Remainder Theorem: Complete Guide with Examples

Introduction

Imagine you are handed a massive mathematical problem: Divide $P(x) = 3x^4 - 5x^3 + 2x^2 - 7x + 10$ by $(x - 4)$. Your teacher asks, “What is the remainder?”

Normally, this would require you to set up a massive polynomial long division bracket, carefully track your negative signs across four different subtraction steps, and pray that you don’t make a simple arithmetic mistake. But what if there was a “cheat code”? What if you could bypass the division entirely and find the exact remainder using basic addition and multiplication in just 20 seconds?

This is exactly what the Remainder Theorem does.

What the Remainder Theorem is: It is a fundamental law of algebra stating that you do not need to divide a polynomial to find its remainder. You simply take the number you are dividing by, plug it into the polynomial’s $x$ variables, and whatever number comes out is your perfect, exact remainder.

Why it is important: It saves a tremendous amount of time. Polynomial evaluation is significantly faster than polynomial division. It reduces complex algebraic structures into simple, primary school arithmetic.

How it simplifies polynomial evaluation: Instead of dividing, you evaluate. If you want the remainder of division by $(x-5)$, you just calculate $P(5)$.

Its role in cubic equations: When solving cubic equations ($x^3$), the goal is to find factors (where the remainder is exactly zero). The Remainder Theorem allows you to rapidly test numbers to see if they produce a 0, rather than endlessly dividing the cubic polynomial by guess after guess.

Learning objectives: This massive, exhaustive guide will teach you exactly how to execute the Remainder Theorem. We will prove why it mathematically works, compare it to synthetic division and the Factor Theorem, explore real-world applications in physics and computer science, and solidify your mastery with 25 complete worked examples and 30 practice problems. Let’s begin.

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What Is the Remainder Theorem?

Before looking at the formula, let’s establish our algebraic vocabulary so the formal definition makes perfect sense.

Defining the Vocabulary

• Polynomial: An algebraic expression containing variables, coefficients, and positive exponents (e.g., $P(x) = 2x^3 - 4x^2 + x + 6$).
• Dividend: The large polynomial that is being divided (the number “inside” the division house).
• Divisor: The smaller piece you are dividing by (usually a linear expression like $x - 3$).
• Quotient: The mathematical answer you get after performing the division.
• Remainder: The “leftover” piece that could not be divided evenly. If you divide 10 by 3, the quotient is 3 and the remainder is 1.
• Polynomial evaluation: The act of plugging a specific number (like 5) into all the $x$‘s in a polynomial and calculating the final total.

The Beginner-Friendly Definition

If you divide a massive polynomial by a small piece like $(x - 2)$, there will probably be a remainder left over at the very end. The Remainder Theorem states that if you just take the number $2$ and plug it into the $x$‘s of the massive polynomial, the final number you calculate will perfectly equal that leftover remainder.

The Formal Theorem

The Remainder Theorem states: If a polynomial $P(x)$ is divided by a linear divisor $(x - a)$, then the remainder of that division is exactly equal to $P(a)$.

A Simple Example Before Cubics

Let’s divide $P(x) = x^2 - 5x + 8$ by the divisor $(x - 3)$. If we did long division, the quotient would be $x - 2$, and there would be a remainder of 2.

Let’s skip the division and use the Remainder Theorem! We are dividing by $(x - 3)$, so the number we plug in is $a = 3$. Plug 3 into $P(x)$: $P(3) = (3)^2 - 5(3) + 8$ $P(3) = 9 - 15 + 8$ $P(3) = -6 + 8 = \mathbf{2}$. The arithmetic gave us exactly the same remainder!

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Understanding the Formula

Let’s break down the mathematical statement: If $P(x)$ is divided by $(x - a)$, the remainder equals $P(a)$.

