Factoring

Section 4.4 Factoring

We often need to factor polynomial or other algebraic expressions. This means to rewrite sums and differences as a product. A monomial is a single term of a polynomial expression. When factoring polynomials, you might first look for the greatest common monomial of each term.

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Example 4.4.1. Factoring out a Monomial Term.

Factor

\begin{equation*} 3x^3 y^3 - 9 x^2 y^5 + 12 x y^7. \end{equation*}

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Solution.

Each term has at least one factor of \(3\text{,}\) one factor of \(x\text{,}\) and three factors of \(y\text{.}\) So we can factor the greatest common factor \(3 x y^3\)

\begin{equation} 3x^3 y^3 - 9 x^2 y^5 + 12 x y^7 = 3xy^3\left(x^2-3xy^2+4y^4\right)\tag{4.4.1} \end{equation}

Mentally redistribute the right side of (4.4.1) to confirm the result is correct.

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A common problem is factoring a quadratic trinomial of the form \(ax^2+bx+c\text{.}\) My recommendation is using a little trial and error while trying to reverse FOIL’ing. (See Example 2.4.4.)

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Example 4.4.2. Factoring a Trinomial.

Factor

\begin{equation*} x^2-2 x-15. \end{equation*}

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Solution.

The product of the first terms must give us \(x^2 = x\cdot x\text{,}\) so we start with

\begin{equation*} x^2-2 x-15 = (x - ??)(x-??). \end{equation*}

The product of the last two must give us \(-15\text{.}\) Put aside the sign for a moment and consider all the integer factorization of \(15\text{:}\)

\begin{equation*} 15 = 1\times 15, \quad 3\times 5. \end{equation*}

Try each one, adjusting the sign of each term. For example, none of

\begin{equation*} (x - 1)(x-15), \quad (x+1)(x-15), \quad (x-1)(x+15), \quad (x+1)(x+15) \end{equation*}

could work. The outer and inner terms all require adding or subtracting \(1x\) and \(15x\) which will never result in our middle term of \(-2x\text{.}\) So let’s try

\begin{equation*} (x\pm 3)(x \pm 5) \end{equation*}

which looks more promising since \(3x\) and \(5x\) will combine with the right sign to get \(-2x\text{.}\) Think about each possibility of sign to conclude

\begin{equation*} x^2-2 x-15 = (x + 3)(x - 5). \end{equation*}

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When the leading coefficient is not \(a=1\text{,}\) the possibilities increase. While there are nice methods for helping you factor, I still recommend trial and error.

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Example 4.4.3. Factoring a Trinomial.

Factor

\begin{equation*} 2 x^2-13 x+15. \end{equation*}

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Solution.

The first term factors as \(2x\cdot x\text{,}\) so we start with

\begin{equation*} (2x - ??)(x - ??). \end{equation*}

You should still consider the possibilities \(15 = 1\times 15\) and \(15 = 3\times 5\text{,}\) but observe now that the factor of \(2\) in the first term will double whatever we put in the second factor. I quickly ruled out using \(15 = 1\times 15\text{.}\) The remaining cases are of the form

\begin{equation*} (2x \pm 3)(x \pm 5), \quad (2x \pm 5)(x \pm 3). \end{equation*}

Try each one by FOIL’ing to find

\begin{equation*} 2 x^2-13 x+15 = (2x - 3)(x-5). \end{equation*}

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Look for special forms, like a difference of squares

\begin{equation} a^2 - b^2 = (a-b)(a+b)\tag{4.4.2} \end{equation}

or the result of expanding a perfect square

\begin{equation} a^2 + 2 ab + b^2 = (a+b)(a+b) = (a+b)^2\tag{4.4.3} \end{equation}

\begin{equation} a^2 - 2 ab + b^2 = (a-b)(a-b) = (a-b)^2\tag{4.4.4} \end{equation}

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Example 4.4.4. Factoring a Difference of Squares.

Factor

\begin{equation*} x^6 - 4y^2. \end{equation*}

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Solution.

Each term is a square

\begin{equation*} x^6 - 4y^2 = (x^3)^2 - (2 y)^2. \end{equation*}

So that

\begin{equation} x^6 - 4y^2 = (x^3-2y)(x^3 + 2y).\tag{4.4.5} \end{equation}

Again, mentally distribute the right side of (4.4.5) to confirm the result is correct.

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Example 4.4.5. Combining Methods.

Factor

\begin{equation*} 3 x^4+3 x^3-36 x^2. \end{equation*}

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Solution.

There is a common monomial factor of \(3x^2\text{:}\)

\begin{equation*} 3 x^4+3 x^3-36 x^2 = 3x^2 (x^2+x-12). \end{equation*}

Next, factor the quadratic

\begin{equation*} 3 x^4+3 x^3-36 x^2 = 3x^2 (x-3)(x+4). \end{equation*}

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Example 4.4.6. Combining Methods.

Factor

\begin{equation*} x^4-10 x^2+9. \end{equation*}

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Solution.

This is like a quadratic if we think of the first term as \(x^4 = (x^2)^2\text{.}\)

\begin{align*} x^4-10 x^2+9 &= (x^2)^2 -10x^2 + 9\\ &= (x^2-9)(x^2-1)\\ &= (x-3)(x+3)(x-1)(x+1). \end{align*}

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By creatively “splitting” the middle term of a quadratic you can sometimes create interesting factoring opportunities. One such method is called factoring by grouping.

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Example 4.4.7. Factor by Grouping.

Factor

\begin{equation*} 9x^2 + 18x+ 8, \end{equation*}

by observing that \(18 x = 6x + 12x\text{.}\)

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Solution.

First, using the advice

\begin{equation*} 9x^2 + 18x+ 8 = 9x^2 + 6x + 12x + 8. \end{equation*}

Now the first two terms have a common factor of \(3x\) and the second two terms have a common factor of \(4\text{.}\)

\begin{equation*} 9x^2 + 18x+ 8 = 3x(3x + 2) + 4(3x + 2). \end{equation*}

Cool! Each term now has a factor \(3x+2\text{,}\) so that

\begin{equation*} 9x^2 + 18x+ 8 = (3x+2)(3x+4). \end{equation*}

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