Mix Examples - Polynomials Questions in English

(A) To find the zeros of $p(x)$,we set $p(x) = 0$.
$\sqrt{3} x^{2} - 8 x + 4 \sqrt{3} = 0$
We split the middle term $-8x$ into $-6x - 2x$ because $(\sqrt{3}) \times (4\sqrt{3}) = 12$,and we need two numbers whose product is $12$ and sum is $-8$.
$\sqrt{3} x^{2} - 6x - 2x + 4 \sqrt{3} = 0$
$\sqrt{3} x(x - 2\sqrt{3}) - 2(x - 2\sqrt{3}) = 0$
$(x - 2\sqrt{3})(\sqrt{3} x - 2) = 0$
Setting each factor to zero:
$x - 2\sqrt{3} = 0 \implies x = 2\sqrt{3}$
$\sqrt{3} x - 2 = 0 \implies x = \frac{2}{\sqrt{3}}$
Thus,the zeros are $\frac{2}{\sqrt{3}}$ and $2\sqrt{3}$.

(B) To find the zeros of $p(x)$,we set $p(x) = 0$.
$4x^2 - 256 = 0$
Divide the entire equation by $4$:
$x^2 - 64 = 0$
This can be written as a difference of squares:
$x^2 - 8^2 = 0$
$(x - 8)(x + 8) = 0$
Setting each factor to zero:
$x - 8 = 0 \implies x = 8$
$x + 8 = 0 \implies x = -8$
Thus,the zeros of the quadratic polynomial are $-8$ and $8$.

(C) To find the zeros of $p(x)$,we set $p(x) = 0$.
$x^{2} + x - 12 = 0$
We factorize the quadratic equation by splitting the middle term:
$x^{2} + 4x - 3x - 12 = 0$
$x(x + 4) - 3(x + 4) = 0$
$(x + 4)(x - 3) = 0$
Setting each factor to zero:
$x + 4 = 0 \implies x = -4$
$x - 3 = 0 \implies x = 3$
Therefore,the zeros of the quadratic polynomial $p(x)$ are $-4$ and $3$.

(D) To find the zeros of the polynomial $p(x)$,we set $p(x) = 0$.
$3x^2 + 15x = 0$
Factor out the common term $3x$:
$3x(x + 5) = 0$
Since $3 \neq 0$,we have:
$x(x + 5) = 0$
This implies either $x = 0$ or $x + 5 = 0$.
Therefore,$x = 0$ or $x = -5$.
Thus,the zeros of the quadratic polynomial $p(x)$ are $0$ and $-5$.

(N/A) To verify that $3$ is a zero of the polynomial $p(x) = 4x - 12$,we need to evaluate the polynomial at $x = 3$.
Substituting $x = 3$ into the given polynomial:
$p(3) = 4(3) - 12$
$p(3) = 12 - 12$
$p(3) = 0$
Since $p(3) = 0$,it is verified that $3$ is a zero of the polynomial $p(x) = 4x - 12$.

(A) For a quadratic polynomial of the form $ax^{2} + bx + c$,the sum of the zeros is given by $-\frac{b}{a}$ and the product of the zeros is given by $\frac{c}{a}$.
Given the polynomial $4x^{2} - 4x + 1$,we have $a = 4$,$b = -4$,and $c = 1$.
Sum of the zeros $= -\frac{b}{a} = -\frac{-4}{4} = 1$.
Product of the zeros $= \frac{c}{a} = \frac{1}{4}$.
Thus,the sum is $1$ and the product is $\frac{1}{4}$.

(A) For a quadratic polynomial of the form $ax^{2} + bx + c$,the sum of the zeros is given by $-\frac{b}{a}$ and the product of the zeros is given by $\frac{c}{a}$.
Given polynomial: $6x^{2} - 7x - 3$.
Here,$a = 6$,$b = -7$,and $c = -3$.
Sum of zeros = $-\frac{b}{a} = -(\frac{-7}{6}) = \frac{7}{6}$.
Product of zeros = $\frac{c}{a} = \frac{-3}{6} = -\frac{1}{2}$.
Thus,the sum and product are $\frac{7}{6}$ and $-\frac{1}{2}$ respectively.

(D) For a quadratic polynomial of the form $ax^2 + bx + c$,the sum of the zeros is given by $-\frac{b}{a}$ and the product of the zeros is given by $\frac{c}{a}$.
Given the polynomial $3x^2 + x - 4$,we have $a = 3$,$b = 1$,and $c = -4$.
Sum of zeros $= -\frac{b}{a} = -\frac{1}{3}$.
Product of zeros $= \frac{c}{a} = \frac{-4}{3} = -\frac{4}{3}$.
Thus,the sum and product are $-\frac{1}{3}$ and $-\frac{4}{3}$ respectively.

(A) For a quadratic polynomial of the form $ax^{2} + bx + c$,the sum of the zeros is given by $-\frac{b}{a}$ and the product of the zeros is given by $\frac{c}{a}$.
Given the polynomial $x^{2} - 7x + 10$,we have $a = 1$,$b = -7$,and $c = 10$.
Sum of the zeros $= -\frac{b}{a} = -\frac{-7}{1} = 7$.
Product of the zeros $= \frac{c}{a} = \frac{10}{1} = 10$.
Therefore,the sum and the product of the zeros are $7$ and $10$ respectively.

