Factor Theorem: Complete Guide with Examples

Introduction

In algebra, one of the most frustrating tasks a student faces is looking at a massive cubic or quartic polynomial and being told to “factor it.” Unlike a simple quadratic equation where you can easily find two numbers that multiply to the end and add to the middle, higher-degree polynomials seem like impenetrable walls of numbers. How do you even know where to begin? How do you know if a piece of the puzzle fits without doing 20 minutes of exhausting long division?

The answer is the Factor Theorem.

What the Factor Theorem is: It is an incredibly elegant mathematical shortcut. It states that if you plug a specific number into an equation and the math works out to exactly zero, you have just found a perfect factor of that equation.

Why it is important: It completely bypasses polynomial long division. Instead of dividing massive equations to see if they fit perfectly together, you simply do basic arithmetic. If the arithmetic equals zero, the division would have been perfect.

How it simplifies solving cubic equations: A cubic equation has three roots. Finding the first root is the hardest part. The Factor Theorem allows you to rapidly test numbers until you hit “zero.” Once you find that first root, the cubic equation instantly crumbles into a much easier quadratic equation.

How it relates to polynomial roots: A root is the numerical answer to an equation. A factor is the algebraic building block of the equation. The Factor Theorem is the unbreakable bridge between these two concepts, proving that if you know one, you instantly know the other.

Learning objectives: This massive, exhaustive guide will teach you exactly how to execute the Factor Theorem. We will prove why it works, compare it to its sibling (the Remainder Theorem), integrate it with synthetic division, and solidify your mastery with 25 complete worked examples and 30 practice problems. Let’s begin.

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What Is the Factor Theorem?

Before looking at the theorem, 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., $P(x) = x^3 - 4x^2 + x + 6$).
• Root: Also known as a zero or a solution. It is a specific numerical value of $x$ (like $x=3$) that makes the entire polynomial equal to zero.
• Factor: A piece of the polynomial. When you multiply all the factors together, you get the original polynomial.
• Linear factor: A factor where $x$ has no exponent higher than 1, usually written in the form $(x - c)$.

The Beginner-Friendly Definition

If you guess a number (let’s call it $c$), and you plug that number into all the $x$‘s in your equation, and the final result is exactly $0$… then the expression $(x - c)$ is a perfect, clean factor of your equation.

The Formal Theorem

The Factor Theorem states: A polynomial $P(x)$ has a factor $(x - c)$ if and only if $P(c) = 0$.

Notice the phrase “if and only if.” This means the theorem works perfectly in both directions:

1. If $P(c) = 0$, then $(x-c)$ is definitely a factor.
2. If you already know $(x-c)$ is a factor, then $P(c)$ will definitely equal 0.

A Simple Example Before Cubics

Look at the quadratic polynomial $P(x) = x^2 - 5x + 6$. Let’s test the number 2. (So, $c = 2$). Plug in 2: $P(2) = (2)^2 - 5(2) + 6 = 4 - 10 + 6 = 0$. Because $P(2)$ exactly equals 0, the Factor Theorem guarantees that $(x - 2)$ is a factor of $x^2 - 5x + 6$. (And we know this is true, because $x^2 - 5x + 6$ factors to $(x-2)(x-3)$).

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Understanding the Relationship Between Roots and Factors

Why does $P(a) = 0$ perfectly translate to $(x - a)$?

A root is a physical point where a graph crosses the x-axis. At that exact point, the y-value is 0. If an equation crosses the x-axis at $x = 5$, we say the root is 5. If $x = 5$, we can use basic algebra to move the 5 to the other side of the equals sign by subtracting it from both sides: $x - 5 = 0$. The expression $(x - 5)$ is now equal to zero, which makes it the fundamental building block (the factor) of the equation!

Easy Examples:

• If $P(7) = 0$, the root is 7. The factor is $(x - 7)$.
• If $P(-4) = 0$, the root is -4. The factor is $(x - (-4))$, which simplifies to $(x + 4)$.
• If $P(1/2) = 0$, the root is $1/2$. The factor is $(x - 1/2)$, which can also be written as $(2x - 1)$.

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Proof of the Factor Theorem

In mathematics, you should never accept a rule just because a textbook says it is true. We can prove the Factor Theorem using the logic of basic division.