Explain every part of the formula:

• $P(x)$: This is your starting equation. It can be any degree: $x^2, x^3, x^{100}$.
• $(x - a)$: This is your divisor. It MUST be linear. This means $x$ cannot have an exponent. You cannot use this theorem to divide by $(x^2 - 4)$.
• $a$: This is the specific numerical value you are plugging in. Notice that the formula says $(x - a)$ but you plug in positive $a$. You must ALWAYS flip the sign. If dividing by $(x - 5)$, plug in $5$. If dividing by $(x + 7)$, plug in $-7$.
• $P(a)$: The act of substituting the number $a$ into the polynomial and doing the math.
• Remainder: The final single number output.

Simple Numerical Example: Find the remainder when $2x^2 + 3x - 1$ is divided by $(x + 2)$.

1. Identify $a$: The divisor is $(x + 2)$, so $a = -2$.
2. Evaluate $P(-2)$: $2(-2)^2 + 3(-2) - 1$.
3. Math: $2(4) - 6 - 1 = 8 - 6 - 1 = 1$.
4. Remainder is 1.

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Proof of the Remainder Theorem

In mathematics, theorems are not magic tricks. They are logical absolutes that can be proven. We can prove the Remainder Theorem using the universal rules of division.

The Intuition: Think about primary school division. If you divide 17 by 5, the answer (quotient) is 3, with a remainder of 2. We can rewrite this division problem as a multiplication problem: $17 = (5 \times 3) + 2$ (Dividend = Divisor $\times$ Quotient + Remainder).

The Step-by-Step Algebraic Proof: We can use this exact same logic for polynomials. If we divide polynomial $P(x)$ by divisor $(x - a)$, we get a Quotient $Q(x)$ and a constant Remainder $R$.

Let’s write this as a multiplication equation: $P(x) = (x - a) \times Q(x) + R$

Now, what happens if we decide to plug the specific number $a$ into this equation? Everywhere there is an $x$, we write $a$: $P(a) = (a - a) \times Q(a) + R$

Look closely at the $(a - a)$ part. Anything subtracted from itself is exactly 0! $P(a) = (0) \times Q(a) + R$

Zero multiplied by anything is zero. Therefore, the entire $Q(a)$ term gets instantly wiped out of existence: $P(a) = R$

The proof is complete. By simply plugging in $a$, we annihilated the quotient and proved that the final result is exactly equal to the remainder $R$.

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How to Use the Remainder Theorem

Here is the flawless, step-by-step workflow for executing the theorem.

Step 1: Write the polynomial

Ensure your equation $P(x)$ is written correctly. (e.g., $P(x) = x^3 - 4x^2 + x + 6$).

Step 2: Identify the divisor

Identify the linear expression you are dividing by, formatted as $(x - a)$.

Step 3: Determine the value of $a$

Flip the sign of the constant in the divisor. If dividing by $(x - 4)$, then $a = 4$. If dividing by $(x + 9)$, then $a = -9$.

Step 4: Evaluate $P(a)$

Substitute the value of $a$ into every $x$ in the equation. CRITICAL: Always calculate exponents first, then multiply by coefficients, then add/subtract.

Step 5: Interpret the remainder

The final number you calculate is the exact remainder.

Step 6: Verify the answer

If needed, you can quickly run synthetic division to verify that the bottom-right number perfectly matches your arithmetic output.

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Remainder Theorem for Cubic Equations

Cubic equations ($ax^3 + bx^2 + cx + d = 0$) are where the Remainder Theorem truly shines, as cubic long division is incredibly tedious.

Polynomial evaluation: If asked to find the remainder of $P(x) = 2x^3 - 5x^2 + 3x - 10$ divided by $(x - 3)$, simply calculate $P(3)$. $P(3) = 2(27) - 5(9) + 3(3) - 10$ $P(3) = 54 - 45 + 9 - 10 = \mathbf{8}$. The remainder is 8.

Checking factors and roots: The ultimate goal of solving a cubic equation is to find its three roots. A root is a factor that divides perfectly, meaning the remainder MUST be zero. The Remainder Theorem allows you to rapidly test root candidates (generated by the Rational Root Theorem) to see if they yield a 0.

