Next Generation Advanced Algebra and Functions Study Guide for the ACCUPLACER Test

Factoring

Factoring is an algebraic method for simplifying equations or expressions. A factored expression is almost always favorable to a non-factored expression, and as a result, you should always factor an expression as much as possible.

In the same way that an expression like \(10\) can be factored into \(2 \cdot 5\), an expression like \(3x^2\) can be factored into \(3 \cdot x \cdot x\), and by extension, an expression like \(3x^2 + 9x\) can be factored into \(3x(x + 3)\).

Notice that in each case, a factored expression can always be verified by distributing each of the factors through to find the original expression; this method can prove useful for checking answers on exam questions.

\(2 \cdot 5 = 10\), \(3 \cdot x \cdot x = 3x^2\), and \(3x(x + 3) = 3x^2 + 9x\)

Quadratics

The standard form of a quadratic equation is: \(Ax^2 + Bx + C = 0\), where \(A\), \(B\), and \(C\) are all constants. Notice that this is presented as an equation, meaning that we can solve for the variable \(x\). In cases where only the \(Ax^2 + Bx + C\) is given, the expression can be factored, but it cannot be solved.

There are a couple of cases to consider- the simpler case, when \(A = 1\), and the more complicated case when \(A \ne 1\).

Consider this expression: \(x^2 + 6x + 9\)

To factor the expression, find two factors, \(r\) and \(s\), that sum to the \(B\) coefficient \(6\) and multiply to the \(C\) coefficient \(9\). The resulting factored expression will be the product of two binomial expressions: \((x + r)(x + s)\).

Begin by finding the factors of \(9\):

\(1 \cdot 9\) and
\(3 \cdot 3\)

Now find which set of factors sums to \(6\):

\(1 + 9 = 10 \ne 6\) and
\(3 + 3 = 6\), so the factors are \(3\) and \(3\), and the binomial expression can be written as:
\((x + 3)(x + 3)\), which can also be written as \((x + 3)^2\).

One more example, with some negative signs: \(x^2 -3x - 10\)

As before, begin by finding the factors of the \(C\) coefficient \(-10\):

\(-1 \cdot 10\) or \(1 \cdot -10\) and
\(-2 \cdot 5\) or \(2 \cdot -5\).

Now find the factor pair that sums to \(-3\):

\(-1 + 10 = 9 \ne -3\), and \(1 + (-10) = -9 \ne -3\) and
\(-2 + 5 = 3\), and \(2 + (-5) = -3\)
So the factors are \(2\) and \(-5\), and the binomial expression can be written as:

\[(x + 2)(x - 5)\]

Notice that by distributing the first binomial through the second binomial results in:

\(x \cdot x + x \cdot -5 + 2 \cdot x + 2 \cdot -5 = x^2 -5x + 2x - 10 = x^2 - 3x - 10\), which is the original expression.

Now let’s consider the case when the \(A\) coefficient is not equal to 1:

\[6x^2 - 7x - 3\]

Just as in the case of \(A = 1\), it is necessary to find the factors of the \(C\) coefficient; however, in the case of \(A \ne 1\), it is also necessary to find the factors of the \(A\) coefficient. When fully factored, the expression will take on the form: \((ax + y)(bx + z)\), where \(a\), \(b\), \(y\), and \(z\) are constants.

In this case, the factors of \(C\):

\(-1 \cdot 3\) and \(1 \cdot -3\).

And the factors of \(A\):

\(1 \cdot 6\) and \(1 \cdot 6\), as well as

\(2 \cdot 3\) and \(2 \cdot 3\).

Also, like in the case of \(A = 1\), a pair of factors that multiply to \(C\) and sum to \(B\) is sought, but in this case, it is also necessary to have the first terms of the binomial expressions multiply to \(A\). It can also be seen that the \(B\) coefficient is negative and the \(C\) coefficient is negative; from this, it can be determined that if \(y\) is positive, then \(z\) is negative, or vice versa.

Let’s substitute some of the factors into the binomial expressions to test the results:

\((2x - 1)(3x + 3)\), and expanding: \(6x^2 + 6x - 3x - 3\)

Notice, here, that the sum of the middle terms gives a positive \(x\) coefficient. This informs us that the larger \(x\) coefficient in the binomial expression, \(3\), should be multiplied with the larger negative constant, \(-3\). Revising our test case:

\((3x + 1)(2x - 3)\), which expands to: \(6x^2 - 9x + 2x - 3 = 6x^2 - 7x - 3\), which is our original expression.

There is a particular style of quadratic expression known as the difference of squares that can be easily factored when the expression is recognized as such. Consider the following two examples:

\[x^2 - 9\]

This should be recognized as a difference of squares: \((x)^2 - (3)^2\), with \(x\) and \(3\) as the values squared. Difference of squares quadratics factor to the following form: \((x + a)(x - a)\), where \(a\) is the square root of the original square (in this case, \(9\)).

