Operations on Polynomials
Polynomial Addition and Subtraction:
Addition or subtraction of polynomials is performed by combining coefficients of like terms (terms with the same degree).
Let $\deg[P(x)] = n$ and $\deg[Q(x)] = m$.
If $n > m$, then:
\[
\deg[P(x) \pm Q(x)] = n
\]
If $n = m$, then:
\[
\deg[P(x) \pm Q(x)] \leq n
\]
Example:
Given the polynomials:
\[
P(x) = 2x^4 + x^2 + 3x + 5,
\]
\[
Q(x) = x^3 – 2x^2 + x – 3
\]
$\bullet \quad P(x) + Q(x) = (2 + 0)x^4 + (0 + 1)x^3 + (1 – 2)x^2 + (3 + 1)x + 5 – 3$
\[
= 2x^4 + x^3 – x^2 + 4x + 2
\]
$\bullet \quad P(x) – Q(x) = (2 – 0)x^4 + (0 – 1)x^3 + (1 – (-2))x^2 + (3 – 1)x + 5 – (-3)$
\[
= 2x^4 – x^3 + 3x^2 + 2x + 8
\]
Notice that:
\[
\deg[P(x) \pm Q(x)] = 4
\]
Example:
Given the polynomials:
\[
P(x) = x^7 + x^5 + 2x^2 + x + 3,
\]
\[
Q(x) = x^7 + x^5 + x^3 + x + 1
\]
$\bullet \quad Q(x) – P(x) = x^3 – 2x^2 – 2$
Here, the leading terms cancel out, resulting in:
\[
\deg[Q(x) – P(x)] = 3
\]
Polynomial Multiplication:
Polynomial multiplication relies on applying the distributive property over addition.
Let $\deg[P(x)] = n$, $\deg[Q(x)] = m$, and $r \in \mathbb{N}$:
\[
\deg[P(x) \cdot Q(x)] = n + m
\]
\[
\deg[P(Q(x))] = m \cdot n
\]
\[
\deg[(P(x))^r] = r \cdot n
\]
Example:
If $P(x) = 2x^2 + x + 1$ and $Q(x) = x^3 + x$, then:
\[
P(x) \cdot Q(x) = (2x^2 + x + 1) \cdot (x^3 + x)
\]
\[
= 2x^5 + 2x^3 + x^4 + x^2 + x^3 + x
\]
\[
= 2x^5 + x^4 + 3x^3 + x^2 + x
\]
\[
\deg[P(x) \cdot Q(x)] = 2 + 3 = 5
\]
Example:
If $\deg[P(x)] = 3$ and $\deg[Q(x)] = 5$, let us determine the value of:
\[
\deg[(P(x))^4 \cdot P(Q(x))]
\]
Using our degree properties:
\[
= \deg[(P(x))^4] + \deg[P(Q(x))]
\]
\[
= 4 \cdot \deg[P(x)] + \deg[P(x)] \cdot \deg[Q(x)]
\]
\[
= 4 \cdot 3 + 3 \cdot 5 = 27
\]
QUESTION 10
\[
\frac{3x^4 + ax^3 – 3x^2 + 2x + b}{x^2 + x – 1}
\]
If the simplified form of the rational expression above is $3x^2 – x + 1$, what is the sum $a + b$?
\[
\text{A)} 0 \quad
\text{B) } 1 \quad
\text{C) } 2 \quad
\text{D) } 3 \quad
\text{E) } 4
\]
Solution:
We can rewrite the equation by clearing the denominator:
\[
3x^4 + ax^3 – 3x^2 + 2x + b = (x^2 + x – 1) \cdot (3x^2 – x + 1)
\]
To isolate $a$ (the coefficient of $x^3$) and $b$ (the constant term), we only need to look at the expansion parts that yield these specific degrees:
– For the $x^3$ term: $(x^2 \cdot (-x)) + (x \cdot 3x^2) = -x^3 + 3x^3 = 2x^3 \Rightarrow a = 2$
– For the constant term: $-1 \cdot 1 = -1 \Rightarrow b = -1$
Thus, $a + b = 2 + (-1) = 1$.
