Mix Examples - Polynomials Questions in English

(A) Let $p(x) = x^{3}+3x^{2}+3x+1$.
According to the Remainder Theorem,when 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,$a = -1$.
To find the remainder,we calculate $p(-1)$:
$p(-1) = (-1)^{3} + 3(-1)^{2} + 3(-1) + 1$
$p(-1) = -1 + 3(1) - 3 + 1$
$p(-1) = -1 + 3 - 3 + 1$
$p(-1) = 0$.
Therefore,the required remainder is $0$.

(D) Given the polynomial $p(x) = x + 3.$
To find $p(-x),$ we substitute $-x$ for $x$ in the expression:
$p(-x) = -x + 3.$
Now,we calculate the sum $p(x) + p(-x)$:
$p(x) + p(-x) = (x + 3) + (-x + 3).$
By simplifying the expression:
$p(x) + p(-x) = x + 3 - x + 3 = 6.$
Therefore,the correct option is $(d)$.

(D) According to the Remainder Theorem,if a polynomial $p(x)$ is divided by $x + a$,the remainder is $p(-a)$.
Here,$p(x) = x^{51} + 51$ is divided by $x + 1$.
Therefore,the remainder is $p(-1) = (-1)^{51} + 51$.
Since $(-1)$ raised to an odd power is $-1$,we have $p(-1) = -1 + 51 = 50$.
Hence,the correct option is $D$.

(C) Given expression: $(25 x^{2}-1)+(1+5 x)^{2}$
Using the identity $a^{2}-b^{2}=(a-b)(a+b)$,we can write $(25 x^{2}-1)$ as $(5 x)^{2}-1^{2}=(5 x-1)(5 x+1)$.
Substituting this into the expression:
$(5 x-1)(5 x+1)+(5 x+1)^{2}$
Taking $(5 x+1)$ as a common factor:
$(5 x+1) \times [(5 x-1)+(5 x+1)]$
Simplifying the terms inside the square bracket:
$(5 x+1) \times (5 x-1+5 x+1) = (5 x+1)(10 x)$
Thus,the factors are $10 x$ and $(5 x+1)$.
Therefore,$(c)$ is the correct answer.

(FALSE) The given statement is false.
Justification: The degree of a polynomial is the highest power of the variable in the polynomial.
Consider two polynomials $P(x) = -x^{5} + 3x^{2} + 4$ and $Q(x) = x^{5} + x^{4} + 2x^{3} + 3$.
Both $P(x)$ and $Q(x)$ have a degree of $5$.
Now,find their sum: $P(x) + Q(x) = (-x^{5} + 3x^{2} + 4) + (x^{5} + x^{4} + 2x^{3} + 3) = x^{4} + 2x^{3} + 3x^{2} + 7$.
The degree of the resulting polynomial is $4$,which is not $5$. Thus,the statement is false.

(B) The given expression is $x^{2}-2xy+y^{2}+1$.
In this expression,the variables present are $x$ and $y$.
Since there are two distinct variables,this is a polynomial in two variables.

(N/A) polynomial having only one term is called a monomial,a polynomial having only two terms is called a binomial,and a polynomial having only three terms is called a trinomial.
$(i)$ $3x$ is a monomial of degree $1$.
$(ii)$ $x^{20} - 7$ is a binomial of degree $20$.
$(iii)$ $5x^2 + 3x - 1$ is a trinomial of degree $2$.

(B) Let $p(z) = az^{3} + 4z^{2} + 3z - 4$ and $q(z) = z^{3} - 4z + a$.
According to the Remainder Theorem,when a polynomial $f(z)$ is divided by $(z - c)$,the remainder is $f(c)$.
Since both polynomials leave the same remainder when divided by $z - 3$,we have $p(3) = q(3)$.
First,calculate $p(3)$:
$p(3) = a(3)^{3} + 4(3)^{2} + 3(3) - 4$
$p(3) = 27a + 36 + 9 - 4 = 27a + 41$.
Next,calculate $q(3)$:
$q(3) = (3)^{3} - 4(3) + a$
$q(3) = 27 - 12 + a = 15 + a$.
Equating the two remainders:
$27a + 41 = 15 + a$
$27a - a = 15 - 41$
$26a = -26$
$a = -1$.
Thus,the value of $a$ is $-1$.