The same method can be applied to more complicated expressions as long as the expression is a difference of squares:

\[81b^2 - 144x^2y^4\]

Notice that each term in the expression can be rewritten as a term squared:

\(81b^2 = (9b)^2\) and \(144x^2y^4 = (12xy^2)^2\)

Following the factored format, the original difference can be written as:

\[(9b^2 + 12xy^2)(9b^2 - 12xy^2)\]

Occasionally, it is most efficient to factor a quadratic equation by employing a method called completing the square.

Consider this quadratic equation: \(x^2 + 8x - 5 = 0\)

The completing the square method works best when the B coefficient is an even value but can be used in other cases as well. The method entails moving the \(C\) coefficient to the other side of the equation, then dividing the \(B\) coefficient by \(2\) and adding the square of the result to both sides of the equation. Doing so enables the left side to be rewritten as a binomial squared, which can then be square rooted, and the \(x\) value can be isolated. Note the steps below:

\(x^2 + 8x - 5 = 0\), add 5 to both sides

\(x^2 + 8x = 5\), divide 8 by 2, square the result, and add it to both sides

\(x^2 + 8x + (\frac{8}{2})^2 = 5 + (\frac{8}{2})^2\), which becomes

\(x^2 + 8x + 16 = 21\), and the left side can now be rewritten as a binomial squared

\((x + 4)^2 = 21\), and taking the square root of both sides

\(\sqrt{(x + 4)^2} = \sqrt{21}\), which becomes

\(x + 4 = \pm \sqrt{21}\)
Recall that the square root of a positive value is both positive and negative.
Set \(x + 4\) equal to both the positive and negative values:

\(x + 4 = \sqrt{21}\) and \(x + 4 = -\sqrt{21}\)

so:

\(x = \sqrt{21} - 4\) and \(x = -\sqrt{21} - 4\)

Not all quadratics will factor nicely, and in such cases, it is necessary to employ a technique called the quadratic formula. However, this method is usually only necessary when asked to find the solutions to a quadratic equation. Remember, an expression like \(x^2 + 4x + 4\) can be factored, but it cannot be solved. On the other hand, if presented as an equation, \(x^2 + 4x + 4 = 0\), there exist values for \(x\) that, when substituted, yield a true statement.

Cubics

The standard form of a cubic equation is: \(Ax^3 + Bx^2 + Cx + D = 0\), where \(A\), \(B\), \(C\), and \(D\) are all constants. As with quadratics, a cubic equation can be solved for \(x\), and a cubic expression can be factored but not solved.

Just as there is a special case for quadratics (the difference of squares), there are some special cases for cubic equations called the difference of cubes and the sum of cubes. The first case is a binomial expression, in which the terms are added, where each term can be written as a value cubed: \(x^3 - 27\), for example. The second case is a binomial expression, in which the terms are added, where each term can be written as a value cubed: \(8a^3 + 64b^6\).

The difference of cubes factors to the following form: \((x^3 - y^3) = (x + y)(x^2 - xy + y^2)\).

The sum of cubes factors to the following form: \((x^3 + y^3) = (x + y)(x^2 - xy + b^2)\).

So, the example cases provided above would factor to the following expressions:

\[x^3 - 27 = (x)^3 - (3)^3 = (x + 3)(x^2 - 3x + 9)\]

and

\[8a^3 + 64b^6 = (2a)^3 + (4b^2)^3 = (2a + 4b^2)(2a - 8ab^2 + 16b^4)\]

In cases where the cubic is of the form \(Ax^3 + Bx^2 + Cx + D = 0\), it is worthwhile to know the factoring method of grouping. An example will illustrate the technique:

\[x^3 + 4x^2 - 9x - 36 = 0\]

Notice that the equation can be rewritten as:

\[(x^3 + 4x^2) - (9x + 36) = 0\]

The expressions within each parentheses can be factored to the following:

\[x^2(x + 4) - 9(x + 4) = 0\]

Notice that the expression \((x + 4)\) is common to both terms, meaning that it can be factored:

\[(x + 4)(x^2 - 9) = 0\]

The cubic has now been factored, but it is always wise to examine the resulting expression for any further possible factoring. In this case, you should recognize the second expression as a difference of squares, so the left side can be fully factored to:

\[(x + 4)(x + 3)(x - 3) = 0\]

and, if necessary, \(x\) can be solved by setting each expression equal to 0:

\(x + 4 = 0\), so \(x = -4\), and

\(x + 3 = 0\), so \(x = -3\), and

\(x - 3 = 0\), so \(x = 3\)

More advanced methods for factoring cubics will not be covered here.

Polynomials

A polynomial is an algebraic expression containing multiple terms; any exponents must be positive integers, and there cannot be division by a term containing a variable. A quadratic equation is a prime example of a polynomial.

When factoring any expression or equation, always begin by looking for a greatest common factor; if a GCF is present, factoring it out simplifies any additional factoring.

Consider the following polynomial:

\[3x^4 + 15x^3 - 9x^2 = 0\]

Notice that a \(3x^2\) is common to each term on the left side; factoring it out produces a quadratic that can then be factored using methods described above:

\[3x^2(x^2 + 5x - 3) = 0\]

and which can be solved by setting each expression equal to \(0\), and applying the quadratic formula to the trinomial expression in parenthesis:

\(3x^2 = 0\) and \(x^2 + 5x - 3 = 0\)