The Intuition: Think about basic division with normal numbers. If you divide 15 by 3, you get 5 with a remainder of 0. Because the remainder is exactly 0, we can confidently say “3 is a factor of 15.” If you divide 16 by 3, you get 5 with a remainder of 1. Because there is a remainder, 3 is NOT a factor of 16.

The Step-by-Step Algebraic Proof: Let’s use Polynomial Division. When you divide a polynomial $P(x)$ by a linear factor $(x - c)$, you get a Quotient $Q(x)$ and a constant Remainder $R$.

We can write this division mathematically as a multiplication problem: $P(x) = (x - c) \times Q(x) + R$

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

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

Zero times anything is zero. Therefore, the entire $Q(c)$ gets wiped out: $P(c) = R$

This algebraic proof (which is actually the Remainder Theorem) proves that plugging a number into a polynomial gives you the exact remainder of the division. If $P(c) = 0$, then the Remainder $R = 0$. And just like regular numbers, if the remainder of division is 0, the thing you divided by $(x - c)$ is a perfect factor! Proof complete.

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How to Use the Factor Theorem

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

Step 1: Write the polynomial

Ensure your equation is written in standard form, $P(x) = ax^3 + bx^2 + cx + d$.

Step 2: Choose a possible root

You need a candidate number to test (the $c$ value). You shouldn’t just guess numbers randomly; use the Rational Root Theorem (factors of the last term divided by factors of the first term) to generate a smart list of candidates.

Step 3: Substitute into the polynomial

Plug your chosen $c$ value into every $x$ in the equation. Calculate the exponents first, then multiply by the coefficients.

Step 4: Evaluate

Add and subtract the resulting numbers.

Step 5: Determine whether it is a factor

• If the final number is exactly 0: You have found a factor! Write down $(x - c)$.
• If the final number is anything else (e.g., 7, -12, 0.5): It is NOT a factor. Cross that candidate off your list and pick a new number.

Step 6: Continue factoring

Once you find your first factor $(x - c)$, divide the original polynomial by $(x - c)$ using synthetic or long division. The result will be a smaller polynomial (a quadratic). You can then easily factor the quadratic.

Step 7: Verify the solution

Multiply your final factored pieces together. If they recreate the original polynomial, you have successfully solved the equation without errors.

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Factor Theorem for Cubic Equations

The Factor Theorem is the ultimate weapon against cubic equations ($x^3$), because cubic equations are usually too large to factor by simple grouping.

Finding one root: A cubic equation has three roots. The Factor Theorem’s job is simply to find the first one. For example, in $P(x) = x^3 - 6x^2 + 11x - 6$, you test $P(1)$. $P(1) = (1)^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0$. Boom. The root is 1. The factor is $(x - 1)$.

Reducing the polynomial: Now that you have $(x - 1)$, you divide the massive cubic equation by $(x - 1)$. Because the Factor Theorem guaranteed a remainder of 0, the division will be clean and perfect. The result of that division will be $x^2 - 5x + 6$.

Finding remaining roots: You have successfully “depressed” the cubic into a quadratic! You no longer need the Factor Theorem. You just factor $x^2 - 5x + 6$ using basic middle-school algebra into $(x - 2)(x - 3)$. The final factorization of the cubic is $(x - 1)(x - 2)(x - 3)$.

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Relationship with the Remainder Theorem

The Factor Theorem and the Remainder Theorem are two sides of the exact same mathematical coin.

Similarities: Both theorems require you to take a number $c$ and plug it into a polynomial $P(x)$ to find $P(c)$. Both theorems bypass polynomial long division.

Differences:

• The Remainder Theorem states that the answer you get when you calculate $P(c)$ is the mathematical remainder you would have gotten if you had divided the polynomial by $(x-c)$.
• The Factor Theorem is simply what happens when that remainder is specifically zero.

Why the Factor Theorem is a special case: The Factor Theorem is literally just the Remainder Theorem applied to the specific scenario where the answer is 0.