\(\textbf{Correct Answer: B} \)
QUESTION 11
What is the coefficient of the $x^4$ term in the expansion of $(2x^3 + x^2 – x + 3)^2$?
\[
\text{A)} -3 \quad
\text{B) } -2 \quad
\text{C) } -1 \quad
\text{D) } 1 \quad
\text{E) } 2
\]
Solution:
Write the product explicitly:
\[ (2x^3 + x^2 – x + 3)^2 = (2x^3 + x^2 – x + 3)(2x^3 + x^2 – x + 3) \]
Instead of expanding fully, collect only the combinations that multiply to $x^4$:
\[
(2x^3 \cdot (-x)) + (x^2 \cdot x^2) + ((-x) \cdot 2x^3) = -2x^4 + x^4 – 2x^4 = -3x^4
\]
The coefficient of the fourth-degree term is $-3$.
\(\textbf{Correct Answer: A} \)
Polynomial Long Division:
Let $P(x)$ and $Q(x)$ be polynomials such that $\deg[P(x)] \geq \deg[Q(x)]$ and $Q(x) \neq 0$.
\[
\begin{array}{r|l}
P(x) & Q(x) \\
B(x) \cdot Q(x) & \rule{25mm}{0.35mm} \\
– \rule{35mm}{0.35mm} & B(x) \\
K(x) &
\end{array}
\]
By the division algorithm:
\[
P(x) = Q(x) \cdot B(x) + K(x) \quad \text{where} \quad \deg[K(x)] < \deg[Q(x)]
\]
In this setup, $P(x)$ is the dividend, $Q(x)$ is the divisor, $B(x)$ is the quotient, and $K(x)$ is the remainder, If $K(x) = 0$, then $P(x)$ is said to be perfectly divisible by $Q(x)$.
Additionally, if $\deg[Q(x)] = n$ and $\deg[B(x)] = m$, then:
\[
\deg[P(x)] = m + n
\]
Example:
If $\deg[P(x)] = 4$ and $\deg[Q(x)] = 3$, let us find:
\[
\deg \left[ \frac{5 P(x)}{P(x) + Q(x)} \right]
\]
Using the properties of polynomial degrees:
\[
\deg[5P(x)] – \deg[P(x) + Q(x)] = 4 – 4 = 0
\]
When performing polynomial long division, execute the following steps systematically:
1. Arrange both the dividend and divisor polynomials in descending order of their exponents.
2. Divide the highest-degree term of the dividend by the highest-degree term of the divisor to get the first term of the quotient.
3. Multiply the entire divisor by this new term, and write the result under the dividend, aligning like terms vertically.
4. Subtract this product from the current dividend to form a new intermediate polynomial.
5. Repeat these steps until the degree of the remaining polynomial becomes strictly less than the degree of the divisor.
Example:
\[
\begin{array}{r|l}
2x^3 + x^2 – 2x + 1 & x^2 + x + 1 \\
2x^3 + 2x^2 + 2x & \rule{35mm}{0.35mm} \\
– \rule{45mm}{0.35mm} & 2x-1 \\
-x^2 – 4x + 1&\\
-x^2-x-1\\
– \rule{45mm}{0.35mm} \\
-3x+2
\end{array}
\]
Here:
– Dividend: $P(x) = 2x^3 + x^2 – 2x + 1 \Rightarrow \deg[P(x)] = 3$
– Divisor: $Q(x) = x^2 + x + 1 \Rightarrow \deg[Q(x)] = 2$
– Quotient: $B(x) = 2x – 1 \Rightarrow \deg[B(x)] = 1$
– Remainder: $K(x) = -3x + 2 \Rightarrow \deg[K(x)] = 1$
Notice that $\deg[P(x)] = \deg[Q(x)] + \deg[B(x)]$ and $\deg[K(x)] < \deg[Q(x)]$.
