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Polynomials and Synthetic Division
The Polynomial Division Algorithm
Polynomial division mirrors numerical division, relating a dividend ($P(x)$) to its divisor ($d(x)$), quotient ($Q(x)$), and remainder ($R$).
$$P(x) = d(x)Q(x) + R$$
The Division Algorithm expresses a polynomial as the product of its divisor and quotient plus any remaining value.
$P(x)$=Dividend polynomial(unitless)
$d(x)$=Divisor polynomial(unitless)
$Q(x)$=Quotient polynomial(unitless)
$R$=Remainder constant (if divisor is linear)(unitless)
$R = 0$
→The divisor is a factor of the dividend.
Standard Form: Both polynomials must be written in descending order of degree before starting.
Degree Constraint: For the algorithm to finish, the degree of the remainder must be less than the degree of the divisor.
Quotient Degree: When dividing a degree-$n$ polynomial by a degree-$m$ divisor, the quotient has degree $n - m$.
Synthetic Division: The Shortcut
Synthetic division is a streamlined method for dividing by a linear divisor of the form $x - c$, focusing solely on coefficients.
$$x - c \implies \text{Use } c \text{ in the synthetic array}$$
The constant $c$ is the value that makes the divisor zero ($x - c = 0$).
$c$=Zero of the divisor(constant)
$x - c$=Linear binomial divisor(unitless)
$x + 3$
→$c = -3$ (Always flip the sign of the constant term).
Placeholder Zero: If a power of $x$ is missing in the dividend, a $0$ must be inserted in the synthetic array.
Leading Coefficient: The first coefficient of the dividend is always dropped straight down to begin the process.
Limitation: Synthetic division only works when the divisor is linear with leading coefficient $1$. For divisors like $2x - 1$, use long division or divide all coefficients by $2$ first.
The Remainder and Factor Theorems
The Remainder Theorem allows calculating $P(c)$ without division, while the Factor Theorem identifies roots where the remainder is zero.
$$P(c) = R$$
Evaluating a function at $c$ gives the same result as the remainder of $P(x) \div (x - c)$.
$P(c)$=Polynomial evaluated at $c$(value)
$R$=Remainder from division(value)
$P(c) = 0$
→$(x - c)$ is a factor and $c$ is a root.
Synthetic Substitution: Using synthetic division to find $P(c)$ is often faster than direct substitution for high-degree polynomials.
Zero of a Function: If $(x - c)$ is a factor, then $x = c$ is an x-intercept on the graph of $P(x)$.
Converse Direction: The Factor Theorem works both ways — $P(c) = 0$ proves $(x - c)$ is a factor, and $(x - c)$ being a factor proves $P(c) = 0$.
Finding Unknown Coefficients: Set $P(c) = R$ (given remainder) to build equations for unknown parameters like $k$ or $p$.
Depressed Equations and Complete Factoring
When a known root $c$ is used in synthetic division, the quotient polynomial (one degree lower) is called the depressed equation. Solving it finds the remaining roots.
$$P(x) = (x - c) \cdot Q(x) \implies \text{solve } Q(x) = 0$$
After removing one factor via synthetic division, the depressed equation $Q(x) = 0$ contains the remaining roots.
$Q(x)$=Depressed polynomial (quotient after division)(unitless)
$(x - c)$=Factor corresponding to known root $c$(unitless)
Multiple known roots
→Apply synthetic division repeatedly — each division reduces the degree by 1.
Depressed equation is quadratic
→Factor directly or use the quadratic formula to find remaining roots.
Degree Reduction: A degree-$n$ polynomial with $k$ known roots requires synthetic division $k$ times, leaving a degree-$(n - k)$ depressed equation.
Order Independence: Multiple synthetic divisions can be performed in any order — the final depressed equation is the same.
Forming Polynomial Equations from Roots
Given all roots of a polynomial, the equation can be reconstructed by multiplying the corresponding linear factors together.
$$P(x) = (x - r_1)(x - r_2) \cdots (x - r_n)$$
Each root $r_i$ contributes a factor $(x - r_i)$. Their product gives the complete polynomial.