(B) To find the zeros of $p(x) = 6x^2 - x - 2$,we set $p(x) = 0$.
$6x^2 - x - 2 = 0$
$6x^2 - 4x + 3x - 2 = 0$
$2x(3x - 2) + 1(3x - 2) = 0$
$(2x + 1)(3x - 2) = 0$
Thus,the zeros are $x = -1/2$ and $x = 2/3$.
Let $\alpha = 2/3$ and $\beta = -1/2$.
For a quadratic polynomial $ax^2 + bx + c$,the sum of zeros is $-b/a$ and the product of zeros is $c/a$.
Here,$a = 6, b = -1, c = -2$.
Sum of zeros: $\alpha + \beta = 2/3 + (-1/2) = (4 - 3)/6 = 1/6$. Also,$-b/a = -(-1)/6 = 1/6$. Thus,$\alpha + \beta = -b/a$.
Product of zeros: $\alpha \cdot \beta = (2/3) \cdot (-1/2) = -2/6 = -1/3$. Also,$c/a = -2/6 = -1/3$. Thus,$\alpha \cdot \beta = c/a$.
Hence,the relationship is verified.

(C) quadratic polynomial is given by the formula $p(x) = k[x^2 - (\text{sum of zeros})x + (\text{product of zeros})]$,where $k$ is a non-zero constant.
Given that the sum of zeros is $\sqrt{2}$ and the product of zeros is $\frac{1}{3}$,we substitute these values into the formula:
$p(x) = k[x^2 - (\sqrt{2})x + \frac{1}{3}]$
Since the condition $a < 0$ is specified,we multiply the expression by $-1$ (setting $k$ to a negative value or simply distributing the negative sign):
$p(x) = k[-x^2 + \sqrt{2}x - \frac{1}{3}]$
Thus,the required polynomial is $k(-x^2 + \sqrt{2}x - \frac{1}{3})$.

(D) To find the zeros of the quadratic polynomial $p(x) = 5x^2 + 8x + 3$,we set $p(x) = 0$.
$5x^2 + 8x + 3 = 0$
We split the middle term $8x$ into $5x + 3x$:
$5x^2 + 5x + 3x + 3 = 0$
Factor by grouping:
$5x(x + 1) + 3(x + 1) = 0$
$(5x + 3)(x + 1) = 0$
Setting each factor to zero:
$5x + 3 = 0 \implies x = -\frac{3}{5}$
$x + 1 = 0 \implies x = -1$
Thus,the zeros are $-1$ and $-\frac{3}{5}$.

(A) To find the zeros of the quadratic polynomial $p(x) = -21x^2 + 16x + 5$,we set $p(x) = 0$.
$-21x^2 + 16x + 5 = 0$
Multiply the entire equation by $-1$ to simplify:
$21x^2 - 16x - 5 = 0$
We need to find two numbers that multiply to $(21 \times -5) = -105$ and add to $-16$. These numbers are $-21$ and $5$.
$21x^2 - 21x + 5x - 5 = 0$
Factor by grouping:
$21x(x - 1) + 5(x - 1) = 0$
$(21x + 5)(x - 1) = 0$
Setting each factor to zero:
$21x + 5 = 0 \implies x = -\frac{5}{21}$
$x - 1 = 0 \implies x = 1$
Thus,the zeros are $1$ and $-\frac{5}{21}$.

(B) To find the zeros of the quadratic polynomial $p(x) = 15x^2 + 16x + 4$,we set $p(x) = 0$.
$15x^2 + 16x + 4 = 0$
We split the middle term $16x$ into two parts such that their sum is $16x$ and their product is $15 \times 4 = 60x^2$. The factors are $10x$ and $6x$.
$15x^2 + 10x + 6x + 4 = 0$
$5x(3x + 2) + 2(3x + 2) = 0$
$(5x + 2)(3x + 2) = 0$
Setting each factor to zero:
$5x + 2 = 0 \implies x = -\frac{2}{5}$
$3x + 2 = 0 \implies x = -\frac{2}{3}$
Thus,the zeros are $-\frac{2}{3}$ and $-\frac{2}{5}$.

(A) quadratic polynomial is given by the formula: $p(x) = k[x^2 - (\text{sum of zeros})x + (\text{product of zeros})]$.
Given:
Sum of zeros $= \frac{1}{4}$
Product of zeros $= -1$
Substituting these values into the formula:
$p(x) = k[x^2 - (\frac{1}{4})x + (-1)]$
$p(x) = k[x^2 - \frac{1}{4}x - 1]$
To simplify,we can take $k = 4$:
$p(x) = 4[x^2 - \frac{1}{4}x - 1] = 4x^2 - x - 4$.
Thus,the polynomial is $k(4x^2 - x - 4)$.