(C) According to the Remainder Theorem,if a polynomial $p(x)$ is divided by $(x - c)$,the remainder is $p(c)$.
Given $p(x) = x^{4} - 2x^{3} + 3x^{2} - ax + 3a - 7$ is divided by $x + 1$,the remainder is $p(-1) = 19$.
Substituting $x = -1$ into $p(x)$:
$p(-1) = (-1)^{4} - 2(-1)^{3} + 3(-1)^{2} - a(-1) + 3a - 7 = 19$
$1 - 2(-1) + 3(1) + a + 3a - 7 = 19$
$1 + 2 + 3 + 4a - 7 = 19$
$4a - 1 = 19$
$4a = 20$
$a = 5$
Now,we find the remainder when $p(x)$ is divided by $x + 2$,which is $p(-2)$ with $a = 5$:
$p(x) = x^{4} - 2x^{3} + 3x^{2} - 5x + 3(5) - 7 = x^{4} - 2x^{3} + 3x^{2} - 5x + 8$
$p(-2) = (-2)^{4} - 2(-2)^{3} + 3(-2)^{2} - 5(-2) + 8$
$p(-2) = 16 - 2(-8) + 3(4) + 10 + 8$
$p(-2) = 16 + 16 + 12 + 10 + 8 = 62$.

(N/A) The expression $x^{3}+3 x^{2} y+3 x y^{2}+y^{3}$ is a polynomial because all the exponents of the variables $x$ and $y$ are non-negative integers. Since it contains two distinct variables,$x$ and $y$,it is a polynomial in two variables,not a polynomial in one variable.

(N/A) $p(x)=21+10 x+x^{2}$
$\therefore p(x)=x^{2}+10 x+21$ and $g(x)=2+x$
$\therefore g(x)=x+2$
$\therefore$ Quotient $=x+8$ and remainder $=5$
Steps of division process:
Step $1:$ We write the dividend $21+10 x+x^{2}$ and the divisor $2+x$ in the standard form,i.e.,after arranging the terms in the descending order of their degrees. So,the dividend is $x^{2}+10 x+21$ and the divisor is $x+2$.
Step $2:$ We divide the first term of the dividend by the first term of the divisor,i.e.,we divide $x^{2}$ by $x$ and get $x$. This gives us the first term of the quotient.
$\frac{x^{2}}{x}=x=$ first term of quotient.
Step $3:$ We multiply the divisor by the first term of the quotient and subtract this product from the dividend,i.e.,we multiply $x+2$ by $x$ and subtract the product $x^{2}+2 x$ from the dividend $x^{2}+10 x+21$. This gives us the remainder as $8 x+21$.
Step $4:$ We treat the remainder $8 x+21$ as the new dividend. The divisor remains the same. We repeat step $2$ to get the next term of the quotient,i.e.,we divide the first term $8 x$ of the (new) dividend by the first term $x$ of the divisor and obtain $8$.
Thus,$8$ is the second term in the quotient.
$\frac{8 x}{x}=8=$ second term of quotient.
New quotient $=x+8$
Step $5:$ We multiply the divisor by the second term of the quotient and subtract the product from the dividend. That is,we multiply $x+2$ by $8$ and subtract the product $8 x+16$ from the dividend $8 x+21$. This gives us $5$ as the remainder.
This process continues till the remainder is $0$ or the degree of the new dividend is less than the degree of the divisor. At this stage,this new dividend becomes the remainder and the sum of the quotients gives us the whole quotient.
Step $6:$ Thus,the quotient in full is $x+8$ and the remainder is $5.$