$r_1, r_2, \ldots, r_n$=The $n$ roots of the polynomial(unitless)
$(x - r_i)$=Linear factor for root $r_i$(unitless)
Quadratic ($n = 2$)
→$x^2 - (r_1 + r_2)x + r_1 r_2 = 0$, i.e., $x^2 - Sx + P = 0$ where $S$ = sum and $P$ = product of roots.
Sum and Product Shortcut: For a quadratic with roots $\alpha, \beta$: the equation is $x^2 - (\alpha + \beta)x + \alpha\beta = 0$.
Scaling: Any constant multiple $k \cdot P(x) = 0$ has the same roots. The leading coefficient can be chosen freely.
Application Strategy
Worked Examples
Problem Statement
Use synthetic division to divide $3x^3 - 2x^2 + 5x - 4$ by $(x - 2)$.
Synthetic Division SetupStandard Form
Strategic Toolkit
$x - c \implies c = 2$
Step 1
Identify Coefficients
Extract coefficients from $3x^3 - 2x^2 + 5x - 4$.
$[3, -2, 5, -4]$
Step 2
Set Divisor Zero
For divisor $(x - 2)$, the value $c$ is 2.
$c = 2$
Step 3
Drop Leading Coefficient
Bring down the first coefficient $3$ to start the bottom row.
$\text{Bottom row: } [3, \_, \_, \_]$
Step 4
Multiply and Add (Column 2)
Multiply $3 \times 2 = 6$, then add $-2 + 6 = 4$.
$\text{Bottom row: } [3, 4, \_, \_]$
Step 5
Multiply and Add (Columns 3–4)
Repeat: $4 \times 2 = 8$, add $5 + 8 = 13$. Then $13 \times 2 = 26$, add $-4 + 26 = 22$.
$$ \begin{array}{c|rrrr} 2 & 3 & -2 & 5 & -4 \\ & & 6 & 8 & 26 \\ \hline & 3 & 4 & 13 & 22 \end{array} $$
Step 6
Interpret Results
The bottom row gives the quotient coefficients (one degree lower) and the remainder.
$Q(x) = 3x^2 + 4x + 13, \quad R = 22$
Problem Statement
Determine if $(x + 3)$ is a factor of $P(x) = x^4 + 3x^3 - x - 3$.
Factor TheoremPlaceholder Zeros
Strategic Toolkit
$(x - c) \text{ is a factor } \iff P(c) = 0$
Step 1
Identify Placeholder
The dividend is missing an $x^2$ term. Add $0x^2$.
$P(x) = 1x^4 + 3x^3 + 0x^2 - 1x - 3$
Step 2
Synthetic Setup
Use $c = -3$ for divisor $(x + 3)$.
$c = -3, \text{ coeffs} = [1, 3, 0, -1, -3]$
Step 3
Execute Division
Process the synthetic array steps.
$$ \begin{array}{c|rrrrr} -3 & 1 & 3 & 0 & -1 & -3 \\ & & -3 & 0 & 0 & 3 \\ \hline & 1 & 0 & 0 & -1 & 0 \end{array} $$
Step 4
Conclusion
Since the remainder is 0, the Factor Theorem confirms the binomial is a factor.
$R = 0 \implies (x + 3) \text{ is a factor}$
Problem Statement
Find $k$ if $P(x) = x^3 + kx^2 - 7x + 6$ has remainder $-4$ when divided by $(x + 2)$.
Remainder TheoremSubstitution
Strategic Toolkit
$$P(c) = R$$
Step 1
Identify $c$
The divisor is $(x + 2) = (x - (-2))$, so $c = -2$.
$c = -2$
Step 2
Apply Remainder Theorem
Set $P(-2) = -4$ and substitute.
$(-2)^3 + k(-2)^2 - 7(-2) + 6 = -4$
Step 3
Simplify
Evaluate each term: $-8 + 4k + 14 + 6 = -4$.
$4k + 12 = -4$
Step 4
Solve for $k$
Isolate $k$.
$4k = -16 \implies k = -4$
Problem Statement
Solve $x^4 - 5x^2 + 4 = 0$ given that $x = -1$ and $x = 2$ are roots.
