5.4 Polynomial Operations, Division & Remainder Theorem

Polynomial Operations, Division & Remainder Theorem

Polynomials form a central pillar of the Advanced Math domain on the Digital SAT. Questions assess operations on polynomial expressions, dividing polynomials to find quotients and remainders, and utilizing the Polynomial Remainder Theorem and Factor Theorem to solve for unknown coefficients.

---

1. Polynomial Operations & Algebraic Expansion

A polynomial in one variable $x$ is an algebraic expression composed of terms in the form $a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$, where exponents are non-negative integers and coefficients are real numbers.

Core Operations:

1. Addition & Subtraction: Group and combine like terms (terms with the same variable raised to the exact same power). Always distribute negative signs across parenthetical subtractions:

   $$(4x^3 - 2x^2 + 5x - 7) - (2x^3 + 3x^2 - 4x + 1) = (4-2)x^3 + (-2-3)x^2 + (5 - (-4))x + (-7-1) = 2x^3 - 5x^2 + 9x - 8$$

2. Multiplication (Distributive / Box Method): Multiply every term of the first polynomial by every term of the second polynomial:

   $$(2x - 3)(x^2 + 4x - 5) = 2x(x^2 + 4x - 5) - 3(x^2 + 4x - 5)$$ $$= 2x^3 + 8x^2 - 10x - 3x^2 - 12x + 15 = 2x^3 + 5x^2 - 22x + 15$$

+-----------------------------------------------------------------------------+
|                      BOX METHOD MULTIPLICATION GRID                         |
|                                                                             |
|   Expression: (2x - 3)(x^2 + 4x - 5)                                        |
|                                                                             |
|                    x^2             +4x             -5                       |
|             +---------------+---------------+---------------+               |
|        2x   |    2x^3       |     +8x^2     |     -10x      |               |
|             +---------------+---------------+---------------+               |
|        -3   |    -3x^2      |     -12x      |     +15       |               |
|             +---------------+---------------+---------------+               |
|                                                                             |
|   Combine Diagonals: 2x^3 + (8x^2 - 3x^2) + (-10x - 12x) + 15               |
|                    = 2x^3 + 5x^2 - 22x + 15                                 |
+-----------------------------------------------------------------------------+

---

2. Polynomial Long Division & The Division Algorithm

When dividing a higher-degree polynomial $P(x)$ (the dividend) by a divisor $D(x)$, the result yields a quotient polynomial $Q(x)$ and a remainder $R(x)$:

$$P(x) = D(x) \cdot Q(x) + R(x) \quad \iff \quad \frac{P(x)}{D(x)} = Q(x) + \frac{R(x)}{D(x)}$$

• The degree of remainder $R(x)$ is strictly less than the degree of divisor $D(x)$.
• If divisor $D(x) = ax + b$ (degree 1), the remainder $R$ is a constant number.

Step-by-Step Long Division Example:

Divide $P(x) = 6x^2 + 7x - 5$ by $D(x) = 2x + 1$:

1. First Term: Divide $6x^2$ by $2x$ to get $3x$.
2. Multiply & Subtract: Multiply $3x(2x + 1) = 6x^2 + 3x$. Subtract: $(6x^2 + 7x) - (6x^2 + 3x) = 4x$.
3. Bring Down: Bring down $-5$ to get $4x - 5$.
4. Second Term: Divide $4x$ by $2x$ to get $+2$.
5. Multiply & Subtract: Multiply $2(2x + 1) = 4x + 2$. Subtract: $(4x - 5) - (4x + 2) = -7$.
6. Final Result: Quotient $Q(x) = 3x + 2$, Remainder $R = -7$. $$\frac{6x^2 + 7x - 5}{2x + 1} = 3x + 2 - \frac{7}{2x + 1}$$

---

3. The Polynomial Remainder Theorem

Polynomial long division can be time-consuming. When the divisor is a linear expression $(x - c)$, the Polynomial Remainder Theorem calculates the remainder in a single evaluation step:

$$\text{If a polynomial } P(x) \text{ is divided by } (x - c), \text{ the remainder is } R = P(c).$$

+-----------------------------------------------------------------------------+
|                    POLYNOMIAL REMAINDER THEOREM LOGIC                       |
|                                                                             |
|   Division Identity:   P(x) = (x - c) · Q(x) + R                            |
|                                                                             |
|   Evaluate at x = c:   P(c) = (c - c) · Q(c) + R                            |
|                        P(c) =   0     · Q(c) + R                            |
|                        P(c) = R                                             |
+-----------------------------------------------------------------------------+

Important Divisor Conversions:

• Divided by $(x - 3) \implies c = 3 \implies R = P(3)$
• Divided by $(x + 5) \implies c = -5 \implies R = P(-5)$
• Divided by $(2x - 1) \implies c = \frac{1}{2} \implies R = P(1/2)$
• Divided by $(ax - b) \implies c = \frac{b}{a} \implies R = P(b/a)$

---

4. The Factor Theorem

The Factor Theorem is the direct corollary of the Remainder Theorem when the remainder equals zero:

$$(x - c) \text{ is a factor of } P(x) \iff P(c) = 0$$

If $P(c) = 0$, then:

1. $(x - c)$ divides evenly into $P(x)$ with zero remainder.
2. $x = c$ is a root / zero of the polynomial equation $P(x) = 0$.
3. The point $(c, 0)$ is an $x$-intercept on the graph of $y = P(x)$.

---

5. Step-by-Step Worked SAT Exam Examples

Worked Example 1: Finding an Unknown Coefficient with Remainder Theorem

Problem: The polynomial $f(x) = 2x^3 - 3x^2 + kx - 8$ is divided by $(x - 2)$. If the remainder is $14$, what is the value of the constant $k$?

Step-by-Step Solution:

1. Identify divisor: Divisor is $(x - 2)$, so $c = 2$.
2. Apply Remainder Theorem: $f(2) = \text{Remainder} = 14$.
3. Substitute $x = 2$ into $f(x)$: $$f(2) = 2(2)^3 - 3(2)^2 + k(2) - 8 = 14$$
4. Simplify numerical terms: $$2(8) - 3(4) + 2k - 8 = 14$$ $$16 - 12 + 2k - 8 = 14 \implies 2k - 4 = 14$$
5. Solve for $k$: $$2k = 18 \implies k = 9$$

Worked Example 2: Equating Coefficients in Polynomial Identity

Problem: In the equation $(x - 3)(2x^2 + ax - 4) = 2x^3 + x^2 - 25x + 12$, where $a$ is a constant, what is the value of $a$?

Step-by-Step Solution:

1. Expand the left-hand side: $$(x - 3)(2x^2 + ax - 4) = x(2x^2 + ax - 4) - 3(2x^2 + ax - 4)$$ $$= 2x^3 + ax^2 - 4x - 6x^2 - 3ax + 12$$ $$= 2x^3 + (a - 6)x^2 + (-3a - 4)x + 12$$
2. Equate corresponding coefficients of $x^2$ from both sides: $$a - 6 = 1 \implies a = 7$$
3. Verify with the $x$-coefficient: $$-3(7) - 4 = -21 - 4 = -25$$ The $x$-coefficient matches $-25$ exactly, confirming $a = 7$.

---

6. Desmos Testing Strategies & Remainder Shortcuts