2 - 2x + 4) = 6x^5 - 4x^4 + 8x^3$. Subtracting this from $p(x)$ gives $9x^4 + 3x^3 - 3x^2 + x + 1$.
$2$. Divide $9x^4$ by $3x^2$ to get $3x^2$. Multiply $3x^2(3x^2 - 2x + 4) = 9x^4 - 6x^3 + 12x^2$. Subtracting this gives $9x^3 - 15x^2 + x + 1$.
$3$. Divide $9x^3$ by $3x^2$ to get $3x$. Multiply $3x(3x^2 - 2x + 4) = 9x^3 - 6x^2 + 12x$. Subtracting this gives $-9x^2 - 11x + 1$.
$4$. Divide $-9x^2$ by $3x^2$ to get $-3$. Multiply $-3(3x^2 - 2x + 4) = -9x^2 + 6x - 12$. Subtracting this gives $-17x + 13$.
The remainder is $r(x) = -17x + 13$.
To make the polynomial divisible,we must add $-r(x) = -(-17x + 13) = 17x - 13$.

(C) To find what must be subtracted,we divide the polynomial $P(x) = 8x^4 + 14x^3 - 2x^2 + 8x - 12$ by $D(x) = 4x^2 + 3x - 2$.
Step $1$: Divide $8x^4$ by $4x^2$ to get $2x^2$. Multiply $2x^2(4x^2 + 3x - 2) = 8x^4 + 6x^3 - 4x^2$. Subtracting this from $P(x)$ gives $(14x^3 - 6x^3) + (-2x^2 + 4x^2) + 8x - 12 = 8x^3 + 2x^2 + 8x - 12$.
Step $2$: Divide $8x^3$ by $4x^2$ to get $2x$. Multiply $2x(4x^2 + 3x - 2) = 8x^3 + 6x^2 - 4x$. Subtracting this from the remainder gives $(2x^2 - 6x^2) + (8x + 4x) - 12 = -4x^2 + 12x - 12$.
Step $3$: Divide $-4x^2$ by $4x^2$ to get $-1$. Multiply $-1(4x^2 + 3x - 2) = -4x^2 - 3x + 2$. Subtracting this from the current remainder gives $(12x + 3x) + (-12 - 2) = 15x - 14$.
Since the remainder is $15x - 14$,subtracting this from the original polynomial will make it exactly divisible by $4x^2 + 3x - 2$.

(D) To find the other polynomial,we divide the product $x^{5}+3x^{4}+5x^{3}+3x^{2}+9x+15$ by the given polynomial $x^{2}+3$.
Performing polynomial long division:
$1$. Divide $x^{5}$ by $x^{2}$ to get $x^{3}$. Multiply $x^{3}(x^{2}+3) = x^{5}+3x^{3}$. Subtracting this from the original polynomial leaves $3x^{4}+2x^{3}+3x^{2}+9x+15$.
$2$. Divide $3x^{4}$ by $x^{2}$ to get $3x^{2}$. Multiply $3x^{2}(x^{2}+3) = 3x^{4}+9x^{2}$. Subtracting this leaves $2x^{3}-6x^{2}+9x+15$.
$3$. Wait,let us re-examine the division: $(x^{5}+3x^{4}+5x^{3}+3x^{2}+9x+15) / (x^{2}+3)$.
Factor by grouping: $x^{4}(x+3) + 5x^{2}(x+3) + 3(x+3) = (x^{4}+5x^{2}+3)(x+3)$.
Actually,let's divide directly: $(x^{2}+3)(x^{3}+3x^{2}+2x+5)$ is not correct. Let's re-divide: $(x^{5}+3x^{4}+5x^{3}+3x^{2}+9x+15) = x^{3}(x^{2}+3) + 3x^{2}(x^{2}+3) + 5(x^{2}+3) = (x^{3}+3x^{2}+5)(x^{2}+3)$.
Therefore,the other polynomial is $x^{3}+3x^{2}+5$.

