Factoring Techniques

Pull out Common Factors

Common Factors

The act of pulling out common factors from a sum can be thought of as applying the distributive property of multiplication in reverse.

\[AB + AC = A(B+C)\]

For example, \(21 + 14 = 7\cdot 3 + 7\cdot 2 = 7(3+2)\).

Video Resource

Factor Polynomials Using the Greatest Common Factor Method (Links to an external site)
A review of identifying and pulling out the greatest common factors when factoring a polynomial.

Video Resource

Factoring Algebraic Expressions with Rational Exponents (Links to an external site)
A review of identifying and pulling out the greatest common factors when factoring expressions with rational exponents.

Example 1

Factoring

Factor \(10x^5 + 15x^4\) by pulling out common factors.

Step 1:   Determine the factors common to both terms.

• Constant factors: \(10\) and \(15\) are both multiples of \(5\).

• Powers of \(x\): the smallest power of \(x\) is \(4\).

Therefore, the factor common to both terms is \(5x^4\).

Step 2:   Pull out the common factor of   \(5x^4\).

\[\begin{align*} 10x^5 + 15x^4 &= 5x^4\cdot 2x + 5x^4\cdot3 && \text{write each term as a multiple of $5x^4$} \\ &= 5x^4(2x + 3) && \text{pull out common factor} \end{align*}\]

Example 2

Factoring

Factor \(4x^3(2x^3+1)^5 + 30x^6(2x^3+1)^4\) by first pulling out common factors.

Step 1:   Determine the factors common to both terms.

• Constant factors: \(4\) and \(30\) are both multiples of \(2\).

• Powers of \(x\): the smallest power of \(x\) is \(3\).

• Powers of \(2x^3+1\): the smallest power of \(2x^3+1\) is \(4\).

Therefore, the factor common to both terms is \(2x^3(2x^3+1)^4\).

Step 2:   Pull out the common factor of   \(2x^3(2x^3+1)^4\).

\[\begin{align*} & 4x^3(2x^3+1)^5 + 30x^6(2x^3+1)^4 \\ &= 2x^3(2x^3+1)^4\cdot 2(2x^3+1) + 2x^3(2x^3+1)^4 \cdot 15x^3 \\%&& \text{write each term as a multiple of $2x^3(2x^3+1)^4$}\\ &= 2x^3(2x^3+1)^4[2(2x^3+1) + 15x^3] && \text{pull out common factor}\\ &= 2x^3(2x^3+1)^4(4x^3+2 + 15x^3) && \text{simplify expression inside $[ ~ ]$}\\ &= 2x^3(2x^3+1)^4(19x^3+2) \end{align*}\]

Example 3

Factoring

Factor \(4x^3(2x^3+1)^{1/2} + 3x^6(2x^3+1)^{-1/2}\) by first pulling out common factors.

Step 1:   Determine the factors common to both terms.

• Constant factors: \(4\) and \(3\) do not have any factors in common.

• Powers of \(x\): the smallest power of \(x\) is \(3\).

• Powers of \(2x^3+1\): the smallest power of \(2x^3+1\) is \(-1/2\).

Therefore, the factor common to both terms is \(x^3(2x^3+1)^{-1/2}\).

Step 2:   Pull out the common factor of   \(x^3(2x^3+1)^{-1/2}\).

\[\begin{align*} & 4x^3(2x^3+1)^{1/2} + 3x^6(2x^3+1)^{-1/2} \\ \\ &= x^3(2x^3+1)^{-1/2}\cdot 4(2x^3+1) + x^3(2x^3+1)^{-1/2} \cdot 3x^3 \\ \\ &= x^3(2x^3+1)^{-1/2}[4(2x^3+1) + 3x^3] && \text{pull out common factor}\\ \\ &= x^3(2x^3+1)^{-1/2}(8x^3+4 + 3x^3) && \text{simplify expression inside $[ ~ ]$}\\ \\ &= x^3(2x^3+1)^{-1/2}(11x^3+4) && \text{combine like terms}\\ \\ &= \frac{x^3(11x^3+4)}{\sqrt{2x^3+1}} \end{align*}\]

Difference of Squares

Applying the FOIL Method to the Product \((A+B)(A-B)\).