Comparison Table

Feature	Remainder Theorem	Factor Theorem
Primary Goal	To find the remainder of division without actually dividing.	To prove if a specific expression is a clean factor of the polynomial.
Calculation	Evaluate $P(c)$.	Evaluate $P(c)$.
If $P(c) = 7$	The remainder of division is 7.	$(x-c)$ is NOT a factor.
If $P(c) = 0$	The remainder of division is 0.	$(x-c)$ IS a perfect factor.

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Relationship with the Rational Root Theorem

The Factor Theorem is useless if you don’t know which numbers to test. The Rational Root Theorem is the engine that provides those numbers.

Finding candidate roots: The Rational Root Theorem generates a list of possible roots using the formula $\pm p/q$, where $p$ is factors of the constant term, and $q$ is factors of the leading coefficient. If the equation ends in a 6, your candidates are $\pm 1, \pm 2, \pm 3, \pm 6$.

Testing candidates efficiently: Instead of dividing the polynomial 8 different times to test those 8 numbers, you simply run them through the Factor Theorem. $P(1) = 4 \rightarrow$ Fail. $P(-1) = -12 \rightarrow$ Fail. $P(2) = 0 \rightarrow$ Success! You have found your factor $(x-2)$.

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Relationship with Synthetic Division

While plugging numbers into $P(x)$ is fast for small numbers like 1 or 2, it becomes a nightmare if you have to test $P(4)$ in the equation $x^3 - 12x^2 + 47x - 60$. Calculating $4^3$ and $12(4^2)$ by hand is tedious and prone to arithmetic errors.

Testing roots with Synthetic Division: Synthetic division is a visual shortcut for the Factor Theorem. Instead of calculating exponents, you perform a rapid series of simple additions and multiplications. The final number in the bottom row of a synthetic division bracket is the remainder.

Verifying factors: Because the final number in synthetic division is the remainder, it is exactly equal to $P(c)$. If the bottom-right number in your synthetic division is 0, the Factor Theorem has been satisfied, and you have proven the factor!

Reducing cubic equations: The massive advantage of using synthetic division to execute the Factor Theorem is that when you hit 0, the rest of the bottom row in your synthetic division bracket is your reduced quadratic equation. You verified the factor AND performed the division simultaneously!

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Relationship with Polynomial Long Division

Verification: If you prefer not to use synthetic division or plugging in numbers, you can physically execute polynomial long division. You set up a bracket, put $x^3 - 4x^2 + x + 6$ inside, and $(x - 2)$ outside.

Factoring: If you execute the entire long division process and the very final subtraction leaves a remainder of $0$, you have visually proven the Factor Theorem. The expression $(x-2)$ is a factor, and the quotient sitting on top of the bracket is the remaining piece of the polynomial.

Applications: Polynomial long division is mandatory when testing non-linear factors (like dividing a quartic $x^4$ by a quadratic $x^2 + 1$). The Factor Theorem and synthetic division only work for linear factors $(x - c)$.

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Factor Theorem and Graphs

How does the algebraic algebra of the Factor Theorem translate to the physical geometry of a graph?

X-intercepts & Roots: If $P(c) = 0$, then $c$ is a root. On a Cartesian graph, this means the physical curve of the polynomial perfectly intersects the x-axis exactly at the coordinate $(c, 0)$.

Factors: Every time a graph touches or crosses the x-axis, there is a hidden algebraic factor $(x - c)$ creating that geometry. A cubic graph that crosses at $1, 2,$ and $3$ is literally the physical embodiment of the factors $(x-1)(x-2)(x-3)$.

Multiplicity: If you test a factor and it works, and you divide it, and then you test the EXACT SAME factor on the remaining polynomial and it works again, you have found a repeated factor (e.g., $(x-5)^2$). The Factor Theorem just proved an even multiplicity, meaning the physical graph will “bounce” off the x-axis at $x=5$ instead of crossing it.

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Advantages and Limitations

Benefits:

• It is exponentially faster than polynomial long division.
• It prevents you from wasting time dividing by a binomial that isn’t actually a factor.
• It is the most reliable way to find the first root of a cubic equation.

Limitations:

• It only tests for specific numbers. If the roots of the equation are irrational decimals (like $\sqrt{3}$) or complex imaginary numbers (like $2i$), guessing numbers to test will be virtually impossible.
• It only tells you IF a factor exists. It does not actually perform the division to give you the remaining pieces of the polynomial.