Reducing calculations: Instead of setting up four different long division brackets to test the candidates $1, -1, 2, -2$, you simply perform four fast lines of arithmetic.

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Relationship with the Factor Theorem

The Remainder Theorem and the Factor Theorem are mathematically identical. The Factor Theorem is simply a specific “special case” of the Remainder Theorem.

Why a remainder of zero means a factor: If you divide 15 by 3, the remainder is 0. Because there is no leftover remainder, 3 fits perfectly into 15, making it a factor. If $P(a) = 0$, the Remainder Theorem states the remainder is 0. The Factor Theorem takes over and states that because the remainder is 0, $(x-a)$ is a perfect factor!

Comparison Table

Feature	Remainder Theorem	Factor Theorem
Purpose	To find the mathematical remainder of a division problem.	To prove whether or not a specific expression is a perfect factor.
Calculation Method	Evaluate $P(a)$.	Evaluate $P(a)$.
If $P(a) = 14$	The remainder is 14.	$(x-a)$ is NOT a factor.
If $P(a) = 0$	The remainder is 0.	$(x-a)$ IS a perfect factor.

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Relationship with Synthetic Division

Synthetic division is an alternative visual algorithm for dividing polynomials. It does exactly what the Remainder Theorem does, but instead of using exponents, it uses a rapid pattern of adding and multiplying.

Providing the same remainder: When you execute synthetic division by $a$, the very last number sitting in the bottom-right corner of the bracket is the mathematical remainder. The Remainder Theorem guarantees that calculating $P(a)$ will result in the exact same number.

Side-by-side example: Divide $P(x) = x^3 - 2x^2 + 5x - 4$ by $(x - 2)$. ($a = 2$).

Remainder Theorem Method: $P(2) = (2)^3 - 2(2)^2 + 5(2) - 4$ $P(2) = 8 - 8 + 10 - 4 = \mathbf{6}$.

Synthetic Division Method:

1. Write coefficients: 1 -2 5 -4
2. Bring down 1. Multiply by 2. Add to -2 = 0.
3. Multiply 0 by 2 = 0. Add to 5 = 5.
4. Multiply 5 by 2 = 10. Add to -4 = 6. The bottom row is 1 0 5 | 6. The remainder is exactly 6.

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Relationship with Polynomial Long Division

Verification: Polynomial long division is the oldest, most traditional way to divide expressions. You set up a division house, multiply, subtract, and bring down terms. At the very end of this long process, you will be left with a constant number that cannot be divided any further. This is the remainder. If you ever want to check your work on a long division test, simply run the Remainder Theorem in your head. If your long division remainder is 14, but $P(a)$ equals 12, you instantly know you made a subtraction error in your long division!

When long division is required: The Remainder Theorem ONLY works for linear divisors $(x - a)$. If you are dividing a polynomial by a quadratic divisor like $(x^2 + 2x + 1)$, the Remainder Theorem cannot help you. You MUST use polynomial long division.

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Applications of the Remainder Theorem

Beyond high school algebra tests, polynomial evaluation is used in high-level applications.

• Computer Algebra Systems: Software like WolframAlpha and Python’s NumPy use polynomial evaluation theorems under the hood. When rendering graphics or calculating intersections, it is computationally cheaper for a CPU to evaluate a polynomial via arithmetic than to run matrix division algorithms.
• Root Testing in Optimization: In economics, profit curves are modeled by polynomials. To find the “break-even” points (roots), economists use the Remainder Theorem to rapidly test variables and check if the remainder profit is exactly 0.
• Physics: When modeling a particle’s trajectory $p(t) = t^3 - 5t^2 + 2$, physicists use the Remainder Theorem to find the exact position (the remainder) at any specific time $t$.
• Graphing: The remainder of $P(x)$ divided by $(x-a)$ is exactly the $y$-coordinate of the graph at $x=a$. If $P(3) = 7$, then the physical coordinate $(3, 7)$ exists on the curve.

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Advantages and Limitations

Benefits:

• Exponentially faster than setting up long division.
• Highly resistant to sign errors (unlike the subtraction steps in long division).
• Instantly verifies whether a candidate is a root or a factor.