Example:
\[
\begin{array}{r|l}
-4x^4+ 2x^3-3x^2+x-1 & 2x^2+x-1 \\
-4x^4-2x^3+2x^2 \phantom{aaaaaaa} & \rule{35mm}{0.35mm} \\
– \rule{65mm}{0.35mm} & -2x^2+2x- \frac{7}{2} \\
4x^3-5x^2+x-1&\\
4x^3+2x^2-2x\phantom{aa}\\
– \rule{45mm}{0.35mm} \\
-7x^2+3x-1\\
-7x^2- \frac{7}{2}x+ \frac{7}{2} \\
– \rule{45mm}{0.35mm} \\
\frac{13}{2} x- \frac{9}{2}
\end{array}
\]
Division by Synthetic Division (Horner’s Method):
Synthetic division is a shorthand method applied specifically when dividing a polynomial $P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_0$ by a first-degree linear divisor of the form $Q(x) = x – c$.
With quotient $B(x) = b_{n-1} x^{n-1} + \dots + b_0$ and remainder $K$, the structured table is constructed using the root $c$ and the coefficients of $P(x)$:
\[
\begin{array}{c|cccccc}
& a_n& a_{n-1} & a_{n-2} & \dots & a_1 & a_0 \\
c & \downarrow & c \cdot b_{n-1} & c \cdot b_{n-2} & \dots & c \cdot b_1 & c \cdot b_0 \\
\hline
& a_n = b_{n-1} & b_{n-2} & b_{n-3} & \dots & b_0 & || \quad K \\
\end{array}
\]
The coefficients are recursively calculated as follows:
– $b_{n-1} = a_n$
– $b_{n-2} = a_{n-1} + c \cdot b_{n-1}$
– $\dots$
– $K = a_0 + c \cdot b_0$
Example:
Let us find the quotient $B(x)$ and remainder $K$ when dividing $P(x) = 3x^4 – x^3 – 2x^2 + x + 12$ by $Q(x) = x – 2$.
The root of the divisor is $x – 2 = 0 \Rightarrow x = 2$. Listing the coefficients in descending power order gives $3, -1, -2, 1, 12$:
\[
\begin{array}{c|ccccc}
& 3 & -1 & -2 & 1 & 12 \\
2 &\downarrow & 6 & 10 & 16 & 34 \\
\hline
& 3 & 5 & 8 & 17 & || \quad 46 \\
\end{array}
\]
The final number ($46$) is our remainder ($K = 46$), and the preceding values form the coefficients of the quotient polynomial:
\[
B(x) = 3x^3 + 5x^2 + 8x + 17
\]
Example:
Let us divide $P(x) = x^5 + x^3 – 2x^2 + 1$ by $Q(x) = x + 1$.
The root is $x = -1$. Remember to include a $0$ coefficient for any missing powers ($x^4$ and $x^1$):
\[
\begin{array}{r|rrrrrr}
& 1 & 0 & 1 & -2 & 0 & 1 \\
-1 & \downarrow & -1 & 1 & -2 & 4 & -4 \\
\hline
& 1 & -1 & 2 & -4 & 4 & || \quad -3 \\
\end{array}
\]
– Remainder: $K = -3$
– Quotient: $B(x) = x^4 – x^3 + 2x^2 – 4x + 4$
Note:
When dividing $P(x)$ by a linear term with a leading coefficient other than one, $Q(x) = ax + b$, we can rewrite it as $a(x + \frac{b}{a})$. First perform synthetic division using the root $-\frac{b}{a}$, then divide all coefficients of the resulting intermediate quotient by $a$ to yield the final quotient $B(x)$. The remainder remains unchanged.
Example:
Find the quotient and remainder for $P(x) = 4x^4 + 5x^2 + 3x – 2$ divided by $Q(x) = 2x – 1$.
Divisor root: $2(x – \frac{1}{2}) = 0 \Rightarrow x = \frac{1}{2}$. Coefficients layout: $4, 0, 5, 3, -2$.
\[
\begin{array}{r|rrrrr}
& 4 & 0 & 5 & 3 & -2 \\
\frac{1}{2} & \downarrow & 2 & 1 & 3 & 3 \\
\hline
& 4 & 2 & 6 & 6 & || \quad 1 \\
\end{array}
\]
Divide the intermediate quotient rows by $2$: $B(x) = 2x^3 + x^2 + 3x + 3$ with a remainder of $K = 1$.