(D) quadratic polynomial is given by the formula $p(x) = k[x^{2} - (\text{sum of zeros})x + (\text{product of zeros})]$,where $k$ is a non-zero real constant.
Given that the sum of the zeros $= -\frac{1}{4}$ and the product of the zeros $= \frac{1}{4}$.
Substituting these values into the formula:
$p(x) = k[x^{2} - (-\frac{1}{4})x + \frac{1}{4}]$
$p(x) = k[x^{2} + \frac{1}{4}x + \frac{1}{4}]$
To simplify,we can take $k = 4$ (or any multiple of $4$):
$p(x) = 4[x^{2} + \frac{1}{4}x + \frac{1}{4}]$
$p(x) = 4x^{2} + x + 1$
Thus,the general form is $k(4x^{2} + x + 1)$.

(A) Given polynomial: $p(x) = 2x^3 + x^2 - 5x + 2$.
Comparing with $ax^3 + bx^2 + cx + d$,we get $a = 2, b = 1, c = -5, d = 2$.
To prove these are zeros,we substitute the values into $p(x)$:
$p(1/2) = 2(1/8) + (1/4) - 5(1/2) + 2 = 1/4 + 1/4 - 5/2 + 2 = 1/2 - 2.5 + 2 = 0$.
$p(1) = 2(1)^3 + (1)^2 - 5(1) + 2 = 2 + 1 - 5 + 2 = 0$.
$p(-2) = 2(-8) + 4 - 5(-2) + 2 = -16 + 4 + 10 + 2 = 0$.
Since $p(1/2) = 0, p(1) = 0, p(-2) = 0$,these are zeros.
Verification:
Sum of zeros: $1/2 + 1 + (-2) = -1/2$. Formula: $-b/a = -1/2$. (Verified)
Sum of product of zeros taken two at a time: $(1/2)(1) + (1)(-2) + (-2)(1/2) = 1/2 - 2 - 1 = -2.5 = -5/2$. Formula: $c/a = -5/2$. (Verified)
Product of zeros: $(1/2)(1)(-2) = -1$. Formula: $-d/a = -2/2 = -1$. (Verified)

(A) Given polynomial: $p(x) = 2x^3 - 5x^2 - 14x + 8$.
Comparing with $ax^3 + bx^2 + cx + d$,we get $a = 2, b = -5, c = -14, d = 8$.
Step $1$: Check zeros.
$p(-2) = 2(-2)^3 - 5(-2)^2 - 14(-2) + 8 = 2(-8) - 5(4) + 28 + 8 = -16 - 20 + 28 + 8 = 0$.
$p(4) = 2(4)^3 - 5(4)^2 - 14(4) + 8 = 2(64) - 5(16) - 56 + 8 = 128 - 80 - 56 + 8 = 0$.
$p(1/2) = 2(1/8) - 5(1/4) - 14(1/2) + 8 = 1/4 - 5/4 - 7 + 8 = -4/4 + 1 = -1 + 1 = 0$.
Thus,$-2, 4, 1/2$ are zeros.
Step $2$: Verify relationships.
Sum of zeros: $\alpha + \beta + \gamma = -2 + 4 + 0.5 = 2.5 = 5/2 = -b/a$.
Sum of product of zeros taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = (-2)(4) + (4)(0.5) + (0.5)(-2) = -8 + 2 - 1 = -7 = -14/2 = c/a$.
Product of zeros: $\alpha\beta\gamma = (-2)(4)(0.5) = -4 = -8/2 = -d/a$.
All relationships are verified.

(A) quadratic polynomial in standard form is given by the expression $p(x) = ax^{2} + bx + c$.
Given the coefficients $a=6$,$b=-7$,and $c=-3$,we substitute these values into the standard form.
Thus,the polynomial is $6x^{2} + (-7)x + (-3)$,which simplifies to $6x^{2} - 7x - 3$.

(A) The standard form of a quadratic polynomial is given by $p(x) = ax^{2} + bx + c$.
Given the coefficients $a=1$,$b=-10$,and $c=25$.
Substituting these values into the standard form:
$p(x) = (1)x^{2} + (-10)x + 25$
$p(x) = x^{2} - 10x + 25$.

(N/A) cubic polynomial in standard form is given by the expression $P(x) = ax^{3} + bx^{2} + cx + d$.
Given the coefficients $a=2, b=5, c=1, d=2$,we substitute these values into the standard form.
Thus,the polynomial is $2x^{3} + 5x^{2} + x + 2$.

(N/A) The standard form of a cubic polynomial is given by $P(x) = ax^3 + bx^2 + cx + d$.
Given the coefficients $a=2, b=3, c=-5$,and $d=0$.
Substituting these values into the standard form,we get:
$P(x) = 2x^3 + 3x^2 + (-5)x + 0$.
Therefore,the required polynomial is $2x^3 + 3x^2 - 5x$.

(A) Given dividend $p(x) = x^{2}+8x+12$ and divisor $s(x) = x+2$.
To divide $x^{2}+8x+12$ by $x+2$,we perform long division:
$1$. Divide the first term of the dividend $(x^{2})$ by the first term of the divisor $(x)$ to get $x$.
$2$. Multiply $x$ by $(x+2)$ to get $x^{2}+2x$.
$3$. Subtract $(x^{2}+2x)$ from $(x^{2}+8x+12)$ to get $6x+12$.
$4$. Divide the first term of the new polynomial $(6x)$ by the first term of the divisor $(x)$ to get $6$.
$5$. Multiply $6$ by $(x+2)$ to get $6x+12$.
$6$. Subtract $(6x+12)$ from $(6x+12)$ to get $0$.
Thus,the quotient is $x+6$ and the remainder is $0$.