Limitations:

• Only works for dividing by linear expressions like $(x - a)$.
• It ONLY gives you the remainder. It does not give you the quotient! If the remainder is 0, you know it’s a factor, but you still have to perform synthetic or long division to find the remaining pieces of the polynomial.

Common misconceptions: Many students think $P(a)$ gives you the quotient. It absolutely does not. The Remainder Theorem obliterates the quotient (as proven earlier: $0 \times Q(a)$) and leaves only the remainder behind.

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Common Mistakes

Students constantly make these errors when executing the Remainder Theorem:

1. Using the wrong value of $a$: Dividing by $(x - 5)$ and plugging in $c = -5$. You MUST flip the sign. The value to plug in is positive $5$.
2. Arithmetic mistakes: Especially with negative numbers. $(-2)^2$ is positive 4. $(-2)^3$ is negative 8. A single dropped negative sign ruins the entire calculation.
3. Confusing quotient with remainder: Trying to write the final answer $P(a)=7$ as the “solution” to the division problem.
4. Incorrect fractional substitution: If the divisor is $(2x - 1)$, you must set it to 0 and solve for $x$. $2x - 1 = 0 \rightarrow 2x = 1 \rightarrow x = 1/2$. The number to plug in is $1/2$.
5. Ignoring standard form: Make sure the polynomial doesn’t have hidden like-terms that need to be combined first.

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Worked Examples

Let’s walk through 25 complete examples using the Remainder Theorem.

Group 1: Simple Polynomials

Example 1: Basic evaluation Find remainder of $x^2 - 3x + 4$ divided by $(x-2)$. $a = 2$. $P(2) = (2)^2 - 3(2) + 4 = 4 - 6 + 4 = 2$. Remainder is 2.

Example 2: Negative numbers Find remainder of $x^2 + 4x - 5$ divided by $(x+3)$. $a = -3$. $P(-3) = (-3)^2 + 4(-3) - 5 = 9 - 12 - 5 = -8$. Remainder is -8.

Example 3: Fractional evaluation Find remainder of $2x^2 - x + 1$ divided by $(x - 1/2)$. $a = 1/2$. $P(1/2) = 2(1/4) - 1/2 + 1 = 1/2 - 1/2 + 1 = 1$. Remainder is 1.

Example 4: Missing terms Find remainder of $x^4 - 16$ divided by $(x-2)$. $a = 2$. $P(2) = (2)^4 - 16 = 16 - 16 = 0$. Remainder is 0. (It is a factor!).

Example 5: Divisor format $(cx - a)$ Find remainder of $4x^2 + 2x - 3$ divided by $(2x - 1)$. $2x - 1 = 0 \rightarrow x = 1/2$. $P(1/2) = 4(1/4) + 2(1/2) - 3 = 1 + 1 - 3 = -1$. Remainder is -1.

Group 2: Cubic Equations

Example 6: Standard cubic remainder Divide $x^3 - 2x^2 + x - 5$ by $(x-3)$. $P(3) = (3)^3 - 2(3)^2 + 3 - 5 = 27 - 18 + 3 - 5 = 7$. Remainder is 7.

Example 7: Checking a root candidate Is $(x+1)$ a factor of $x^3 + 4x^2 + x - 6$? $a = -1$. $P(-1) = (-1)^3 + 4(-1)^2 + (-1) - 6 = -1 + 4 - 1 - 6 = -4$. Remainder is -4. It is NOT a factor.

Example 8: Large coefficients Find remainder of $5x^3 - 10x^2 + 15x - 20$ divided by $(x-2)$. $P(2) = 5(8) - 10(4) + 15(2) - 20 = 40 - 40 + 30 - 20 = 10$. Remainder is 10.

Example 9: Negative cubic Find remainder of $-x^3 + 3x^2 - 4x + 2$ divided by $(x-1)$. $P(1) = -(1) + 3(1) - 4(1) + 2 = -1 + 3 - 4 + 2 = 0$. Remainder is 0.

