🍬Honors Algebra II Unit 1 Review

Algebraic expressions and factoring form the foundation for nearly everything else in Algebra II. Factoring, in particular, is a skill you'll use constantly when solving equations, simplifying rational expressions, and analyzing functions. This section covers polynomial classification, operations, factoring techniques, and solving quadratics by factoring.

Polynomial classification

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Polynomial definition and components

A polynomial is an expression made up of variables and coefficients, using only addition, subtraction, multiplication, and non-negative integer exponents. So $3x^2 + 5x - 7$ is a polynomial, but $x^{-1} + 2$ is not (because of the negative exponent).

A few key terms to know:

The degree of a term is the sum of the exponents on its variables. The degree of the polynomial is the highest degree among all its terms. For example, in $4x^3 + 2x - 1$ , the degree is 3.
The leading coefficient is the coefficient attached to the highest-degree term. In $4x^3 + 2x - 1$ , the leading coefficient is 4.
Standard form means writing the terms in descending order of degree, from left to right. Always rewrite polynomials this way before classifying them.

Classification by degree and number of terms

Polynomials get classified two ways: by their degree and by how many terms they have.

By degree:

Linear: degree 1 ( $ax + b$ )
Quadratic: degree 2 ( $ax^2 + bx + c$ )
Cubic: degree 3 ( $ax^3 + bx^2 + cx + d$ )

By number of terms:

Monomial: one term ( $3x^2$ , $-5y$ )
Binomial: two terms ( $x + 2$ , $3x^2 - 4y$ )
Trinomial: three terms ( $2x^2 + 3x - 1$ )

You can combine both labels. For instance, $x^2 - 9$ is a "quadratic binomial."

Polynomial operations

Polynomial definition and components, Identifying the Degree and Leading Coefficient of Polynomials | College Algebra

Addition and subtraction

Adding polynomials means combining like terms, which are terms that have the same variable(s) raised to the same exponent(s). For subtraction, distribute the negative sign to every term in the polynomial being subtracted, then combine like terms.

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

A common mistake on subtraction problems is forgetting to distribute the negative to every term in the second polynomial. Write it out explicitly if you need to.

Multiplication and division

To multiply polynomials, distribute each term in the first polynomial to every term in the second. For two binomials specifically, you can use FOIL (First, Outer, Inner, Last) as a shortcut.

$(2x + 3)(x - 1)$ : First $= 2x^2$ , Outer $= -2x$ , Inner $= 3x$ , Last $= -3$ . Combine: $2x^2 + x - 3$ .

The degree of the product equals the sum of the degrees of the factors. So a degree-2 polynomial times a degree-1 polynomial gives a degree-3 result.

Polynomial long division works similarly to numerical long division. You divide the leading term of the dividend by the leading term of the divisor, multiply back, subtract, and repeat.

$(x^2 + 3x - 4) \div (x + 4)$ : $x^2 \div x = x$ , then $x(x+4) = x^2 + 4x$ . Subtract to get $-x - 4$ . Then $-x \div x = -1$ , and $-1(x+4) = -x - 4$ . Subtract to get remainder 0. Result: $x - 1$ .

Polynomial factoring

Factoring reverses multiplication. You're taking a polynomial and rewriting it as a product of simpler polynomials. This is one of the most important skills in this course.

Polynomial definition and components, PCK Map for Algebraic Expressions - Mathematics for Teaching

Greatest common factor and grouping

Always start by pulling out the greatest common factor (GCF), the largest factor shared by every term.

$12x^3 + 18x^2 = 6x^2(2x + 3)$

Factoring by grouping is useful for four-term polynomials. Here's the process:

Split the polynomial into two groups of two terms.
Factor the GCF out of each group.
If both groups now share a common binomial factor, factor that out.

$2x^2 + 3x + 6x + 9$ $2 x^{2} + 3 x + 6 x + 9$
- Group: $(2x^2 + 3x) + (6x + 9)$
- Factor each group: $x(2x + 3) + 3(2x + 3)$
- Factor out the common binomial: $(x + 3)(2x + 3)$

If the two groups don't produce a common binomial, try rearranging the terms before grouping.

Special factoring patterns

Memorize these patterns. They show up constantly.

Difference of squares: $a^2 - b^2 = (a + b)(a - b)$ $a^{2} - b^{2} = (a + b) (a - b)$
- $x^2 - 9 = (x + 3)(x - 3)$
Perfect square trinomial: $a^2 \pm 2ab + b^2 = (a \pm b)^2$ $a^{2} \pm 2 ab + b^{2} = (a \pm b)^{2}$
- $x^2 + 6x + 9 = (x + 3)^2$
Sum/difference of cubes: $a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$ $a^{3} \pm b^{3} = (a \pm b) (a^{2} \mp ab + b^{2})$
- $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$

For the sum/difference of cubes, remember the pattern SOAP: Same sign, Opposite sign, Always Positive. The first factor uses the same sign as the original, the second factor's middle term uses the opposite sign, and the last term is always positive.

Factoring quadratic trinomials ( $ax^2 + bx + c$ ):

When $a = 1$ , find two numbers $m$ and $n$ such that $m + n = b$ and $mn = c$ . Then the factored form is $(x + m)(x + n)$ .

$x^2 + 5x + 6$ : You need two numbers that add to 5 and multiply to 6. That's 2 and 3, so $(x + 2)(x + 3)$ .

When $a \neq 1$ , you can use grouping: multiply $a \cdot c$ , find two numbers that add to $b$ and multiply to $ac$ , then split the middle term and factor by grouping.

Solving quadratic equations by factoring

Quadratic equations and the zero-product property

A quadratic equation has the standard form $ax^2 + bx + c = 0$ , where $a \neq 0$ .

The zero-product property says: if $AB = 0$ , then $A = 0$ or $B = 0$ (or both). This is what makes factoring useful for solving equations.

To solve a quadratic by factoring:

Rearrange the equation into standard form (everything on one side, zero on the other).
Factor the non-zero side completely.
Set each factor equal to zero and solve.
Check your solutions by substituting back into the original equation.

$x^2 - 5x + 6 = 0 \rightarrow (x - 2)(x - 3) = 0 \rightarrow x = 2$ or $x = 3$
$x^2 - 2x + 1 = 0 \rightarrow (x - 1)^2 = 0 \rightarrow x = 1$ (a repeated root)

Solutions and graphical representation

The solutions (or roots) of $ax^2 + bx + c = 0$ correspond to the x-intercepts of the parabola $y = ax^2 + bx + c$ . The discriminant, $b^2 - 4ac$ , tells you how many real solutions to expect:

Positive discriminant ( $b^2 - 4ac > 0$ ): two distinct real solutions. The parabola crosses the x-axis twice.
Zero discriminant ( $b^2 - 4ac = 0$ ): one repeated real solution. The parabola touches the x-axis at its vertex.
Negative discriminant ( $b^2 - 4ac < 0$ ): no real solutions (two complex solutions). The parabola doesn't touch the x-axis at all.

You won't always use the discriminant when factoring, but it's a quick way to check whether a quadratic can be factored over the reals before you spend time trying.

🍬Honors Algebra II Unit 1 Review