(N/A) Given dividend $p(x) = x^{3}-3x^{2}+5x-3$ and divisor $s(x) = x^{2}-2$.
Step $1$: Divide the first term of the dividend $(x^{3})$ by the first term of the divisor $(x^{2})$ to get $x$. Multiply $x$ by $(x^{2}-2)$ to get $x^{3}-2x$. Subtract this from the dividend: $(x^{3}-3x^{2}+5x-3) - (x^{3}-2x) = -3x^{2}+7x-3$.
Step $2$: Divide the first term of the new polynomial $(-3x^{2})$ by the first term of the divisor $(x^{2})$ to get $-3$. Multiply $-3$ by $(x^{2}-2)$ to get $-3x^{2}+6$. Subtract this from the current polynomial: $(-3x^{2}+7x-3) - (-3x^{2}+6) = 7x-9$.
Thus,the quotient is $q(x) = x-3$ and the remainder is $r(x) = 7x-9$.

(N/A) The dividend polynomial is $p(x) = x^{4} + 0x^{3} - 3x^{2} + 4x + 5$ and the divisor polynomial is $s(x) = x^{2} - x + 1$.
Performing polynomial long division:
$1$. Divide the first term of the dividend $(x^{4})$ by the first term of the divisor $(x^{2})$ to get $x^{2}$.
$2$. Multiply $x^{2}$ by $(x^{2} - x + 1)$ to get $x^{4} - x^{3} + x^{2}$. Subtract this from the dividend to get $x^{3} - 4x^{2} + 4x + 5$.
$3$. Divide the first term of the new polynomial $(x^{3})$ by $x^{2}$ to get $x$. Multiply $x$ by $(x^{2} - x + 1)$ to get $x^{3} - x^{2} + x$. Subtract to get $-3x^{2} + 3x + 5$.
$4$. Divide the first term $(-3x^{2})$ by $x^{2}$ to get $-3$. Multiply $-3$ by $(x^{2} - x + 1)$ to get $-3x^{2} + 3x - 3$. Subtract to get the remainder $8$.
Thus,the quotient is $q(x) = x^{2} + x - 3$ and the remainder is $r(x) = 8$.

(A) The dividend is $p(t) = 2t^4 + 3t^3 - 2t^2 - 9t - 12$ and the divisor is $s(t) = t^2 - 3$.
Step $1$: Divide the first term of the dividend $(2t^4)$ by the first term of the divisor $(t^2)$ to get $2t^2$. Multiply $2t^2$ by $(t^2 - 3)$ to get $2t^4 - 6t^2$. Subtract this from the dividend.
Step $2$: The result is $3t^3 + 4t^2 - 9t - 12$. Divide $3t^3$ by $t^2$ to get $3t$. Multiply $3t$ by $(t^2 - 3)$ to get $3t^3 - 9t$. Subtract this from the current expression.
Step $3$: The result is $4t^2 - 12$. Divide $4t^2$ by $t^2$ to get $4$. Multiply $4$ by $(t^2 - 3)$ to get $4t^2 - 12$. Subtract this to get a remainder of $0$.
Thus,the quotient is $2t^2 + 3t + 4$ and the remainder is $0$.

(N/A) To divide $x^{3}-6x^{2}+11x-6$ by $x^{2}-8x+27$,we use polynomial long division:
$1$. Divide the first term of the dividend $(x^{3})$ by the first term of the divisor $(x^{2})$: $x^{3} / x^{2} = x$.
$2$. Multiply the divisor $(x^{2}-8x+27)$ by $x$: $x(x^{2}-8x+27) = x^{3}-8x^{2}+27x$.
$3$. Subtract this from the dividend: $(x^{3}-6x^{2}+11x-6) - (x^{3}-8x^{2}+27x) = 2x^{2}-16x-6$.
$4$. Divide the first term of the new polynomial $(2x^{2})$ by the first term of the divisor $(x^{2})$: $2x^{2} / x^{2} = 2$.
$5$. Multiply the divisor $(x^{2}-8x+27)$ by $2$: $2(x^{2}-8x+27) = 2x^{2}-16x+54$.
$6$. Subtract this from the current polynomial: $(2x^{2}-16x-6) - (2x^{2}-16x+54) = -60$.
Thus,the quotient is $x+2$ and the remainder is $-60$.

(N/A) To divide $x^{4}+4 x^{3}-2 x^{2}-12 x+9$ by $x^{2}-2 x+1$,we use long division:
$1$. Divide the first term of the dividend $(x^4)$ by the first term of the divisor $(x^2)$ to get $x^2$.
$2$. Multiply $x^2$ by $(x^2-2x+1)$ to get $x^4-2x^3+x^2$. Subtract this from the dividend to get $6x^3-3x^2-12x+9$.
$3$. Divide $6x^3$ by $x^2$ to get $6x$. Multiply $6x$ by $(x^2-2x+1)$ to get $6x^3-12x^2+6x$. Subtract this to get $9x^2-18x+9$.
$4$. Divide $9x^2$ by $x^2$ to get $9$. Multiply $9$ by $(x^2-2x+1)$ to get $9x^2-18x+9$. Subtract this to get a remainder of $0$.
Thus,the quotient is $x^{2}+6 x+9$ and the remainder is $0$.