Example 10: All positive terms Find remainder of $x^3 + x^2 + x + 1$ divided by $(x+2)$. $P(-2) = (-2)^3 + (-2)^2 + (-2) + 1 = -8 + 4 - 2 + 1 = -5$. Remainder is -5.

Group 3: Finding Unknown Constants

Example 11: Finding $k$ with a known remainder When $x^3 - kx^2 + 2x - 4$ is divided by $(x-2)$, the remainder is 8. Find $k$. $P(2) = 8$. $P(2) = 8 - 4k + 4 - 4 = 8 - 4k$. $8 - 4k = 8 \rightarrow -4k = 0 \rightarrow k = 0$.

Example 12: Finding $k$ with a negative divisor When $x^3 + kx - 5$ is divided by $(x+3)$, the remainder is -20. $P(-3) = -20$. $P(-3) = -27 - 3k - 5 = -32 - 3k$. $-32 - 3k = -20 \rightarrow -3k = 12 \rightarrow k = -4$.

Example 13: Finding $k$ for a perfect factor For what value of $k$ does $(x-1)$ perfectly divide $kx^3 - 4x^2 + 2x + 1$? Perfect division means remainder = 0. $P(1) = 0$. $k(1) - 4(1) + 2 + 1 = 0 \rightarrow k - 1 = 0 \rightarrow k = 1$.

Example 14: Equating two remainders $P(x) = x^3 + ax^2 - 5x + 6$. The remainder when divided by $(x-1)$ is the same as the remainder when divided by $(x+2)$. Find $a$. $P(1) = P(-2)$. $P(1) = 1 + a - 5 + 6 = a + 2$. $P(-2) = -8 + 4a + 10 + 6 = 4a + 8$. $a + 2 = 4a + 8 \rightarrow -3a = 6 \rightarrow a = -2$.

Example 15: Two unknowns $P(x) = x^3 + ax^2 + bx - 2$. Divisible by $(x-1)$ (meaning $R=0$) and has a remainder of $-12$ when divided by $(x+2)$. Find $a$ and $b$. $P(1) = 1 + a + b - 2 = 0 \rightarrow a + b = 1$. $P(-2) = -8 + 4a - 2b - 2 = -12 \rightarrow 4a - 2b = -2 \rightarrow 2a - b = -1$. Add equations: $(a+b) + (2a-b) = 1 - 1 \rightarrow 3a = 0 \rightarrow a = 0$. Plug in $a$: $0 + b = 1 \rightarrow b = 1$.

Group 4: Higher Degree and Complex Polynomials

Example 16: Quartic evaluation Find remainder of $x^4 - 2x^3 + x^2 - 4$ divided by $(x-3)$. $P(3) = 81 - 54 + 9 - 4 = 32$.

Example 17: Massive exponents Find remainder of $x^{100} - 1$ divided by $(x-1)$. $P(1) = (1)^{100} - 1 = 1 - 1 = 0$.

Example 18: Testing $x$ as a divisor Find remainder of $x^3 - 4x^2 + 5x + 7$ divided by $x$. $x$ is the same as $(x-0)$. So $a = 0$. $P(0) = 0 - 0 + 0 + 7 = 7$. (The remainder of dividing by $x$ is always just the constant!).

Example 19: Symbolic remainders Find remainder of $x^3 - y^3$ divided by $(x+y)$. Plug in $-y$ for $x$: $P(-y) = (-y)^3 - y^3 = -y^3 - y^3 = -2y^3$.

Example 20: Complex divisors Find remainder of $x^3 - 2x^2 + x - 1$ divided by $(x-i)$. Plug in $i$: $P(i) = i^3 - 2i^2 + i - 1 = -i - 2(-1) + i - 1 = -i + 2 + i - 1 = 1$. Remainder is 1.

Group 5: Real-World and Graphical Connections

Example 21: Finding coordinates If $P(x) = 2x^3 - x^2 + 4x$, what is the physical coordinate on the graph when $x=2$? Evaluate $P(2)$: $16 - 4 + 8 = 20$. The coordinate is $(2, 20)$.