Depressed EquationRepeated Synthetic Division
Strategic Toolkit
$$P(x) = (x - c_1)(x - c_2) \cdot Q(x)$$
Step 1
First Division ($c = -1$)
Divide $x^4 + 0x^3 - 5x^2 + 0x + 4$ by $(x + 1)$.
$$ \begin{array}{c|rrrrr} -1 & 1 & 0 & -5 & 0 & 4 \\ & & -1 & 1 & 4 & -4 \\ \hline & 1 & -1 & -4 & 4 & 0 \end{array} $$
Step 2
Second Division ($c = 2$)
Divide the depressed polynomial $x^3 - x^2 - 4x + 4$ by $(x - 2)$.
$$ \begin{array}{c|rrrr} 2 & 1 & -1 & -4 & 4 \\ & & 2 & 2 & -4 \\ \hline & 1 & 1 & -2 & 0 \end{array} $$
Step 3
Solve Depressed Quadratic
Factor $x^2 + x - 2 = 0$.
$(x + 2)(x - 1) = 0 \implies x = -2, 1$
Step 4
Complete Solution
Combine all roots.
$x \in \{-2, -1, 1, 2\}$
Memorize formulas
Division Algorithm
Relationship between dividend, divisor, quotient, and remainder.
Formula
$P(x) = (x - c)Q(x) + R$
Remainder Theorem
Find the remainder without performing full division.
Formula
$R = P(c)$
Factor Theorem
The condition for a binomial to be a factor.
Formula
$(x - c) \text{ is a factor } \iff P(c) = 0$
Quadratic from Roots
Form a quadratic equation from sum and product of roots.
Formula
$x^2 - (\alpha + \beta)x + \alpha\beta = 0$
Review
1
Divide $P(x) = x^3 - 4x^2 + 2x + 5$ by $(x - 2)$ using synthetic division. What is the resulting quotient?
Ans
$x^2 - 2x - 2$
Review
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Set up synthetic division with the divisor root
For $(x-2)$, use $c=2$ and list the polynomial coefficients in descending order. Then carry the coefficients through the synthetic division table.
$$[1, -4, 2, 5] \xrightarrow{c=2}$$
Read off the quotient coefficients
The synthetic division produces quotient coefficients $1, -2, -2$, with the final entry as the remainder. The quotient is formed from the non-remainder entries.
$$x^2 - 2x - 2$$
2
If $P(x) = 2x^4 - 5x^2 + 3x - 1$ is divided by $(x + 1)$, what is the remainder according to the Remainder Theorem?
Ans
$-7$
Review
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Use the Remainder Theorem
A divisor of the form $(x+1)$ means the remainder is $P(-1)$. Substitute $x=-1$ directly into the polynomial.
$$P(-1)=2(-1)^4-5(-1)^2+3(-1)-1$$
3
A student performs synthetic division on $P(x) = x^3 - 7x + 6$ divided by $(x - 2)$ and gets an error. Their top row of the synthetic array is $[1, -7, 6]$. What did they do wrong?
Ans
They forgot the placeholder zero for the $x^2$ term.
Review
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Insert a zero for the missing power
Synthetic division must include every power of $x$ in order. Because the $x^2$ term is missing, its coefficient is $0$, so the correct row is $[1, 0, -7, 6]$.
$$x^3 + 0x^2 - 7x + 6$$
4
Given that $(x - 3)$ is a factor of $P(x) = x^3 - kx^2 + 2x - 6$, find the value of $k$.
Ans
$k = 3$
Review
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Apply the factor theorem
If $(x-3)$ is a factor, then the polynomial must vanish at $x=3$. Substitute $3$ into $P(x)$ and set the result equal to zero.
$$P(3)=27-9k+6-6=0$$
5
If $(x - 1)$ and $(x + 3)$ are both factors of $g(x) = x^3 + px^2 + qx - 6$, what are the values of $p$ and $q$?