(N/A) To divide $x^{4}-5x+6$ by $2-x^{2}$,we first rewrite the polynomials in descending order of their degrees:
Dividend: $x^{4}+0x^{3}+0x^{2}-5x+6$
Divisor: $-x^{2}+2$
Step $1$: Divide the first term of the dividend $(x^{4})$ by the first term of the divisor $(-x^{2})$: $x^{4} / (-x^{2}) = -x^{2}$. This is the first term of the quotient.
Step $2$: Multiply $-x^{2}$ by the divisor $(-x^{2}+2)$: $-x^{2}(-x^{2}+2) = x^{4}-2x^{2}$.
Step $3$: Subtract this from the dividend: $(x^{4}+0x^{3}+0x^{2}-5x+6) - (x^{4}-2x^{2}) = 2x^{2}-5x+6$.
Step $4$: Divide the first term of the new dividend $(2x^{2})$ by the first term of the divisor $(-x^{2})$: $2x^{2} / (-x^{2}) = -2$. This is the second term of the quotient.
Step $5$: Multiply $-2$ by the divisor $(-x^{2}+2)$: $-2(-x^{2}+2) = 2x^{2}-4$.
Step $6$: Subtract this from the current dividend: $(2x^{2}-5x+6) - (2x^{2}-4) = -5x+10$.
Thus,the quotient is $-x^{2}-2$ and the remainder is $-5x+10$.

(N/A) To divide $14x^3 - 5x^2 + 9x - 1$ by $2x - 1$,we use polynomial long division:
$1$. Divide the first term of the dividend $(14x^3)$ by the first term of the divisor $(2x)$: $14x^3 / 2x = 7x^2$. This is the first term of the quotient.
$2$. Multiply $7x^2$ by $(2x - 1)$ to get $14x^3 - 7x^2$. Subtract this from the dividend: $(14x^3 - 5x^2) - (14x^3 - 7x^2) = 2x^2$. Bring down the next term $(9x)$ to get $2x^2 + 9x$.
$3$. Divide the first term of the new expression $(2x^2)$ by the first term of the divisor $(2x)$: $2x^2 / 2x = x$. This is the second term of the quotient.
$4$. Multiply $x$ by $(2x - 1)$ to get $2x^2 - x$. Subtract this from the current expression: $(2x^2 + 9x) - (2x^2 - x) = 10x$. Bring down the last term $(-1)$ to get $10x - 1$.
$5$. Divide the first term $(10x)$ by the first term of the divisor $(2x)$: $10x / 2x = 5$. This is the third term of the quotient.
$6$. Multiply $5$ by $(2x - 1)$ to get $10x - 5$. Subtract this from the current expression: $(10x - 1) - (10x - 5) = 4$.
Thus,the quotient is $7x^2 + x + 5$ and the remainder is $4$.

(N/A) To divide the polynomial $p(x) = -x^{3} + 3x^{2} - 3x + 5$ by $g(x) = -x^{2} + x - 1$:
Step $1$: Divide the first term of the dividend $(-x^{3})$ by the first term of the divisor $(-x^{2})$,which gives $x$.
Step $2$: Multiply $x$ by the divisor $(-x^{2} + x - 1)$ to get $-x^{3} + x^{2} - x$. Subtract this from the dividend: $(-x^{3} + 3x^{2} - 3x + 5) - (-x^{3} + x^{2} - x) = 2x^{2} - 2x + 5$.
Step $3$: Divide the first term of the new dividend $(2x^{2})$ by the first term of the divisor $(-x^{2})$,which gives $-2$.
Step $4$: Multiply $-2$ by the divisor $(-x^{2} + x - 1)$ to get $2x^{2} - 2x + 2$. Subtract this from the current dividend: $(2x^{2} - 2x + 5) - (2x^{2} - 2x + 2) = 3$.
Thus,the quotient is $x - 2$ and the remainder is $3$.

(N/A) To divide $x^{3}-6 x^{2}+11 x-6$ by $x^{2}+x+1$,we use the long division method:
$1$. Divide the first term of the dividend $(x^{3})$ by the first term of the divisor $(x^{2})$ to get $x$.
$2$. Multiply $x$ by the divisor $(x^{2}+x+1)$ to get $x^{3}+x^{2}+x$.
$3$. Subtract this from the dividend: $(x^{3}-6 x^{2}+11 x-6) - (x^{3}+x^{2}+x) = -7 x^{2}+10 x-6$.
$4$. Divide the first term of the new polynomial $(-7 x^{2})$ by the first term of the divisor $(x^{2})$ to get $-7$.
$5$. Multiply $-7$ by the divisor $(x^{2}+x+1)$ to get $-7 x^{2}-7 x-7$.
$6$. Subtract this from the current polynomial: $(-7 x^{2}+10 x-6) - (-7 x^{2}-7 x-7) = 17 x+1$.
Thus,the quotient is $x-7$ and the remainder is $17 x+1$.