Example 22: Physics limits A particle’s position is $p(t) = t^3 - 7t + 6$. What is the position at $t=3$? Evaluate $p(3)$: $27 - 21 + 6 = 12$. Position is 12.

Example 23: Economics Profit is $P(x) = 2x^3 - 50x$. What is the profit at 4 units? $P(4) = 2(64) - 50(4) = 128 - 200 = -72$. The remainder is -72 (A loss!).

Example 24: Verifying Synthetic Division A student does synthetic division of $x^3 - 2x + 5$ by $x-2$ and gets a remainder of 10. Check their work. $P(2) = 8 - 4 + 5 = 9$. The remainder is 9. The student made a mistake in their synthetic division.

Example 25: Competition Math Trick Find the remainder when $P(x) = (x-2)^{200} + (x-1)^{100} - 5$ is divided by $(x-2)$. Plug in $a=2$: $P(2) = (0)^{200} + (1)^{100} - 5 = 0 + 1 - 5 = -4$. The remainder is -4. (Try doing that with long division!).

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Practice Problems

Test your mastery of the Remainder Theorem. Solutions are provided below.

Beginner Level

1. What value should you plug in to find the remainder of division by $(x-5)$?
2. What value should you plug in for $(x+8)$?
3. Find the remainder when $x^2 - 4x + 3$ is divided by $(x-1)$.
4. Find the remainder when $x^2 + 2x - 5$ is divided by $(x+2)$.
5. What does the remainder represent geometrically on a graph?
6. If $P(-3) = 14$, what is the remainder when divided by $(x+3)$?
7. If $P(4) = 0$, what does this mean about the divisor?
8. Find the remainder of $x^3 - 1$ divided by $(x-1)$.
9. Find the remainder of $x^3 + x$ divided by $x$.
10. Is the Remainder Theorem faster than long division?

Intermediate Level

11. Use the theorem to find the remainder of $2x^3 - 5x^2 - 4x + 3$ divided by $(x-3)$.
12. Find the remainder of $4x^3 + 4x^2 - x - 1$ divided by $(2x+1)$.
13. If dividing $x^3 + kx - 6$ by $(x-2)$ yields a remainder of 4, find $k$.
14. Use the theorem to check if $(x-2)$ is a factor of $x^3 - x^2 - 4x + 4$.
15. If the remainder of $x^3 - 2x^2 + x - a$ divided by $(x-1)$ is 5, find $a$.
16. Find the value of $m$ if $x^3 - mx^2 + 2x - 4$ has a remainder of 8 when divided by $(x-2)$.
17. Evaluate $P(x) = -x^3 + 5x^2 - 2x$ at $x=-1$.
18. True or False: The Remainder Theorem gives you the quotient of the division.
19. Find the remainder of $x^4 - 81$ divided by $(x+3)$.
20. Why do we ignore the quotient $Q(x)$ in the theorem’s proof?

Advanced Level

21. Find $a$ and $b$ if $x^3 + ax^2 + bx - 3$ has a remainder of 2 when divided by $(x-1)$ and a remainder of -6 when divided by $(x+1)$.
22. Prove that the remainder of $(x^n - a^n)$ divided by $(x-a)$ is always 0.
23. The remainders when $P(x) = x^3 - ax^2 + bx - 1$ is divided by $(x-2)$ and $(x-3)$ are equal. Find an equation linking $a$ and $b$.
24. Explain algebraically why the Remainder Theorem works for $(cx - a)$ by substituting $a/c$.
25. Find the remainder when $(x-1)^5 + (x+1)^3 - x$ is divided by $(x-1)$.
26. If $P(x) = x^3 - 3x^2 + 4$, find the remainder when divided by $(x-i)$.
27. Without doing division, find the remainder when $x^{99} + 1$ is divided by $(x+1)$.
28. Can the Remainder Theorem be used to test if $(x^2 - 1)$ is a divisor of $x^4 - 1$?
29. A polynomial leaves a remainder of 4 when divided by $x-2$, and a remainder of 6 when divided by $x-3$. What is the remainder when divided by $(x-2)(x-3)$?
30. If $P(c) = R$ for a polynomial with purely positive coefficients, what must be true about $R$ if $c$ is positive?