When applying the FOIL technique to a product of the form \((A+B)(A-B)\), we get the following result.

\[\begin{align*} (A+B)(A-B) &= A\cdot A - A\cdot B + B\cdot A - B\cdot B \\ &= A^2 \cancel{- AB} \cancel{+ AB} - B^2\\ &= A^2 - B^2 \end{align*}\]

We refer to \(A^2-B^2\) as a difference of squares, which can always be factored in the following manner.

\[A^2-B^2 = (A+B)(A-B)\]

Example 4

Applying difference of squares

Factor \(x^2-25\).

Step 1:   Rewrite the expression as a difference of squares.

\[x^2 - 25 = x^2 - 5^2\]

Step 2:   Apply the difference of squares formula with   \(A=x\)   and   \(B=5\).

\[\begin{align*} x^2 - 25 &= x^2 - 5^2 && \text{Step 1} \\ &= (x+5)(x-5) && \text{difference of squares} \end{align*}\]

Example 5

Applying difference of squares

Factor \(9x^3-4x^5\).

Step 1:   Determine the factors common to both terms.

• Constant factors: \(9\) and \(-4\) do not have any factors in common.

• Powers of \(x\): the smallest power of \(x\) is \(3\).

Therefore, the factor common to both terms is \(x^3\).

Step 2:   Pull out the common factor of   \(x^3\).

\[\begin{align*} 9x^3-4x^5 &= x^3(9) - x^3(4x^2)\\ &= x^3(9-4x^2) && \text{pull out common factor} \end{align*}\]

Step 3:   Rewrite   \(9-4x^2\)   as a difference of squares.

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

Step 4:   Apply difference of squares with   \(A=3\)   and   \(B=2x\).

\[\begin{align*} 9x^3-4x^5 &= x^3(9-4x^2) && \text{Step 2}\\ &= x^3[3^2-(2x)^2] && \text{Step 3} \\ &= x^3(3+2x)(3-2x) && \text{difference of squares} \end{align*}\]

AC Grouping Method

Steps of the AC Grouping Method.

The AC grouping method is a technique for factoring certain quadratic expressions of the form

\[ax^2 + bx + c.\]

The process is broken down into the following five steps.

1. Find two integers, \(r\) and \(s\), that multiply to \(a\times c\) and sum to \(b\).

2. Replace \(bx\) with \(rx\) + \(sx\).

3. Group the first two terms and the last two terms of \(ax^2 + rx + sx + c\).

4. Pull out common factors from each group.

5. Pull out common factor.

Example 6

Applying AC grouping

Factor \(6x^2 + 7x - 5\) using the AC grouping method.

Step 1:   Find two integers that multiply to   \(6(-5) = -30\)   and sum to   \(7\).

Since the product is negative, the two numbers must have opposite signs. And since the sum is positive, the larger number in absolute value, must be positive.

Product equals \(-30\)	Sum equals \(7\)?
\(-1 \times 30 = -30\)	\(-1 + 30 = 29\)     NO
\(-2 \times 15 =-30\)	\(-2 + 15 = 13\)     NO
\(-3 \times 10 =-30\)	\(-3 + 10 = 7\)     YES

Step 2:   Since   \(7 = -3+10\), replace the linear term,   \(7x\), with   \(-3x+10x\).

\[6x^2 - 3x + 10x - 5\]

Step 3:   Group the first two terms and the last two terms.

\[(6x^2 - 3x) + (10x - 5)\]

Step 4:   Pull out common factors from each group.

\[3x(2x-1) + 5(2x-1)\]

Step 5:   Pull out common factor of   \(2x-1\).

\[(2x-1)(3x+5)\]

Therefore,

\[6x^2 + 7x - 5 = (2x-1)(3x+5)\]

Check Our Work.

After factoring a polynomial, it’s always a good idea to check our work by expanding the product and making sure we get back what we started with.

\[\begin{align*} (2x-1)(3x+5) &= (2x)(3x) + (2x)(5) + (-1)(3x) + (-1)(5) && \text{FOIL}\\ &= 6x^2 + 10x - 3x - 5 && \text{simplify}\\ &= 6x^2 + 7x -5 && \text{combine like terms} \end{align*}\]

Example 7

Applying AC grouping

Factor \(x^2 - 13x + 36\) using the AC grouping method.