(A) To divide $x^{4}+2x^{3}+3x^{2}+2x+20$ by $x^{2}+2x+2$,we use long division:
$1$. Divide the first term of the dividend $(x^{4})$ by the first term of the divisor $(x^{2})$ to get $x^{2}$.
$2$. Multiply $x^{2}$ by $(x^{2}+2x+2)$ to get $x^{4}+2x^{3}+2x^{2}$.
$3$. Subtract this from the dividend: $(x^{4}+2x^{3}+3x^{2}+2x+20) - (x^{4}+2x^{3}+2x^{2}) = x^{2}+2x+20$.
$4$. Divide the first term of the new polynomial $(x^{2})$ by the first term of the divisor $(x^{2})$ to get $1$.
$5$. Multiply $1$ by $(x^{2}+2x+2)$ to get $x^{2}+2x+2$.
$6$. Subtract this from the current remainder: $(x^{2}+2x+20) - (x^{2}+2x+2) = 18$.
Thus,the quotient is $x^{2}+1$ and the remainder is $18$.

(N/A) To divide $x^{3}-3x^{2}-3x+1$ by $x+1$,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 $(x+1)$ to get $x^{3}+x^{2}$. Subtract this from the dividend: $(x^{3}-3x^{2}-3x+1) - (x^{3}+x^{2}) = -4x^{2}-3x+1$.
$3$. Divide $-4x^{2}$ by $x$ to get $-4x$. Multiply $-4x$ by $(x+1)$ to get $-4x^{2}-4x$. Subtract this: $(-4x^{2}-3x+1) - (-4x^{2}-4x) = x+1$.
$4$. Divide $x$ by $x$ to get $1$. Multiply $1$ by $(x+1)$ to get $x+1$. Subtract this: $(x+1) - (x+1) = 0$.
Thus,the quotient is $x^{2}-4x+1$ and the remainder is $0$.

(N/A) To divide $3 x^{4}+5 x^{3}-7 x^{2}+2 x+2$ by $x^{2}+3 x+1$,we use long division:
$1$. Divide the first term of the dividend $(3 x^{4})$ by the first term of the divisor $(x^{2})$ to get $3 x^{2}$.
$2$. Multiply $3 x^{2}$ by $(x^{2}+3 x+1)$ to get $3 x^{4}+9 x^{3}+3 x^{2}$.
$3$. Subtract this from the dividend: $(3 x^{4}+5 x^{3}-7 x^{2}+2 x+2) - (3 x^{4}+9 x^{3}+3 x^{2}) = -4 x^{3}-10 x^{2}+2 x+2$.
$4$. Divide the first term of the new polynomial $(-4 x^{3})$ by $x^{2}$ to get $-4 x$.
$5$. Multiply $-4 x$ by $(x^{2}+3 x+1)$ to get $-4 x^{3}-12 x^{2}-4 x$.
$6$. Subtract this: $(-4 x^{3}-10 x^{2}+2 x+2) - (-4 x^{3}-12 x^{2}-4 x) = 2 x^{2}+6 x+2$.
$7$. Divide $2 x^{2}$ by $x^{2}$ to get $2$.
$8$. Multiply $2$ by $(x^{2}+3 x+1)$ to get $2 x^{2}+6 x+2$.
$9$. Subtracting this results in a remainder of $0$.
Thus,the quotient is $3 x^{2}-4 x+2$ and the remainder is $0$.

(N/A) To divide $x^{6}+5 x^{3}+7 x+3$ by $x^{2}+2$,we use long division:
$1$. Divide the first term $x^{6}$ by $x^{2}$ to get $x^{4}$. Multiply $x^{4}(x^{2}+2) = x^{6}+2 x^{4}$. Subtracting this from the dividend gives $-2 x^{4}+5 x^{3}+7 x+3$.
$2$. Divide $-2 x^{4}$ by $x^{2}$ to get $-2 x^{2}$. Multiply $-2 x^{2}(x^{2}+2) = -2 x^{4}-4 x^{2}$. Subtracting this gives $5 x^{3}+4 x^{2}+7 x+3$.
$3$. Divide $5 x^{3}$ by $x^{2}$ to get $5 x$. Multiply $5 x(x^{2}+2) = 5 x^{3}+10 x$. Subtracting this gives $4 x^{2}-3 x+3$.
$4$. Divide $4 x^{2}$ by $x^{2}$ to get $4$. Multiply $4(x^{2}+2) = 4 x^{2}+8$. Subtracting this gives $-3 x-5$.
Thus,the quotient is $x^{4}-2 x^{2}+5 x+4$ and the remainder is $-3 x-5$.

(A) Let the given polynomial be $P(x) = x^{3}-3x^{2}-x+3$ and the known factor be $g(x) = x+1$. To find the other polynomial $q(x)$,we divide $P(x)$ by $g(x)$.
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 $(x^{3}-3x^{2}-x+3) - (x^{3}+x^{2}) = -4x^{2}-x+3$.
$4$. Divide $-4x^{2}$ by $x$ to get $-4x$.
$5$. Multiply $-4x(x+1) = -4x^{2}-4x$.
$6$. Subtract $(-4x^{2}-x+3) - (-4x^{2}-4x) = 3x+3$.
$7$. Divide $3x$ by $x$ to get $3$.
$8$. Multiply $3(x+1) = 3x+3$.
$9$. Subtract $(3x+3) - (3x+3) = 0$.
The quotient is $x^{2}-4x+3$.

