• Degree: The highest power of the variable
• Leading coefficient: The coefficient of the highest-degree term
• Constant term: The term with no variable ($a_0$)

Types of Polynomials

Factoring Techniques

1. Greatest Common Factor (GCF)

$6x^3 + 9x^2 = 3x^2(2x + 3)$

2. Difference of Squares

$a^2 - b^2 = (a + b)(a - b)$ Example: $x^2 - 25 = (x+5)(x-5)$

3. Perfect Square Trinomials

$a^2 + 2ab + b^2 = (a + b)^2$ $a^2 - 2ab + b^2 = (a - b)^2$ Example: $x^2 + 10x + 25 = (x+5)^2$

4. Trinomial Factoring ($x^2 + bx + c$)

Find two numbers that multiply to $c$ and add to $b$. Example: $x^2 + 7x + 12 = (x+3)(x+4)$

5. Trinomial Factoring ($ax^2 + bx + c$, $a \neq 1$)

Use the AC method or trial and error. Example: $2x^2 + 7x + 3 = (2x+1)(x+3)$

6. Sum and Difference of Cubes

$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

7. Factor by Grouping

$x^3 + 3x^2 + 2x + 6 = x^2(x+3) + 2(x+3) = (x^2+2)(x+3)$

Polynomial Operations

Addition/Subtraction

Combine like terms (same variable, same exponent).

Multiplication

Use FOIL for binomials, or distribute each term. $(2x + 3)(x - 4) = 2x^2 - 8x + 3x - 12 = 2x^2 - 5x - 12$

Division

Polynomial long division or synthetic division (for dividing by $x - c$).

Remainder Theorem

When polynomial $P(x)$ is divided by $(x - c)$, the remainder is $P(c)$.

Example: If $P(x) = x^3 - 2x + 1$, the remainder when divided by $(x - 3)$ is: $P(3) = 27 - 6 + 1 = 22$

Factor Theorem

$(x - c)$ is a factor of $P(x)$ if and only if $P(c) = 0$.

Example: Is $(x - 2)$ a factor of $x^3 - 4x^2 + x + 6$? $P(2) = 8 - 16 + 2 + 6 = 0$ → Yes!

SAT Question Types

Type 1: Factor a Polynomial

"Factor $x^2 - 5x - 6$" → $(x-6)(x+1)$

Type 2: Find Zeros from Factored Form

"If $f(x) = (x-2)(x+3)(x-5)$, what are the zeros?" → $x = 2, -3, 5$

Type 3: Polynomial Division

"What is the remainder when $x^3 + 2x - 5$ is divided by $x - 1$?" $P(1) = 1 + 2 - 5 = -2$ (use the Remainder Theorem!)

Type 4: Equivalent Expressions

"Which expression is equivalent to $(x+2)^3$?" $= x^3 + 3(x^2)(2) + 3(x)(4) + 8 = x^3 + 6x^2 + 12x + 8$

Common SAT Mistakes

1. Sign errors when factoring — double-check by FOILing your answer
2. Forgetting the GCF before trying other methods
3. Confusing $(x+3)^2$ with $x^2 + 9$ — it's $x^2 + 6x + 9$!
4. Not using the Remainder Theorem — much faster than long division
5. Dropping terms when subtracting polynomials — distribute the negative sign

Solution:

Use FOIL (First, Outer, Inner, Last):

F: $x \cdot x = x^2$ O: $x \cdot (-2) = -2x$ I: $6 \cdot x = 6x$ L: $6 \cdot (-2) = -12$

Combine: $x^2 - 2x + 6x - 12$ $= x^2 + 4x - 12$

Answer: $x^2 + 4x - 12$

SAT Tip: Don't forget to combine like terms after FOIL-ing!

Here: $a = 2x$, $b = 3$

$(2x)^2 = 4x^2$ $-2(2x)(3) = -12x$ $3^2 = 9$

Result: $4x^2 - 12x + 9$

Answer: C

Common trap: A is $(2x+3)(2x-3)$ (difference of squares, not square!)

SAT Tip: $(a-b)^2$ has THREE terms, not two! Don't forget the middle term.

Answer: The remainder is 11.

SAT Tip: The Remainder Theorem saves enormous time compared to polynomial long division!

Answer: The remainder is 11.

SAT Tip: The Remainder Theorem saves enormous time compared to polynomial long division!

$1 | 1 \quad -6 \quad 11 \quad -6$ $\quad | \quad 1 \quad -5 \quad 6$ $\quad 1 \quad -5 \quad 6 \quad 0$

Quotient: $x^2 - 5x + 6$

Step 3: Factor the quadratic: $x^2 - 5x + 6 = (x - 2)(x - 3)$

Answer: $f(x) = (x-1)(x-2)(x-3)$

The zeros are $x = 1, 2, 3$.

$1 | 1 \quad -6 \quad 11 \quad -6$ $\quad | \quad 1 \quad -5 \quad 6$ $\quad 1 \quad -5 \quad 6 \quad 0$

Quotient: $x^2 - 5x + 6$

Step 3: Factor the quadratic: $x^2 - 5x + 6 = (x - 2)(x - 3)$

Answer: $f(x) = (x-1)(x-2)(x-3)$

The zeros are $x = 1, 2, 3$.

Step 2: Use the point $(0, -16)$ to find $a$: $f(0) = a(0 + 1)(0 - 2)(0 - 4) = a(1)(-2)(-4) = 8a$ $-16 = 8a$ $a = -2$

Step 3: Write the final polynomial: $f(x) = -2(x+1)(x-2)(x-4)$

Check: $f(0) = -2(1)(-2)(-4) = -2(8) = -16$ ✓

Answer: $f(x) = -2(x+1)(x-2)(x-4)$

Expanded: $f(x) = -2x^3 + 10x^2 - 4x - 16$

Step 2: Use the point $(0, -16)$ to find $a$: $f(0) = a(0 + 1)(0 - 2)(0 - 4) = a(1)(-2)(-4) = 8a$ $-16 = 8a$ $a = -2$

Step 3: Write the final polynomial: $f(x) = -2(x+1)(x-2)(x-4)$

Check: $f(0) = -2(1)(-2)(-4) = -2(8) = -16$ ✓

Answer: $f(x) = -2(x+1)(x-2)(x-4)$

Expanded: $f(x) = -2x^3 + 10x^2 - 4x - 16$