Note:
To execute synthetic division on a composite quadratic divisor $Q(x) = (ax + b)(cx + d)$, apply the method sequentially: first divide $P(x)$ by the first linear factor, then divide the resulting quotient expression by the second linear factor.
Example:
Divide $P(x) = 2x^5 – x^4 – x^3 + x^2 + 1$ by $Q(x) = x^2 – 3x + 2 = (x – 1)(x – 2)$.
Roots are $x = 1$ and $x = 2$.
\[
\begin{array}{r|rrrrrr}
& 2 & -1 & -1 & 1 & 0 & 1 \\
1 & \downarrow & 2 & 1 & 0 & 1 & 1 \\
\hline
& 2 & 1 & 0 & 1 & 1 & || \quad 2 = K_1\\
2 & \downarrow& 4 & 10 & 20 & 42 & \\
\hline
& 2 & 5 & 10 & 21 & || \quad 43 = K_2 & \\
\end{array}
\]
– Quotient: $B(x) = 2x^3 + 5x^2 + 10x + 21$
– Combined Remainder Equation: $K(x) = K_2(x – 1) + K_1 = 43(x – 1) + 2 = 43x – 41$
QUESTION 12
If $B(x)$ is the quotient obtained by dividing $P(x) = 2x^6 + x^4 – 2x + 3$ by $x^2 – 1$, what is the value of $B(1)$?
\[
\text{A)} 11 \quad
\text{B) } 10 \quad
\text{C) } 9 \quad
\text{D) } 8 \quad
\text{E) } 7
\]
Solution:
Factor the divisor: $x^2 – 1 = (x – 1)(x + 1)$. Apply sequential synthetic division:
\[
\begin{array}{r|rrrrrr}
& 2 & 0 & 1 & 0 & 0 & -2 & 3 \\
x = 1 & \downarrow & 2 & 2 & 3 & 3 & 3 & 1 \\
\hline
& 2 & 2 & 3 & 3 & 3 & 1& || \quad 4 \\
x = -1 & \downarrow & -2 & 0 & -3 & 0 & -3 \\
\hline
& 2 & 0 & 3 & 0 & 3 & || \quad -2 \\
\end{array}
\]
The row coefficients for the quotient $B(x)$ are $2, 0, 3, 0, 3$. Since evaluating $B(1)$ corresponds precisely to the sum of its coefficients:
\[
B(1) = 2 + 0 + 3 + 0 + 3 = 8
\]
\(\textbf{Correct Answer: D} \)
Finding the Remainder Directly (The Remainder Theorem):
Instead of working through full division steps to find a remainder, we use the basic statement:
\[ P(x) = Q(x) \cdot B(x) + K(x) \]
To isolate $K(x)$ without knowing $B(x)$, choose values that set the divisor term equal to zero ($Q(x) = 0$).
\[ P(x) = \underbrace{0 \cdot B(x)}_{0} + K(x) \Rightarrow P(x) = K(x) \]
1. Finding the Remainder when Dividing by $ax + b$:
Set the divisor to zero: $ax + b = 0 \Rightarrow x = -\frac{b}{a}$. Substituting this root into the dividend gives the constant remainder:
\[
K = P\left(-\frac{b}{a}\right)
\]
Example:
Let us establish the remainder calculation models when dividing various related polynomials by $x + 1 = 0 \Rightarrow x = -1$:
– Remainder of $P(x)$ divided by $(x+1)$ is $K = P(-1)$
– Remainder of $P(x+2)$ divided by $(x+1)$ is $K = P(-1 + 2) = P(1)$
– Remainder of $P(3x+1)$ divided by $(x+1)$ is $K = P(3(-1) + 1) = P(-2)$
– Remainder of $P(2x^2 – 3x + 1)$ divided by $(x+1)$ is $K = P(2(-1)^2 – 3(-1) + 1) = P(6)$
Example:
Find the remainder when $P(x) = 8x^3 + 4x^2 + 2x – 1$ is divided by $2x – 1$.