(B) Let the given product be $P(x) = 3x^3 - x^2 - 3x + 1$ and one polynomial be $A(x) = 3x^2 + 2x - 1$.
To find the other polynomial $B(x)$,we divide $P(x)$ by $A(x)$:
$B(x) = \frac{3x^3 - x^2 - 3x + 1}{3x^2 + 2x - 1}$
Performing polynomial long division:
$1$. Divide $3x^3$ by $3x^2$ to get $x$.
$2$. Multiply $x$ by $(3x^2 + 2x - 1)$ to get $3x^3 + 2x^2 - x$.
$3$. Subtract this from $P(x)$: $(3x^3 - x^2 - 3x + 1) - (3x^3 + 2x^2 - x) = -3x^2 - 2x + 1$.
$4$. Divide $-3x^2$ by $3x^2$ to get $-1$.
$5$. Multiply $-1$ by $(3x^2 + 2x - 1)$ to get $-3x^2 - 2x + 1$.
$6$. Subtracting this gives a remainder of $0$.
Thus,the other polynomial is $x - 1$.

(C) Given that $1$ and $3$ are zeros of $p(x)$,$(x - 1)$ and $(x - 3)$ are factors of $p(x)$.
Therefore,$(x - 1)(x - 3) = x^2 - 4x + 3$ is a factor of $p(x)$.
Dividing $p(x) = 2x^4 - 7x^3 - 13x^2 + 63x - 45$ by $(x^2 - 4x + 3)$:
$2x^4 - 7x^3 - 13x^2 + 63x - 45 = (x^2 - 4x + 3)(2x^2 + x - 15)$.
Now,factorize the quadratic $2x^2 + x - 15$:
$2x^2 + 6x - 5x - 15 = 2x(x + 3) - 5(x + 3) = (2x - 5)(x + 3)$.
Setting these factors to zero,we get $x = -3$ and $x = \frac{5}{2}$.
Thus,the remaining zeros are $-3$ and $\frac{5}{2}$.

(N/A) To find the number of chocolates received by each student and the remainder,we perform polynomial division of $(x^{3}-3x^{2}+5x-3)$ by $(x^{2}-2)$.
Step $1$: Divide $x^{3}$ by $x^{2}$ to get $x$.
Step $2$: Multiply $x$ by $(x^{2}-2)$ to get $x^{3}-2x$.
Step $3$: Subtract $(x^{3}-2x)$ from $(x^{3}-3x^{2}+5x-3)$ to get $-3x^{2}+7x-3$.
Step $4$: Divide $-3x^{2}$ by $x^{2}$ to get $-3$.
Step $5$: Multiply $-3$ by $(x^{2}-2)$ to get $-3x^{2}+6$.
Step $6$: Subtract $(-3x^{2}+6)$ from $(-3x^{2}+7x-3)$ to get $7x-9$.
Thus,each student receives $x-3$ chocolates,and the number of chocolates left undistributed is $7x-9$.

(A) To find the price of one mobile,we divide the total cost by the number of pieces bought.
Price of one mobile = $\frac{x^{3}+6x^{2}+11x+12}{x^{2}+2x+3}$.
Performing polynomial long division:
$1$. Divide the first term of the dividend $(x^3)$ by the first term of the divisor $(x^2)$ to get $x$.
$2$. Multiply $x$ by the divisor: $x(x^{2}+2x+3) = x^{3}+2x^{2}+3x$.
$3$. Subtract this from the dividend: $(x^{3}+6x^{2}+11x+12) - (x^{3}+2x^{2}+3x) = 4x^{2}+8x+12$.
$4$. Divide the first term of the remainder $(4x^2)$ by the first term of the divisor $(x^2)$ to get $4$.
$5$. Multiply $4$ by the divisor: $4(x^{2}+2x+3) = 4x^{2}+8x+12$.
$6$. Subtract this from the remainder: $(4x^{2}+8x+12) - (4x^{2}+8x+12) = 0$.
Therefore,the price of one mobile is $x+4$.

(N/A) To divide $p(x) = x^{3} - 3x^{2} - 3x + 1$ by $x + 1$ using synthetic division:
$1$. Identify the coefficients of the dividend $p(x)$: $1, -3, -3, 1$.
$2$. Determine the root of the divisor $x + 1 = 0$,which gives $x = -1$.
$3$. Perform the synthetic division:
$\begin{array}{c|cccc} -1 & 1 & -3 & -3 & 1 \\ & & -1 & 4 & -1 \\ \hline & 1 & -4 & 1 & 0 \end{array}$
$4$. The bottom row represents the coefficients of the quotient and the remainder. The quotient is $q(x) = x^{2} - 4x + 1$ and the remainder is $r(x) = 0$.