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Solutions to Practice Problems

Beginner Solutions:

1. $c = 5$.
2. $c = -8$.
3. $P(1) = 1 - 4 + 3 = 0$.
4. $P(-2) = 4 - 4 - 5 = -5$.
5. The y-coordinate at that specific x-value.
6. The remainder is 14.
7. It means $(x-4)$ is a perfect factor (Remainder is 0).
8. $P(1) = 1^3 - 1 = 0$.
9. $x$ is $(x-0)$. $P(0) = 0 + 0 = 0$.
10. Yes, exponentially faster.

Intermediate Solutions: 11. $P(3) = 2(27) - 5(9) - 4(3) + 3 = 54 - 45 - 12 + 3 = 0$. 12. Plug in $-1/2$. $P(-1/2) = 4(-1/8) + 4(1/4) - (-1/2) - 1 = -1/2 + 1 + 1/2 - 1 = 0$. 13. $P(2) = 8 + 2k - 6 = 4 \rightarrow 2 + 2k = 4 \rightarrow 2k = 2 \rightarrow k = 1$. 14. $P(2) = 8 - 4 - 8 + 4 = 0$. Yes, it is a factor. 15. $P(1) = 1 - 2 + 1 - a = 5 \rightarrow -a = 5 \rightarrow a = -5$. 16. $P(2) = 8 - 4m + 4 - 4 = 8 \rightarrow 8 - 4m = 8 \rightarrow -4m = 0 \rightarrow m = 0$. 17. $P(-1) = -(-1) + 5(1) - 2(-1) = 1 + 5 + 2 = 8$. 18. False. It only gives the remainder. It destroys the quotient. 19. $P(-3) = (-3)^4 - 81 = 81 - 81 = 0$. 20. Because $(a-a) = 0$, and 0 multiplied by the quotient obliterates it.

Advanced Solutions: 21. $P(1) = 1 + a + b - 3 = 2 \rightarrow a + b = 4$. $P(-1) = -1 + a - b - 3 = -6 \rightarrow a - b = -2$. Add equations: $2a = 2 \rightarrow a = 1$. Plug in: $1 + b = 4 \rightarrow b = 3$. 22. Plug $a$ in for $x$: $P(a) = a^n - a^n = 0$. The remainder is always 0. 23. $P(2) = 8 - 4a + 2b - 1 = 7 - 4a + 2b$. $P(3) = 27 - 9a + 3b - 1 = 26 - 9a + 3b$. Set equal: $7 - 4a + 2b = 26 - 9a + 3b \rightarrow 5a - b = 19$. 24. Dividing by $(cx - a)$ means setting $cx - a = 0 \rightarrow cx = a \rightarrow x = a/c$. The algebraic proof holds the exact same way. 25. Plug in $a=1$: $(0)^5 + (2)^3 - 1 = 0 + 8 - 1 = 7$. 26. Plug in $i$: $i^3 - 3(i^2) + 4 = -i - 3(-1) + 4 = -i + 3 + 4 = 7 - i$. 27. Plug in $a=-1$: $(-1)^{99} + 1 = -1 + 1 = 0$. 28. No, the Remainder Theorem only works for linear divisors $(x-c)$, not quadratics $(x^2-1)$. You must test $(x-1)$ and $(x+1)$ separately. 29. The remainder of a quadratic divisor is linear: $ax+b$. $P(2) = 2a+b=4$. $P(3) = 3a+b=6$. Subtract: $a=2$. Plug in: $2(2)+b=4 \rightarrow b=0$. Remainder is $2x$. 30. If all coefficients are positive, and $c$ is positive, adding them all up will result in a massive positive number. The remainder $R$ MUST be a positive number.

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Frequently Asked Questions