Best use cases: The Factor Theorem is best used in tandem with the Rational Root Theorem to rapidly scan a list of integer and fraction candidates until a hit (a 0) is found.

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Common Mistakes

Students constantly make these errors when executing the Factor Theorem:

1. Sign errors with $c$: If testing the factor $(x + 3)$, the value to plug in is $c = -3$. Students mistakenly plug in $+3$ and get the wrong answer. Always flip the sign!
2. Arithmetic errors: When plugging negative numbers into exponents. $(-2)^2$ is $+4$. $(-2)^3$ is $-8$. Missing these negatives will ruin the evaluation.
3. Ignoring standard form: Applying the theorem when the equation isn’t set to zero or isn’t fully simplified.
4. Stopping before complete factorization: Finding $P(2)=0$, writing down $(x-2)$, and moving to the next problem. You must divide the equation and factor the remaining quadratic!
5. Misinterpreting results: Getting $P(c) = 1$ and assuming $c$ is a factor. Only exactly zero counts!

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Worked Examples

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

Group 1: Simple Polynomial Verification

Example 1: Proving a factor Is $(x-1)$ a factor of $P(x) = x^2 - 3x + 2$? Test $c = 1$. $P(1) = (1)^2 - 3(1) + 2 = 1 - 3 + 2 = 0$. Result: Yes, it is a factor.

Example 2: Proving a non-factor Is $(x+2)$ a factor of $P(x) = x^2 + 5x + 4$? Test $c = -2$. (Remember to flip the sign!). $P(-2) = (-2)^2 + 5(-2) + 4 = 4 - 10 + 4 = -2$. Result: No, $-2 \neq 0$. It is not a factor.

Example 3: Fractional factors Is $(2x - 1)$ a factor of $P(x) = 2x^2 + x - 1$? Set factor to 0: $2x - 1 = 0 \rightarrow x = 1/2$. Test $c = 1/2$. $P(1/2) = 2(1/4) + (1/2) - 1 = 1/2 + 1/2 - 1 = 0$. Result: Yes, it is a factor.

Example 4: Missing terms Is $(x-3)$ a factor of $P(x) = x^3 - 27$? Test $c = 3$. $P(3) = (3)^3 - 27 = 27 - 27 = 0$. Result: Yes, it is a factor.

Example 5: Negative exponents check Is $(x+1)$ a factor of $P(x) = x^4 + x^3 + x^2 + x$? Test $c = -1$. $P(-1) = (-1)^4 + (-1)^3 + (-1)^2 + (-1) = 1 - 1 + 1 - 1 = 0$. Result: Yes, it is a factor.

Group 2: Solving Cubic Equations

Example 6: Using Rational Root Theorem first Solve $x^3 - 6x^2 + 11x - 6 = 0$. Candidates: $\pm 1, \pm 2, \pm 3, \pm 6$. Test $c = 1$: $P(1) = 1 - 6 + 11 - 6 = 0$. Success! Factor is $(x-1)$.

Example 7: Dividing the cubic Following Example 6, divide $x^3 - 6x^2 + 11x - 6$ by $(x-1)$. Result is $x^2 - 5x + 6$. Factor the quadratic: $(x-2)(x-3)$. Final factorization: $(x-1)(x-2)(x-3)=0$.

Example 8: Finding a negative root Solve $x^3 + 4x^2 + x - 6 = 0$. Test $c = 1$: $P(1) = 1 + 4 + 1 - 6 = 0$. Factor is $(x-1)$. Divide by $(x-1)$ to get $x^2 + 5x + 6$. Factor quadratic to $(x+2)(x+3)$. Roots are $1, -2, -3$.

Example 9: Repeated roots Solve $x^3 - 3x + 2 = 0$. Test $c = 1$: $P(1) = 1 - 3 + 2 = 0$. Factor is $(x-1)$. Divide by $(x-1)$ to get $x^2 + x - 2$. Factor quadratic to $(x-1)(x+2)$. Notice $(x-1)$ appears twice! Roots are $1, 1, -2$.