Set $2x – 1 = 0 \Rightarrow x = \frac{1}{2}$. Evaluate $P(\frac{1}{2})$:
\[
K = P\left(\frac{1}{2}\right) = 8\left(\frac{1}{2}\right)^3 + 4\left(\frac{1}{2}\right)^2 + 2\left(\frac{1}{2}\right) – 1 = 1 + 1 + 1 – 1 = 2
\]
Example:
Find the remainder when $P(x+2)$ is divided by $x – 1$, given that $P(x) = x^3 + x^2 + x + 5$.
Set $x – 1 = 0 \Rightarrow x = 1$. Plug this into the target expression $P(x+2)$ to see that we need to find $K = P(1+2) = P(3)$. Now evaluate $P(3)$ using our original definition:
\[
K = P(3) = 3^3 + 3^2 + 3 + 5 = 27 + 9 + 3 + 5 = 44
\]
Example:
Find the remainder when $P(x,y) = (x – y + 1)^3 + (x – y – 1)^2 + x^2 + y^2 – 2xy + 1$ is divided by $x – y + 2$.
Set the divisor to zero: $x – y + 2 = 0 \Rightarrow x – y = -2$. Grouping our polynomial terms using $(x-y)$ simplifies the statement:
\[
P(x, y) = (x – y + 1)^3 + (x – y – 1)^2 + (x – y)^2 + 1
\]
Now replace every instance of $(x – y)$ with $-2$:
\[
K = (-2 + 1)^3 + (-2 – 1)^2 + (-2)^2 + 1 = -1 + 9 + 4 + 1 = 13
\]
QUESTION 13
What is the remainder when $P(x) = (x^4 – 6x^2 + 9)^{19} + x^2 – 3$ is divided by $3x – 6$?
\[
\text{A)} 1 \quad
\text{B) } 2 \quad
\text{C) } 3 \quad
\text{D) } 4 \quad
\text{E) } 5
\]
Solution:
Set $3x – 6 = 0 \Rightarrow x = 2$, so our target remainder is $K = P(2)$. Notice that the core term can be factored into a perfect square: $x^4 – 6x^2 + 9 = (x^2 – 3)^2$.
\[
P(x) = [(x^2 – 3)^2]^{19} + x^2 – 3 = (x^2 – 3)^{38} + x^2 – 3
\]
Evaluating at $x = 2$:
\[
K = P(2) = (2^2 – 3)^{38} + 2^2 – 3 = (1)^{38} + 1 = 2
\]
\(\textbf{Correct Answer: B} \)
QUESTION 14
Given $P(x+1) = (x – 7)^5 + x^2 – 35$, find the remainder when $P(x^2 + x + 1)$ is divided by $x – 2$.
\[
\text{A)} -2 \quad
\text{B) } -1 \quad
\text{C) } 0 \quad
\text{D) } 1 \quad
\text{E) } 2
\]
Solution:
Set $x – 2 = 0 \Rightarrow x = 2$. Substituting this into the dividend expression yields $K = P(2^2 + 2 + 1) = P(7)$.
To calculate $P(7)$ using the given equation $P(x+1)$, set the input $x+1 = 7 \Rightarrow x = 6$:
\[
K = P(7) = (6 – 7)^5 + 6^2 – 35 = -1 + 36 – 35 = 0
\]
\(\textbf{Correct Answer: C} \)
QUESTION 15
Given the equation:
\[
\frac{P(x+3)}{x + Q(x-2)} = x^2 + x + 1
\]
If the remainder when dividing $P(x)$ by $x – 4$ is $6$, what is the remainder when dividing $Q(x+1)$ by $x + 2$?