(A) According to the division algorithm for polynomials, the dividend $p(x)$ is given by the formula: $p(x) = (\text{divisor} \times \text{quotient}) + \text{remainder}$.
Given:
Divisor $d(x) = x^{2}+4x+2$
Quotient $q(x) = x^{2}-4x+14$
Remainder $r(x) = 9x-13$
Substituting these values into the formula:
$p(x) = (x^{2}+4x+2)(x^{2}-4x+14) + (9x-13)$
First, multiply the divisor and the quotient:
$(x^{2}+4x+2)(x^{2}-4x+14) = x^{2}(x^{2}-4x+14) + 4x(x^{2}-4x+14) + 2(x^{2}-4x+14)$
$= (x^{4}-4x^{3}+14x^{2}) + (4x^{3}-16x^{2}+56x) + (2x^{2}-8x+28)$
Combine like terms:
$= x^{4} + (-4x^{3}+4x^{3}) + (14x^{2}-16x^{2}+2x^{2}) + (56x-8x) + 28$
$= x^{4} + 0x^{3} + 0x^{2} + 48x + 28$
Now, add the remainder $r(x) = 9x-13$:
$p(x) = (x^{4}+48x+28) + (9x-13)$
$p(x) = x^{4} + (48x+9x) + (28-13)$
$p(x) = x^{4} + 57x + 15$.

(C) polynomial is classified based on its degree,which is the highest power of the variable $x$ present in the expression.
In the given polynomial $P(x) = 4x^{3} + 3x^{2} + 2x + 1$,the highest power of $x$ is $3$.
$A$ polynomial of degree $3$ is called a cubic polynomial.
Therefore,$P(x)$ is a cubic polynomial.

(B) The degree of a polynomial is defined as the highest power of the variable present in the polynomial expression.
Given the polynomial $p(x) = x^2 - x^3 + x + 1$.
The powers of the variable $x$ are $2, 3, 1,$ and $0$ (for the constant term $1$).
The highest power among these is $3$.
Therefore,the degree of the polynomial $p(x)$ is $3$.

(D) To find the factors of the polynomial $p(x) = x^{3} - 3x^{2} + 9x - 27$,we can use the method of grouping.
Step $1$: Group the terms of the polynomial: $p(x) = (x^{3} - 3x^{2}) + (9x - 27)$.
Step $2$: Factor out the common terms from each group: $p(x) = x^{2}(x - 3) + 9(x - 3)$.
Step $3$: Factor out the common binomial $(x - 3)$: $p(x) = (x - 3)(x^{2} + 9)$.
Thus,the factors are $(x - 3)$ and $(x^{2} + 9)$.
Comparing this with the given options,$(x - 3)$ is one of the factors.

(B) To find the factors of $p(x) = x^{3} + 2x^{2} + 3x + 2$,we can use the Factor Theorem.
If $(x + 1)$ is a factor,then $p(-1)$ must be $0$.
$p(-1) = (-1)^{3} + 2(-1)^{2} + 3(-1) + 2$
$p(-1) = -1 + 2(1) - 3 + 2$
$p(-1) = -1 + 2 - 3 + 2 = 0$
Since $p(-1) = 0$,by the Factor Theorem,$(x + 1)$ is a factor of $p(x)$.

(A) The general form of a cubic polynomial is given by $p(x) = ax^{3} + bx^{2} + cx + d$.
Given the coefficients:
$a = 3$
$b = -5$
$c = -11$
$d = -3$
Substituting these values into the general form,we get:
$p(x) = (3)x^{3} + (-5)x^{2} + (-11)x + (-3)$
$p(x) = 3x^{3} - 5x^{2} - 11x - 3$
Therefore,the correct option is $A$.

(D) Let the two polynomials be $P(x)$ and $Q(x)$.
Given that $P(x) \times Q(x) = x^{2}-x-72$.
One polynomial is $P(x) = x+8$.
To find the other polynomial $Q(x)$,we divide the product by the given polynomial:
$Q(x) = \frac{x^{2}-x-72}{x+8}$.
We can factorize the quadratic expression $x^{2}-x-72$ by splitting the middle term:
$x^{2}-9x+8x-72 = x(x-9)+8(x-9) = (x+8)(x-9)$.
Therefore,$Q(x) = \frac{(x+8)(x-9)}{x+8} = x-9$.
Thus,the other polynomial is $x-9$.

(D) polynomial of degree $2$ is called a quadratic polynomial.
In option $A$,$p(x) = 2 - 4x$ is a linear polynomial (degree $1$).
In option $B$,$p(x) = 2x^3 - 1$ is a cubic polynomial (degree $3$).
In option $C$,$p(x) = 2x - 5$ is a linear polynomial (degree $1$).
In option $D$,$p(x) = 1 - x^2$ is a quadratic polynomial because the highest power of the variable $x$ is $2$.