Example 10: Imaginary leftover roots Solve $x^3 - x^2 + x - 1 = 0$. Test $c = 1$: $P(1) = 1 - 1 + 1 - 1 = 0$. Factor is $(x-1)$. Divide by $(x-1)$ to get $x^2 + 1$. The factor $(x^2+1)$ yields roots $\pm i$. Final roots are $1, i, -i$.

Group 3: Finding Unknown Constants

Example 11: Finding $k$ with a known factor Find $k$ if $(x-2)$ is a factor of $P(x) = x^3 - kx^2 + 2x - 4$. Since it is a factor, $P(2) = 0$. $P(2) = (2)^3 - k(2)^2 + 2(2) - 4 = 0$. $8 - 4k + 4 - 4 = 0$. $8 - 4k = 0 \rightarrow 4k = 8 \rightarrow k = 2$.

Example 12: Finding $k$ with a negative factor Find $k$ if $(x+3)$ is a factor of $x^3 + kx - 15$. $P(-3) = (-3)^3 + k(-3) - 15 = 0$. $-27 - 3k - 15 = 0$. $-42 - 3k = 0 \rightarrow -3k = 42 \rightarrow k = -14$.

Example 13: Finding $k$ in a quadratic Find $k$ if $(x-5)$ is a factor of $kx^2 - 11x + 5$. $P(5) = k(25) - 11(5) + 5 = 0$. $25k - 55 + 5 = 0 \rightarrow 25k - 50 = 0 \rightarrow 25k = 50 \rightarrow k = 2$.

Example 14: Two unknowns, two factors Find $a$ and $b$ if $(x-1)$ and $(x+2)$ are factors of $ax^3 + bx^2 - 5x + 2$. Condition 1 ($c=1$): $a(1) + b(1) - 5 + 2 = 0 \rightarrow a + b = 3$. Condition 2 ($c=-2$): $a(-8) + b(4) - 5(-2) + 2 = 0 \rightarrow -8a + 4b = -12 \rightarrow -2a + b = -3$. Subtract equations: $(a+b) - (-2a+b) = 3 - (-3) \rightarrow 3a = 6 \rightarrow a = 2$. Plug in $a=2$: $2 + b = 3 \rightarrow b = 1$.

Example 15: Finding the remainder of a non-factor Find $k$ if $(x-1)$ leaves a remainder of 5 in $x^3 + kx + 1$. (This is Remainder Theorem, but uses the same logic). $P(1) = 5 \rightarrow (1)^3 + k(1) + 1 = 5 \rightarrow 2 + k = 5 \rightarrow k = 3$.

Group 4: Higher Degree Polynomials

Example 16: Quartic evaluation Is $(x-2)$ a factor of $x^4 - 2x^3 + x^2 - 4$? $P(2) = 16 - 16 + 4 - 4 = 0$. Yes.

Example 17: Quintic evaluation Is $(x+1)$ a factor of $x^5 + 1$? $P(-1) = (-1)^5 + 1 = -1 + 1 = 0$. Yes.

Example 18: Testing 0 as a factor Is $x$ a factor of $x^3 - 4x^2 + 5x$? Testing $x$ means testing $c=0$. $P(0) = 0 - 0 + 0 = 0$. Yes, $x$ is a factor.

Example 19: Testing $(x-a)$ symbolically Is $(x-y)$ a factor of $x^3 - y^3$? Plug in $y$ for $x$: $P(y) = y^3 - y^3 = 0$. Yes, it is a factor.

Example 20: Factoring completely Factor $x^4 - 5x^2 + 4$. Test $c=1$: $1 - 5 + 4 = 0$. $(x-1)$ is a factor. Test $c=-1$: $1 - 5 + 4 = 0$. $(x+1)$ is a factor. Test $c=2$: $16 - 20 + 4 = 0$. $(x-2)$ is a factor. Test $c=-2$: $16 - 20 + 4 = 0$. $(x+2)$ is a factor. Final factorization: $(x-1)(x+1)(x-2)(x+2)$.

Group 5: Advanced and Real-World Scenarios

Example 21: Competition Math If $P(x)$ is a cubic polynomial where $P(1)=0, P(2)=0, P(3)=0$, and $P(0)=6$. Find $P(x)$. Because 1, 2, 3 are roots, $P(x) = a(x-1)(x-2)(x-3)$. Plug in 0: $P(0) = a(-1)(-2)(-3) = -6a = 6 \rightarrow a = -1$. Equation: $P(x) = -1(x-1)(x-2)(x-3)$.