\[
\text{A)} 1 \quad
\text{B) } 2 \quad
\text{C) } 3 \quad
\text{D) } 4 \quad
\text{E) } 5
\]
Solution:
From the problem statement:
– $x – 4 = 0 \Rightarrow P(4) = 6$
– $x + 2 = 0 \Rightarrow K = Q(-2 + 1) = Q(-1)$
To connect these target values, pick a value for $x$ in our main identity equation that outputs inputs of $4$ for $P$ and $-1$ for $Q$. Setting $x = 1$ balances both terms perfectly:
\[
\frac{P(1+3)}{1 + Q(1-2)} = 1^2 + 1 + 1 \Rightarrow \frac{P(4)}{1 + Q(-1)} = 3
\]
Substitute $P(4) = 6$:
\[
\frac{6}{1 + K} = 3 \Rightarrow 1 + K = 2 \Rightarrow K = 1
\]
\(\textbf{Correct Answer: A} \)
2. Finding the Remainder when Dividing by $(ax + b)(cx + d)$:
When dividing by a quadratic product, the remainder can have a degree of at most 1, written as:
\[
K(x) = Ax + B
\]
Set up the general division expression:
\[
P(x) = (ax + b)(cx + d)B(x) + Ax + B
\]
By evaluating this statement at the roots of both linear components ($x = -\frac{b}{a}$ and $x = -\frac{d}{c}$), you generate a solvable system of two linear equations to find parameters $A$ and $B$.
Example:
The remainder of $P(x)$ divided by $x – 2$ is $3$, and the remainder of $P(x+1)$ divided by $x + 3$ is $1$. Let us calculate the remainder $K(x)$ when $P(x)$ is divided by $x^2 – 4$.
Extract the root statements:
– $x – 2 = 0 \Rightarrow P(2) = 3$
– $x + 3 = 0 \Rightarrow P(-3+1) = P(-2) = 1$
Set up the division model for the divisor $x^2 – 4 = (x-2)(x+2)$:
\[
P(x) = (x^2 – 4)B(x) + Ax + B
\]
Evaluate at our known roots:
– For $x = 2 \Rightarrow 2A + B = 3$
– For $x = -2 \Rightarrow -2A + B = 1$
Solving this system yields $4A = 2 \Rightarrow A = \frac{1}{2}$ and $B = 2$. Therefore, the remainder is:
\[
K(x) = \frac{1}{2}x + 2
\]
QUESTION 16
Let $n \in \mathbb{N}$. What is the remainder when $P(x) = (x – 3)^{2n} – (x – 2)^n + 1$ is divided by $x^2 – 5x + 6$?
\[
\text{A)} 2x-5 \quad
\text{B) } -2x+5 \quad
\text{C) } 2x+6 \quad
\text{D) } 2x-6 \quad
\text{E) } -2x+6
\]
Solution:
Factor the quadratic divisor: $x^2 – 5x + 6 = (x-2)(x-3)$. The roots are $x=2$ and $x=3$. Evaluate $P(x)$ at these values:
– $P(2) = (2-3)^{2n} – (2-2)^n + 1 = (-1)^{2n} – 0 + 1 = 1 + 1 = 2$ (since $2n$ is even)
– $P(3) = (3-3)^{2n} – (3-2)^n + 1 = 0 – 1^n + 1 = 0$
Using the form $K(x) = Ax + B$:
– $2A + B = 2$
– $3A + B = 0$
Subtracting the equations gives $A = -2$, which means $B = 6$. The remainder expression is $-2x + 6$.
\(\textbf{Correct Answer: E} \)
3. Finding the Remainder when Dividing by $ax^n + b$:
To determine the remainder when dividing by non-linear single-variable power packages, use the constraint equation:
\[
ax^n + b = 0 \Rightarrow x^n = -\frac{b}{a}
\]
Rearrange the higher-power structures of $P(x)$ into expressions of $x^n$, then substitute $-\frac{b}{a}$ directly to extract the remainder.
Example:
Find the remainder $K(x)$ when $P(x) = x^{16} + x^7 + x^5 + x^2 + 1$ is divided by $x^5 – 1$.
