# Page 2 College-Level Mathematics Study Guide for the ACCUPLACER® test

#### Factoring

Factoring is the process of breaking down a unified whole into the product of its pieces. For example:

$24$ can be factored into $2 \cdot 2 \cdot 2 \cdot 3$

This particular case is also the prime factorization of the number. (Prime factorization is useful for finding both the greatest common factor and the least common multiple of a set of numbers).

When factoring an expression, always begin by looking for a greatest common factor among the term(s). The greatest common factor (GCF) can be “pulled out” of every term to create the product of the GCF with the remaining terms:

The GCF is $3x$, and when pulled out (divided) from each term, the resulting product is: $3x(x^2 + 2x + 3)$. Notice that when the $3x$ is distributed through the parenthesis, the original expression is attained.

Polynomials can sometimes be factored, and there are a few common forms that should be memorized:

$x^2 - y^2 = (x + y)(x - y)$, known as a difference of squares

$x^3 + y^3 = (x + y)(x^2 - xy + y^2)$, known as a sum of cubes

$x^3 - y^3 = (x - y)(x^2 + xy + y^2)$, known as a difference of cubes

Again notice that the distribution of the products on the right hand side yields the polynomials on the left hand side.

When provided a trinomial of the form: $ax^2 + bx + c$, where $a = 1$, the factors will be composed of binomials that contain 2 integers that multiply to $c$ and combine to $b$. The signs of the coefficients $b$ and $c$ can help in quickly narrowing the possibilities of the integer values.

If $c$ is negative, only one of the integers can be negative. Otherwise, both of the integers are negative or both of the integers are positive.

If $b$ is negative, the absolute value of the negative integer is larger than the positive integer. If $b$ is positive, the positive integer is larger than the absolute value of the negative integer.

For example: Factor $x^2 - 2x -15$

Because both $b$ and $c$ are negative, there will be one positive integer and one negative integer. First look for the factors of $15$: $1, 3, 5, 15$. Only $3$ and $5$ can combine to produce a $-2$, so the factored form of the expression is: $(x + 3)(x - 5)$. Apply the distributive property to verify the factors.

In cases where the $a$ coefficient is not equal to $0$ or $1$, use this method of factoring but consider the factors of the $a$ coefficient as well. The product of the first terms of the binomials must equal $ax^2$.

For example: Factor $6x^2 + 7x + 2$

Notice again that the $b$ and $c$ coefficients are positive, so both of the integer factors will be positive. The factors of $a = 6$: $1, 2, 3, 6$ and the factors of $c = 2$: $1, 2$.

Try $(x + 1)(6x + 2)$ but notice that $2x + 6x$ does not equal $7x$ so this cannot be the factored form.

Try $(3x + 2)(2x + 1) = 6x^2 + 3x + 4x + 2 = 6x^2 + 7x + 2$, which works out to be the answer.

In cases involving a 3rd degree polynomial that contains 4 terms, factoring by grouping might be necessary.

Consider: $x^3 + x^2 + 4x + 4$

This expression can be rewritten as $(x^3 + x^2) + (4x + 4)$, and the binomials inside the parentheses can be factored into $x^2(x + 1) + 4(x + 1)$. Notice that the $(x + 1)$ expression is common to both terms. Consequently, it can be factored out of both terms to yield $(x + 1)(x^2 + 4)$.