Example 22: Physics limits A particle’s position is $p(t) = t^3 - 7t + 6$. Does the particle pass through the origin at $t=2$? Test $c=2$: $p(2) = 8 - 14 + 6 = 0$. Yes, $p(2)=0$, so $(t-2)$ is a factor and the particle passes through the origin.

Example 23: Economics break-even Profit is $P(x) = 2x^3 - 50x$. Do we break even at 5 units? Test $c=5$: $P(5) = 2(125) - 50(5) = 250 - 250 = 0$. Yes, profit is exactly 0.

Example 24: Complex roots via Factor Theorem Is $(x-i)$ a factor of $x^3 + x^2 + x + 1$? Test $c=i$: $(i)^3 + (i)^2 + i + 1 = -i - 1 + i + 1 = 0$. Yes!

Example 25: Geometric representation If you know $P(4) = 0$, what coordinates MUST exist on the graph of $P(x)$? The coordinate $(4, 0)$ must exist, proving an x-intercept.

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Practice Problems

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

Beginner Level

1. What value of $c$ should you plug in to test if $(x-4)$ is a factor?
2. What value of $c$ should you plug in to test if $(x+7)$ is a factor?
3. Is $(x-1)$ a factor of $x^2 - 1$?
4. Is $(x+2)$ a factor of $x^2 - 4$?
5. If $P(3) = 0$, what is the factor?
6. If $P(-5) = 0$, what is the root?
7. If $P(2) = 7$, is $(x-2)$ a factor?
8. Is $x$ a factor of $x^2 + x$?
9. What is the remainder when a polynomial is divided by a perfect factor?
10. Is $(x-1)$ a factor of $x^3 + x^2 + x + 1$?

Intermediate Level

11. Use the Factor Theorem to determine if $(x-3)$ is a factor of $2x^3 - 5x^2 - 4x + 3$.
12. Is $(2x+1)$ a factor of $4x^3 + 4x^2 - x - 1$?
13. If $(x-2)$ is a factor of $x^3 + kx - 6$, find $k$.
14. Use the Rational Root Theorem to list candidates for $x^3 - x^2 - 4x + 4$, then use the Factor Theorem to find the first factor.
15. If $(x+1)$ and $(x-2)$ are factors of a cubic equation with a leading coefficient of 1, and the constant is 4, what is the third factor?
16. Find the value of $m$ if $x^3 - mx^2 + 2x - 4$ is perfectly divisible by $(x-2)$.
17. A polynomial $P(x)$ has roots at $-1, 2, 4$. Write the factored form of $P(x)$ assuming $a=1$.
18. True or False: If $(x-c)$ is a factor of $P(x)$, then synthetic division by $c$ will have a remainder of 0.
19. Is $(x+3)$ a factor of $x^4 - 81$?
20. Why do we flip the sign of the number in the parentheses when testing for $P(c)$?

Advanced Level

21. Find $a$ and $b$ if $(x-1)$ and $(x+3)$ are factors of $x^3 + ax^2 + bx - 3$.
22. Prove that $(x-a)$ is always a factor of $(x^n - a^n)$ for any positive integer $n$.
23. If a cubic polynomial has factors $(x-2)$ and $(x+3)$, and passes through the point $(1, -8)$, what is the equation of the polynomial?
24. Explain algebraically why the Remainder Theorem proves the Factor Theorem.
25. Use the Factor Theorem to prove that $(x-1)$ is a factor of $nx^{n+1} - (n+1)x^n + 1$.
26. If $P(x) = x^3 - 3x^2 + 4$, prove that $(x-2)$ is a repeated factor (multiplicity of 2).
27. Factor $x^3 - 7x - 6$ completely using the Factor Theorem and division.
28. Can the Factor Theorem be used to test if $(x^2 - 1)$ is a factor of $x^4 - 1$? How?
29. A bridge’s structural integrity follows $S(x) = x^3 - 10x^2 + 31x - 30$. Does it fail (equal zero) at $x=5$?
30. If $P(c) = 0$ for a polynomial with purely positive coefficients, what must be true about the number $c$?