(N/A) To divide $p(x) = x^{3} + 7x^{2} + 14x + 1$ by $x + 3$,we use polynomial long division:
$1$. Divide the first term of the dividend $(x^{3})$ by the first term of the divisor $(x)$ to get $x^{2}$.
$2$. Multiply $x^{2}$ by $(x + 3)$ to get $x^{3} + 3x^{2}$. Subtract this from the dividend to get $4x^{2} + 14x + 1$.
$3$. Divide $4x^{2}$ by $x$ to get $4x$. Multiply $4x$ by $(x + 3)$ to get $4x^{2} + 12x$. Subtract this to get $2x + 1$.
$4$. Divide $2x$ by $x$ to get $2$. Multiply $2$ by $(x + 3)$ to get $2x + 6$. Subtract this to get $-5$.
Thus,the quotient is $x^{2} + 4x + 2$ and the remainder is $-5$.

(N/A) To find the quotient and remainder,we perform polynomial long division of $2x^2 - 7x - 15$ by $x + 1$.
$1$. Divide the first term of the dividend $(2x^2)$ by the first term of the divisor $(x)$: $2x^2 / x = 2x$. This is the first term of the quotient.
$2$. Multiply the divisor $(x + 1)$ by $2x$: $2x(x + 1) = 2x^2 + 2x$.
$3$. Subtract this from the dividend: $(2x^2 - 7x - 15) - (2x^2 + 2x) = -9x - 15$.
$4$. Divide the first term of the new expression $(-9x)$ by the first term of the divisor $(x)$: $-9x / x = -9$. This is the second term of the quotient.
$5$. Multiply the divisor $(x + 1)$ by $-9$: $-9(x + 1) = -9x - 9$.
$6$. Subtract this from the current expression: $(-9x - 15) - (-9x - 9) = -9x - 15 + 9x + 9 = -6$.
Thus,the quotient is $2x - 9$ and the remainder is $-6$.

(A) According to the Remainder Theorem,if $p(x)$ is divided by $(x + 1)$,the remainder is $p(-1)$.
Given $p(-1) = 16$,we have:
$2(-1)^3 - 3(-1)^2 + a(-1) - 3a + 9 = 16$
$-2 - 3 - a - 3a + 9 = 16$
$4 - 4a = 16$
$-4a = 12$
$a = -3$.
Now,substitute $a = -3$ into $p(x)$:
$p(x) = 2x^3 - 3x^2 - 3x - 3(-3) + 9 = 2x^3 - 3x^2 - 3x + 18$.
To find the remainder when $p(x)$ is divided by $(x + 2)$,we calculate $p(-2)$:
$p(-2) = 2(-2)^3 - 3(-2)^2 - 3(-2) + 18$
$p(-2) = 2(-8) - 3(4) + 6 + 18$
$p(-2) = -16 - 12 + 6 + 18 = -4$.
Thus,$a = -3$ and the remainder is $-4$.

(A) According to the Remainder Theorem,if a polynomial $f(x)$ is divided by $(x - c)$,the remainder is $f(c)$.
For $p(x) = x^{3} + 2x^{2} - 5ax - 7$ divided by $(x + 1)$,$R_{1} = p(-1) = (-1)^{3} + 2(-1)^{2} - 5a(-1) - 7 = -1 + 2 + 5a - 7 = 5a - 6$.
For $q(x) = x^{3} + ax^{2} - 12x + 6$ divided by $(x - 2)$,$R_{2} = q(2) = (2)^{3} + a(2)^{2} - 12(2) + 6 = 8 + 4a - 24 + 6 = 4a - 10$.
Given $2R_{1} + R_{2} = 6$,substitute the expressions for $R_{1}$ and $R_{2}$:
$2(5a - 6) + (4a - 10) = 6$
$10a - 12 + 4a - 10 = 6$
$14a - 22 = 6$
$14a = 28$
$a = 2$.

(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 $62$.

(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 $250$.