Step 1:   Find two integers that multiply to   \(1\times 36 = 36\)   and sum to   \(-13\).

Since the product is positive, the two numbers must have the same sign. And since the sum is negative, both numbers must be negative.

Product equals \(36\)	Sum equals \(-13\)?
\(-1 \times -36\)	NO
\(-2 \times -18\)	NO
\(-3 \times -12\)	NO
\(-4 \times -9\)	YES

Step 2:   Since   \(-13 = -4-9\), replace the linear term,   \(-13x\), with   \(-4x-9x\).

\[x^2 - 4x - 9x + 36\]

Step 3:   Group the first two terms and the last two terms.

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

Step 4:   Pull out common factors from each group.

\[x(x-4) + (-9)(x-4)\]

Step 5:   Pull out common factor of   \(x-4\).

\[(x-4)(x-9)\]

Therefore,

\[x^2 - 13x + 36 = (x-4)(x-9)\]

Check Our Work.

\[\begin{align*} (x-4)(x-9) &= x^2 -9x -4x + 36 && \text{FOIL}\\ &= x^2 - 13x + 36 && \text{combine like terms} \end{align*}\]

A Special Case of the AC Grouping Method

Important Observation

If the coefficient of \(x^2\) is one (i.e., \(a=1\) in \(ax^2 + bx + c\)), then Step 1 of the \(AC\) grouping method is to find two numbers that multiply to \(c\) and sum to \(b\). If those numbers exist, call them \(r\) and \(s\), then the factorization can be written as

\[x^2 + bx + c = (x+r)(x+s)\]

where \(r+s = b\) and \(rs = c\).

Example 8

AC grouping

Factor \(x^2 - 4x - 12\).

Step 1:   Find two integers that multiply to   \(-12\)   and sum to   \(-4\).

Since the product is negative, the two numbers must have opposite signs. And since the sum is negative, the larger number in absolute value, must be negative.

Product equals \(-12\)	Sum equals \(-4\)?
\(1 \times -12\)	NO
\(2 \times -6\)	YES

Step 2:   Apply special case of AC Grouping.

Since the coefficient of \(x^2\) is one, then the factorization is given by

\[x^2 - 4x - 12 = (x+2)(x-6).\]

Check Our Work.

\[\begin{align*} (x+2)(x-6) &= x^2 -6x +2x - 12 && \text{FOIL}\\ &= x^2 - 4x - 12 && \text{combine like terms} \end{align*}\]

Example 9

AC grouping and pulling out common factors

Factor \(7x^4 + 35x^3 + 42x^2\).

Step 1:   Determine the factors common to all three terms.

• Constant factors: \(7\), \(35\), and \(42\) are all multiples of \(7\).

• Powers of \(x\): the smallest power of \(x\) is \(2\).

Therefore, the factor common to all three terms is \(7x^2\).

Step 2:   Pull out the common factor of   \(7x^2\).

\[\begin{align*} 7x^4 + 35x^3 + 42x^2 &= 7x^2(x^2) + 7x^2(5x) + 7x^2(6) \\ &= 7x^2(x^2 + 5x + 6) && \text{pull out common factor} \end{align*}\]

Step 3:   Factor   \(x^2 + 5x + 6\)   by finding two integers that multiply to   \(6\)   and sum to   \(5\).

Product equals \(6\)	Sum equals \(5\)?
\(1 \times 6\)	NO
\(2 \times 3\)	YES

Therefore,

\[\begin{align*} 7x^4 + 35x^3 + 42x^2 &= 7x^2(x^2 + 5x + 6) && \text{Step 2} \\ &= 7x^2(x+2)(x+3) && \text{AC grouping} \end{align*}\]

Check Our Work.

\[\begin{align*} 7x^2(x+2)(x+3) &= 7x^2(x^2 + 3x + 2x + 6) && \text{FOIL}\\ &= 7x^2(x^2 + 5x + 6) && \text{combine like terms}\\ &= 7x^2(x^2) + 7x^2(5x) + 7x^2(6) && \text{distribute $7x^2$}\\ &= 7x^4 + 35x^3 + 42x^2 && \text{simplify} \end{align*}\]