Set $x^5 – 1 = 0 \Rightarrow x^5 = 1$. Rewrite $P(x)$ explicitly using power bases of $x^5$:
\[
P(x) = (x^5)^3 \cdot x + (x^5) \cdot x^2 + (x^5) + x^2 + 1
\]
Substitute $1$ for $x^5$:
\[
K(x) = (1)^3 \cdot x + (1) \cdot x^2 + 1 + x^2 + 1 = 2x^2 + x + 2
\]
QUESTION 17
Given the polynomial relationship $P(3x+2) = (x – 1)^{10} + x^2 + x + 1$, what is the remainder when $P(x^4 + x^2 + 2)$ is divided by $x^2 – 2$?
\[
\text{A) } 6\quad
\text{B) } 7 \quad
\text{C) } 8 \quad
\text{D) }9 \quad
\text{E) } 10
\]
Solution:
Set the divisor to zero: $x^2 – 2 = 0 \Rightarrow x^2 = 2$. Replace $x^2$ with $2$ inside our target dividend expression:
\[
\text{Remainder} = P((x^2)^2 + x^2 + 2) = P(2^2 + 2 + 2) = P(8)
\]
To evaluate $P(8)$ using our initial equation model, set the inner expression $3x + 2 = 8 \Rightarrow x = 2$:
\[
P(8) = (2 – 1)^{10} + 2^2 + 2 + 1 = 1 + 4 + 2 + 1 = 8
\]
\(\textbf{Correct Answer: C} \)
4. Finding the Remainder when Dividing by an Arbitrary Polynomial $Q(x)$:
For generalized expressions, isolate the highest-degree term of the divisor: $Q(x) = 0$. Substitute this definition across $P(x)$. Repeat this substitution process recursively until the remaining expression’s degree is strictly lower than $\deg[Q(x)]$.
Example:
Find the remainder $K(x)$ when $P(x) = 2x^4 – x^3 + x + 1$ is divided by $Q(x) = x^2 – x – 1$.
Set $x^2 – x – 1 = 0 \Rightarrow x^2 = x + 1$. Structure $P(x)$ using $x^2$:
\[
P(x) = 2(x^2)^2 – (x^2)x + x + 1
\]
Substitute the expression $(x + 1)$ for $x^2$:
\[
K_1(x) = 2(x+1)^2 – (x+1)x + x + 1 = 2(x^2 + 2x + 1) – x^2 – x + x + 1 = x^2 + 4x + 3
\]
Since the degree of $K_1(x)$ is not yet strictly less than the divisor’s degree, substitute $x^2 = x + 1$ once more:
\[
K(x) = (x + 1) + 4x + 3 = 5x + 4
\]
Example:
Find the remainder $K(x)$ when $P(x) = x^5 + x^3 + x^2 + x – 2$ is divided by $Q(x) = x^2 + x$.
Set $x^2 + x = 0 \Rightarrow x^2 = -x$. Organize our powers:
\[
P(x) = (x^2)^2 \cdot x + (x^2)x + x^2 + x – 2
\]
Substitute:
\[
K_1(x) = (-x)^2 \cdot x + (-x)x + (-x) + x – 2 = x^3 – x^2 – 2 = (x^2)x – x^2 – 2
\]
Continue substituting $x^2 = -x$:
\[
K_2(x) = (-x)x – (-x) – 2 = -x^2 + x – 2
\]
One final substitution for $x^2$:
\[
K(x) = -(-x) + x – 2 = 2x – 2
\]
Example:
Find the remainder when $P(x) = (x^3 + x^2 – x + 2)^9 + (x^3 + x^2 + 1)^2$ is divided by $x^3 + x^2 – x + 1$.
Set the divisor to zero: $x^3 + x^2 – x + 1 = 0$, which gives us two useful substitution forms:
1. $x^3 + x^2 – x = -1$
2. $x^3 + x^2 + 1 = x$
Substitute these directly into our grouped polynomial blocks:
\[
K(x) = (-1 + 2)^9 + (x)^2 = 1^9 + x^2 = x^2 + 1
\]
Example:
Find the remainder when $P(x, y) = x^6 y^3 – 3x^5 y^2 + 3x^4 y + x^3 y – x^3 + 1$ is divided by $x^2 y – x$.