(D) According to the Factor Theorem,$(x+1)$ is a factor of a polynomial $p(x)$ if and only if $p(-1) = 0$.
For option $A$: $p(-1) = (-1)^3 + 2(-1)^2 + 2(-1) + 1 = -1 + 2 - 2 + 1 = 0$. Thus,$(x+1)$ is a factor.
For option $B$: $p(-1) = (-1)^3 - 2(-1) - 1 = -1 + 2 - 1 = 0$. Thus,$(x+1)$ is a factor.
For option $C$: $p(-1) = (-1)^3 + 2(-1)^2 - 1 = -1 + 2 - 1 = 0$. Thus,$(x+1)$ is a factor.
For option $D$: $p(-1) = (-1)^3 - 1 = -1 - 1 = -2$. Since $p(-1) \neq 0$,$(x+1)$ is not a factor of $p(x) = x^3 - 1$.

(C) For a cubic polynomial of the form $p(x) = ax^{3} + bx^{2} + cx + d$,the product of the zeros $\alpha \beta \gamma$ is given by the formula $\alpha \beta \gamma = -\frac{d}{a}$.
Comparing the given polynomial $p(x) = x^{3} - 3x^{2} - 6x + 8$ with the standard form,we have $a = 1$,$b = -3$,$c = -6$,and $d = 8$.
Substituting these values into the formula,we get $\alpha \beta \gamma = -\frac{8}{1} = -8$.

(A) polynomial of degree $n$ has at most $n$ zeros.
For $A$: $p(x) = x^3 + 8$. This is a cubic polynomial (degree $3$),so it has $3$ zeros. Thus,this statement is true.
For $B$: $p(x) = x^2 - 36$. This is a quadratic polynomial (degree $2$),so it has at most $2$ zeros. This statement is false.
For $C$: $p(x) = 4x^3 + x = x(4x^2 + 1)$. The zeros are $x = 0$ and $4x^2 = -1$ (no real zeros). It has only $1$ real zero,though $3$ complex zeros. However,in the context of standard school mathematics,we often look for real roots. Even if we consider complex roots,$A$ is a more direct example of a cubic polynomial having $3$ roots.
For $D$: $p(x) = 4x^3 - x = x(4x^2 - 1) = x(2x - 1)(2x + 1)$. The zeros are $x = 0, 1/2, -1/2$. This also has $3$ zeros.
Note: Both $A, C,$ and $D$ have $3$ zeros. However,$p(x) = x^3 + 8$ is a standard cubic form. Given the options,$A$ is a correct statement.

(C) To find the degree of the polynomial $p(x) = x^2(1 + x + x^2) + 5$,we first expand the expression.
By distributing $x^2$ into the parentheses,we get:
$p(x) = x^2(1) + x^2(x) + x^2(x^2) + 5$
$p(x) = x^2 + x^3 + x^4 + 5$
Rearranging the terms in descending order of their powers,we have:
$p(x) = x^4 + x^3 + x^2 + 5$
The degree of a polynomial is defined as the highest power of the variable $x$ present in the expression.
In this polynomial,the highest power of $x$ is $4$.
Therefore,the degree of $p(x)$ is $4$.

(D) To find the cost of one camera,we divide the total cost by the number of cameras.
Cost of one camera = $\frac{x^{3}+3x^{2}+5x+3}{x+1}$.
Using polynomial long division:
$1$. Divide $x^{3}$ by $x$ to get $x^{2}$.
$2$. Multiply $x^{2}(x+1) = x^{3}+x^{2}$.
$3$. Subtract $(x^{3}+3x^{2}+5x+3) - (x^{3}+x^{2}) = 2x^{2}+5x+3$.
$4$. Divide $2x^{2}$ by $x$ to get $2x$.
$5$. Multiply $2x(x+1) = 2x^{2}+2x$.
$6$. Subtract $(2x^{2}+5x+3) - (2x^{2}+2x) = 3x+3$.
$7$. Divide $3x$ by $x$ to get $3$.
$8$. Multiply $3(x+1) = 3x+3$.
$9$. Subtract $(3x+3) - (3x+3) = 0$.
The quotient is $x^{2}+2x+3$. Thus,the cost of one camera is ₹ $(x^{2}+2x+3)$.

(B) two-digit number with $x$ at the ten's place and $y$ at the unit's place is represented as $10x + y$.
When a zero is inserted between these two digits,the new number becomes a three-digit number.
The digit $x$ moves to the hundred's place,the digit $0$ moves to the ten's place,and the digit $y$ remains at the unit's place.
Therefore,the value of the new number is $100(x) + 10(0) + y = 100x + y$.

(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) 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) 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: $p(x) = (x-1)(x-1) = (x-1)^{2}$.