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Solutions to Practice Problems

Beginner Solutions:

1. $c = 4$.
2. $c = -7$.
3. Yes. $(1)^2 - 1 = 0$.
4. Yes. $(-2)^2 - 4 = 0$.
5. $(x-3)$.
6. Root is $-5$.
7. No. The remainder is 7, not 0.
8. Yes. $P(0) = 0^2 + 0 = 0$.
9. Exactly 0.
10. No. $P(1) = 1+1+1+1 = 4$.

Intermediate Solutions: 11. $P(3) = 2(27) - 5(9) - 4(3) + 3 = 54 - 45 - 12 + 3 = 0$. Yes. 12. Test $c = -1/2$. $P(-1/2) = 4(-1/8) + 4(1/4) - (-1/2) - 1 = -1/2 + 1 + 1/2 - 1 = 0$. Yes. 13. $P(2) = 8 + 2k - 6 = 0 \rightarrow 2 + 2k = 0 \rightarrow k = -1$. 14. Candidates: $\pm 1, \pm 2, \pm 4$. Test 1: $1 - 1 - 4 + 4 = 0$. First factor is $(x-1)$. 15. $(x+1)(x-2)(x-r_3) = x^3... - 2(r_3)$. If constant is 4, then $-2(r_3) = 4 \rightarrow r_3 = -2$. Third factor is $(x+2)$. 16. $P(2) = 8 - 4m + 4 - 4 = 0 \rightarrow 8 - 4m = 0 \rightarrow 4m = 8 \rightarrow m = 2$. 17. $P(x) = (x+1)(x-2)(x-4)$. 18. True. That is the definition of the Remainder/Factor theorem. 19. Test $c = -3$. $(-3)^4 - 81 = 81 - 81 = 0$. Yes. 20. Because the root $x = c$ comes from solving $(x - c) = 0$. To move $c$ to the other side of the equals sign, its sign must change.

Advanced Solutions: 21. $P(1) = 1 + a + b - 3 = 0 \rightarrow a + b = 2$. $P(-3) = -27 + 9a - 3b - 3 = 0 \rightarrow 9a - 3b = 30 \rightarrow 3a - b = 10$. Add equations: $4a = 12 \rightarrow a = 3$. Plug in: $3 + b = 2 \rightarrow b = -1$. 22. Plug $a$ in for $x$: $P(a) = a^n - a^n = 0$. Since it equals 0, $(x-a)$ is a factor. 23. $P(x) = a(x-2)(x+3)(x-r_3)$. More information is needed to find both $a$ and $r_3$. If it’s a standard parabola (quadratic), $P(x) = a(x-2)(x+3)$. $-8 = a(1-2)(1+3) \rightarrow -8 = a(-1)(4) \rightarrow -4a = -8 \rightarrow a=2$. Eq: $2(x-2)(x+3)$. 24. The Remainder Theorem states $P(x) = (x-c)Q(x) + R$, and $P(c) = R$. If $P(c) = 0$, then $R=0$, making $P(x) = (x-c)Q(x)$, which is the exact algebraic definition of a factor. 25. Test $c=1$: $n(1)^{n+1} - (n+1)(1)^n + 1 = n(1) - (n+1)(1) + 1 = n - n - 1 + 1 = 0$. 26. Test $c=2$: $8 - 12 + 4 = 0$. Factor is $(x-2)$. Divide by $(x-2)$ using synthetic division to get $x^2 - x - 2$. Factor that quadratic to get $(x-2)(x+1)$. Notice $(x-2)$ appeared twice! 27. Test $c=-1$: $-1 + 7 - 6 = 0$. Factor is $(x+1)$. Divide to get $x^2 - x - 6$. Factor quadratic to $(x-3)(x+2)$. Final: $(x+1)(x-3)(x+2)$. 28. Yes. Factor $(x^2-1)$ into $(x-1)(x+1)$. Then use the Factor Theorem twice, testing $c=1$ and $c=-1$. If both equal 0, the overall quadratic is a factor. 29. Test $c=5$: $125 - 250 + 155 - 30 = 280 - 280 = 0$. Yes, it fails at $x=5$. 30. The number $c$ must be negative. Adding positive coefficients with a positive $c$ value will never equal 0.

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