Ans
$p = 4, q = 1$
Review
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Translate each factor into a root equation
Because $(x-1)$ and $(x+3)$ are factors, the polynomial must satisfy $g(1)=0$ and $g(-3)=0$. This gives two equations in the two unknowns.
$$p+q=5 \quad \text{and} \quad 3p-q=11$$
Solve the system
Add the equations to eliminate $q$, then substitute back to find the second value. That yields the coefficient pair that matches both roots.
$$p=4, \quad q=1$$
6
Using synthetic division, solve $x^3 - 6x^2 + 11x - 6 = 0$ given that $x = 1$ is a root. What is the complete solution set?
Ans
$\{1, 2, 3\}$
Review
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Deflate the cubic with the known root
Since $x=1$ is a root, divide by $(x-1)$ using synthetic division. The remainder should be zero, leaving a quadratic quotient.
$$x^3-6x^2+11x-6=(x-1)(x^2-5x+6)$$
Factor the quadratic
Now solve the remaining quadratic by factoring. Its roots complete the solution set of the original cubic.
$$x^2-5x+6=(x-2)(x-3) \Rightarrow \{1,2,3\}$$
7
The polynomial $P(x) = x^3 - kx^2 + 5x - 3$ has remainder $7$ when divided by $(x - 1)$. What is $k$?
Ans
$k = -4$
Review
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Evaluate the polynomial at the divisor root
A remainder of $7$ for division by $(x-1)$ means $P(1)=7$. Substitute $x=1$ and solve the resulting linear equation in $k$.
$$1-k+5-3=7$$
8
If $-2$ and $2$ are roots of $P(x) = x^3 - 4x^2 + ax + b$, find $a + b$.
Ans
$12$
Review
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Use both roots to build two equations
Because $2$ and $-2$ are roots, substitute each value into the polynomial and set both expressions equal to zero. This produces a linear system in $a$ and $b$.
$$2a+b=8 \quad \text{and} \quad -2a+b=24$$
Solve for the sum
Add the equations to isolate $b$, then substitute back to get $a$. Once both coefficients are known, their sum is immediate.
$$a=-4, \quad b=16, \quad a+b=12$$
9
What is the polynomial equation of least degree with integer coefficients that has roots $2, -1$, and $4$?
Ans
$x^3 - 5x^2 + 2x + 8 = 0$
Review
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Write a factor for each root
Roots $2$, $-1$, and $4$ correspond to factors $(x-2)$, $(x+1)$, and $(x-4)$. Multiply these factors to get the least-degree polynomial.
$$(x-2)(x+1)(x-4)$$
10
If $(x - 2)$ and $(x + 2)$ are factors of $x^4 - 13x^2 + 36$, what are the other two factors?
Ans
$(x + 3)$ and $(x - 3)$
Review
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Factor the remaining quadratic form
After removing the known factors $(x-2)$ and $(x+2)$, the leftover polynomial is a difference of squares in $x^2$. Factor it to expose the remaining linear factors.
$$x^4-13x^2+36=(x^2-4)(x^2-9)$$
11
When the polynomial $x^3 + 2x^2 + kx + 4$ is divided by $(x - 2)$, the remainder is $14$. What is $k$?
Ans
$k = -3$
Review
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Substitute the divisor root
A divisor of $(x-2)$ means the remainder equals $P(2)$. Plug in $x=2$ and use the given remainder to form the equation.
$$8+8+2k+4=14$$
12
Solve $x^4 - 10x^2 + 9 = 0$ given that $x = 1$ and $x = -3$ are roots. Which of the following is NOT a root?
Ans
$x = 2$
Review
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Factor the biquadratic form
Rewrite the polynomial as a product of two quadratics in $x^2$. This makes the root set easy to list and compare with the answer choices.
$$x^4-10x^2+9=(x^2-1)(x^2-9)$$
List the roots and eliminate the false choice
The factors give roots $x=\pm 1$ and $x=\pm 3$. Since $2$ is not in that set, it is the value that is not a root.
$$\{-3,-1,1,3\} \Rightarrow x=2\text{ is not a root}$$
Question 1 of 12
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1
Divide $P(x) = x^3 - 4x^2 + 2x + 5$ by $(x - 2)$ using synthetic division. What is the resulting quotient?