(N/A) To divide $p(x) = x^{4} - 1$ by $x - 1$ using synthetic division:
$1$. Write the dividend $p(x)$ in standard form including all missing terms with zero coefficients: $p(x) = 1x^{4} + 0x^{3} + 0x^{2} + 0x - 1$.
$2$. The divisor is $x - 1$,so set $x - 1 = 0$,which gives $x = 1$. Use $1$ for the synthetic division.
$3$. Set up the synthetic division table:
$\begin{array}{c|ccccc} 1 & 1 & 0 & 0 & 0 & -1 \\ & & 1 & 1 & 1 & 1 \\ \hline & 1 & 1 & 1 & 1 & 0 \end{array}$
$4$. The bottom row represents the coefficients of the quotient and the remainder. The last value is the remainder.
Thus,the quotient is $q(x) = x^{3} + x^{2} + x + 1$ and the remainder is $r(x) = 0$.

(A) To find the number of pens received by each student and the remainder,we divide the polynomial $p(x) = x^{3}-3x^{2}+4x-4$ by the divisor $s(x) = x-2$ using synthetic division.
Setting the divisor $x-2 = 0$ gives $x = 2$.
Using synthetic division:
$\begin{array}{c|cccc} 2 & 1 & -3 & 4 & -4 \\ & & 2 & -2 & 4 \\ \hline & 1 & -1 & 2 & 0 \end{array}$
The quotient is $q(x) = x^{2}-x+2$ and the remainder is $0$.
Therefore,each student receives $x^{2}-x+2$ pens,and there are $0$ pens left undistributed.

(N/A) To divide $p(x) = x^{3} - 4x^{2} + 5x + 3$ by $x - 2$ using synthetic division:
$1$. Identify the coefficients of $p(x)$,which are $1, -4, 5, 3$.
$2$. The divisor is $x - 2$,so we use $x = 2$ for synthetic division.
$3$. Set up the synthetic division table:
$2 | 1, -4, 5, 3$
| $2$,-$4$,$2$
-----------------
| $1$,-$2$,$1$,$5$
$4$. The coefficients of the quotient are $1, -2, 1$,which corresponds to the polynomial $x^{2} - 2x + 1$.
$5$. The remainder is the last value,which is $5$.
Therefore,the quotient is $x^{2} - 2x + 1$ and the remainder is $5$.

(A) To divide $p(x) = x^{4} + 0x^{3} + 0x^{2} + 0x + 1$ by $x + 1$,we use the synthetic division method.
The zero of the divisor $x + 1$ is $x = -1$.
Setting up the synthetic division table:
$-1$ | $1$ $0$ $0$ $0$ $1$
| $-1$ $1$ $-1$ $1$
-----------------------
$1$ $-1$ $1$ $-1$ $2$
The coefficients of the quotient are $1, -1, 1, -1$,which corresponds to the polynomial $x^{3} - x^{2} + x - 1$.
The remainder is the last value,which is $2$.
Thus,the quotient is $x^{3} - x^{2} + x - 1$ and the remainder is $2$.

(A) To divide $p(t) = 2t^2 + 3t + 1$ by $t + 2$ using synthetic division:
$1$. Identify the root of the divisor $t + 2 = 0$,which is $t = -2$.
$2$. Write the coefficients of the dividend $p(t) = 2t^2 + 3t + 1$,which are $2, 3, 1$.
$3$. Perform synthetic division:
- Bring down the first coefficient: $2$.
- Multiply by the root: $2 \times (-2) = -4$.
- Add to the next coefficient: $3 + (-4) = -1$.
- Multiply by the root: $(-1) \times (-2) = 2$.
- Add to the next coefficient: $1 + 2 = 3$.
$4$. The resulting coefficients are $2$ and $-1$,representing the quotient $2t - 1$.
$5$. The final value $3$ is the remainder.
Quotient polynomial: $2t - 1$;
Remainder polynomial: $3$.

(A) To divide $p(x) = x^{4} - a^{4}$ by $x - a$ using synthetic division:
$1$. The coefficients of $p(x) = 1x^{4} + 0x^{3} + 0x^{2} + 0x - a^{4}$ are $1, 0, 0, 0, -a^{4}$.
$2$. The divisor is $x - a$,so we use $a$ for synthetic division.
$3$. Setting up the synthetic division table:
$a$ | $1$ $0$ $0$ $0$ $-a^{4}$
| $a$ $a^{2}$ $a^{3}$ $a^{4}$
---------------------------
$1$ $a$ $a^{2}$ $a^{3}$ $0$
$4$. The quotient coefficients are $1, a, a^{2}, a^{3}$,which corresponds to the polynomial $x^{3} + ax^{2} + a^{2}x + a^{3}$.
$5$. The remainder is $0$.

(N/A) To divide $p(x) = x^{3} - 2x^{2} + x - 2$ by $x + 1$ using synthetic division,we identify the root of the divisor $x + 1 = 0$,which is $x = -1$.
The coefficients of $p(x)$ are $1, -2, 1, -2$.
Setting up the synthetic division:
$-1$ | $1$ $-2$ $1$ $-2$
| $-1$ $3$ $-4$
-------------------
$1$ $-3$ $4$ $-6$
The coefficients of the quotient polynomial are $1, -3, 4$,which corresponds to $x^{2} - 3x + 4$.
The remainder is $-6$.

(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.