Set $x^2 y – x = 0 \Rightarrow x^2 y = x$. Notice that the first three terms resemble a binomial cube expansion:
\[
P(x, y) = (x^2 y – x)^3 + (x^2 y) \cdot x + 1
\]
Substitute our value constraint:
\[
\text{Remainder} = (x – x)^3 + (x)x + 1 = x^2 + 1
\]
Properties of Remainders:
1. If $K(x) = 0$ in our division model statement, then $P(x)$ is perfectly divisible by $Q(x)$. This means $Q(x)$ is a factor of $P(x)$, allowing us to write $P(x) = Q(x) \cdot B(x)$.
Example:
Given $P(x) = x^4 + 2x^2 – ax + b = (x+1)^2 \cdot Q(x)$, let us calculate parameters $a$ and $b$.
Since $(x+1)^2$ is a factor, dividing $P(x)$ by it leaves a remainder of zero. Expand the factor: $(x+1)^2 = x^2 + 2x + 1 = 0 \Rightarrow x^2 = -2x – 1$. Substitute this into $P(x)$:
\[
K_1(x) = (-2x – 1)^2 + 2(-2x – 1) – ax + b = 4x^2 + 4x + 1 – 4x – 2 – ax + b = 4x^2 – ax + b – 1
\]
Substitute for $x^2$ once more to complete the division:
\[
K(x) = 4(-2x – 1) – ax + b – 1 = (-a – 8)x + (b – 5)
\]
For the remainder to equal zero identically, both components must vanish independently:
– $-a – 8 = 0 \Rightarrow a = -8$
– $b – 5 = 0 \Rightarrow b = 5$
2. If a polynomial is perfectly divisible by $Q(x)$, it is also perfectly divisible by any individual factor of $Q(x)$.
Example:
Find the parameter $a$ if $P(x) = ax^5 – x^4 + 3x^2 – bx – 24$ is perfectly divisible by $x^2 – 3x + 2$.
Factor the divisor: $x^2 – 3x + 2 = (x-1)(x-2)$. This means $P(1) = 0$ and $P(2) = 0$.
– $P(1) = a – 1 + 3 – b – 24 = 0 \Rightarrow a – b = 22$
– $P(2) = 32a – 16 + 12 – 2b – 24 = 0 \Rightarrow 32a – 2b = 28 \Rightarrow 16a – b = 14$
Subtracting the first equation from the second isolates $a$:
\[
15a = -8 \Rightarrow a = -\frac{8}{15}
\]
3. Let $K(x)$ be the remainder of $P(x)$ divided by $Q(x)$, and $M(x)$ be the remainder of $R(x)$ divided by $Q(x)$. For any $n \in \mathbb{N}$:
a) The remainder of $P(x) \pm R(x)$ divided by $Q(x)$ is $K(x) \pm M(x)$
b) The remainder of $P(x) \cdot R(x)$ divided by $Q(x)$ is $K(x) \cdot M(x)$
c) The remainder of $P^n(x)$ divided by $Q(x)$ is $K^n(x)$
Note: If the degree of any resulting combination exceeds or equals $\deg[Q(x)]$, simplify it further by dividing that combination by $Q(x)$ to get the final remainder.
Example:
Let the remainder of $P(x)$ divided by $x^2 + 1$ be $2x + 1$, and the remainder of $Q(x)$ divided by $x^2 + 1$ be $x + 1$. Let us find the remainders for $P(x) \cdot Q(x)$ and $x^2 + x + P^2(x)$ when divided by $x^2 + 1$.
$\to$ Remainder for the product $P(x) \cdot Q(x)$:
\[
K(x) \cdot M(x) = (2x + 1)(x + 1) = 2x^2 + 3x + 1
\]
Apply the divisor constraint $x^2 + 1 = 0 \Rightarrow x^2 = -1$:
\[
= 2(-1) + 3x + 1 = 3x – 1
\]
$\to$ Remainder for the expression $x^2 + x + P^2(x)$:
Substitute the remainders and apply $x^2 = -1$:
\[
-1 + x + (2x + 1)^2 = -1 + x + 4x^2 + 4x + 1 = 4x^2 + 5x
\]
Substitute $x^2 = -1$ one last time to yield the final remainder:
\[
4(-1) + 5x = 5x – 4
\]
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