0.3: Expanding and Factoring

Goals:

Revisit like terms and how to combine them
Revisit how to use the distributive law to expand
Revisit the special product formulas to expand
Learn the four basic ways of factoring

Revisit the GCF method of factoring
Learn an intuitive method to factor $ax^2 + bx + c$
Revisit the three special factoring formulas
Discuss the grouping method of factoring

Determine how the number of terms informs the factoring method to try

Algebraic Expressions

Algebraic expressions are just operations, numbers, and variables mixed together. For example \[2x^2 - 3x + 4 \qquad \sqrt{x} + 10 \qquad (x + 1)(x + 2) \qquad (a + b + c)^2(x^2y + 3)\] are all algebraic expressions.

Adding and Subtracting

To add and subtract algebraic expressions, look for like terms.

Like terms
Like terms are terms with the same factors except for the numerical factor (called the coefficient).

Which are like terms?

$x^3$ and $3x^3$

These are like terms; they have the same factors $x^3$.
$-4x^2y^3$ and $3y^2x^2$

These are not like terms; left term has $y^3$ while right term has $y^2$.
$-(x+2)(x+1)$ and $3(x+1)(x+2)$

These are like terms; they have the same factors $(x+2)(x+1)$.

Remember: identify the factors. This is your guiding light.

To add like terms, we undo the distributive law: \[x^3 + 2x^3 = 1\cdot x^3 + 2\cdot x^3 = x^3(1 + 2) = 3x^3\]

In essence, add together the coefficients for like terms.

Add and subtract the following:

$x^3 + 2x^2 + 3x^3$

Combining like terms: \[x^3 + 2x^2 + 3x^3 = \boxed{4x^3 + 2x^2}\]
$(4x^2y + 3xy^2) - (3x^2y^2 - 2xy^2)$

Distribute the $-$ and combine like terms: \begin{align} (4x^2y + 3xy^2) - (3x^2y^2 - 2xy^2) &= 4x^2y + 3xy^2 - 3x^2y^2 + 2xy^2 \\&= \boxed{4x^2y+5xy^2 - 3x^2y^2} \end{align}
Suppose \[a = 4x^3 - x^2 \qquad b = - 2x^2 + x^3 - x\] Calculate $a - b$.

$-b$ is subtracting three terms. Remember the parentheses. \begin{align} a - b &= 4x^3 - x^2 - (-2x^2 + x^3 - x) \\&= 4x^3 - x^2 + 2x^2 - x^3 + x \\&= \boxed{3x^3 + x^2 + x} \end{align}

Expanding (Multiplying)

Expanding is just multiplying algebraic expressions. You will apply the distributive law multiple times.

In expanding problems, you convert factors into terms.

Look at the Distibutive Property: \[a\cdot(b + c) = a\cdot b + a \cdot c\]

The Property is saying for the factor $a$ outside the parenthesis, distribute it to each term within the parenthesis.

Expand the following:

$(2x + 1)(3x - 5)$

Using the distributive law twice: \begin{align*} \underbrace{(2x + 1)}_{a}(\underbrace{3x}_b - \underbrace{5}_c) &= (2x+1)3x - (2x+1)5 \\&= 6x^2 + 3x - 10x - 5 \\&= \boxed{6x^2 - 7x - 5} \end{align*} Don't forget parentheses!!
$(x^2 + 2x + 1)(x + 1)$

Using the distributive law twice: \begin{align*} (x^2 + 2x + 1)(x + 1) &= (x^2 + 2x + 1)x + (x^2 + 2x + 1)1 \\&= x^3 + 2x^2 + x + x^2 + 2x + 1 \\&= \boxed{x^3 + 3x^2 + 3x + 1} \end{align*}
Simplify the following: $3(x+1)^2 - (x - 2)x$

Structure is a little more complex than previous problems. Identify terms to break down a bigger problem into two smaller problems.
$3(x+1)^2$ is one problem, while $-(x-2)x$ is another problem. On each term, apply distributive law. \begin{align*} 3(x+1)^2 - (x - 2)x &= 3(x+1)(x+1) - (x-2)x && \text{rewriting for clarity} \\&= (3x+3)(x+1)-x^2+2x && \text{Dist. Prop. three times} \\&= (3x+3)x + (3x+3)1 - x^2 + 2x && \text{Dist. Prop.} \\&= 3x^2 + 3x + 3x + 3 - x^2 + 2x \\&= \boxed{2x^2 + 8x + 3} \end{align*}

This last problem was a lot of work. Fortunately, we have special formulas for dealing with $(x+1)^2$ quickly, instead of having to use distributive property twice.

Special Product Formulas

Here are three special factor structures that quickly expand.

$(A + B)(A - B) = A^2 - B^2$
$(A + B)^2 = A^2 + 2AB + B^2$
$(A - B)^2 = A^2 - 2AB + B^2$

Make sure to identify the local terms $A$ and $B$.

Expand the following using the formulas:

$(2x - 3)(2x + 3)$

Local terms $A = 2x$ and $B = 3$. Following formula 1: \[(2x-3)(2x+3) = (2x)^2 - 3^2 = \boxed{4x^2 - 9}\]
$2(2x + 1)^2$

Local terms $A = 2x$ and $B = 1$. Following formula 2: \begin{align*} 2(2x+1)^2 &= 2((2x)^2 + 2(2x)1 + 1^2) && \text{don't forget parentheses (multiplying into 3 terms)} \\&= 2(4x^2 + 4x + 1) && \text{Laws of Exponents} \\&= \boxed{8x^2 + 8x + 2} \end{align*}

This is incorrect: $(2x + 1)^2 = (2x)^2 + 1^2$
Exponents do not interact with terms.

Factoring

Consider this expanding problem: \[(x+4)(x+3) = x^2 + 7x + 12\] The expression structure on the left is all factors.

On the right, it is all terms.

If we start from the right and move leftwards instead, this process is called factoring.

Factoring converts terms into factors and is the reverse process of expanding.

Method 1: Greatest Common Factor (among terms)

GCF factoring works on two or more terms.

GCF Factoring

Identify all the terms (usually global).
Identify the factors the terms have in common (the GCF).
Factor out the common factor and write in the factor structure of the distributive property.

Factor each expression:

$3x^2 - 6x$

Global terms are $3x^2 = 3\cdot x \cdot x$ and $6x = 3 \cdot 2 \cdot x$.
The GCF is $3x$.
Factoring out $3x$ and writing in distributive property form: \[3x^2 - 6x = \boxed{3x(x - 2)}\]
$x^6 - 3x^4 + 20x^2$

Global terms are $x^6$, $3x^4$ and $20x^2$.
The GCF is $x^2$ because $20x^2$ doesn't have any more $x$'s to give.
Factoring out $x^2$ and writing in distributive property form: \[x^6 - 3x^4 + 20x^2 = \boxed{x^2(x^4 - 3x^2 + 20)}\]
$(2x + 4)(x-3) - 5(x-3)$

Global terms are $(2x+4)(x-3)$ and $5(x-3)$.
The GCF is $(x-3)$.
Factoring out $(x-3)$ and writing in distributive property form: \[(2x+4)(x-3)-5(x-3) = (x-3)((2x+4) - 5) = \boxed{(x-3)(2x-1)}\]

Once you do enough of these problems, you'll understand it's all about the fundamentals. The distributive law is key to understanding expanding/factoring.

Method 2: Factoring the form $ax^2 + bx + c$.

This method works on three terms. Some examples include \begin{align} &x^2 + 4x + 3 \\&x^2 - 4 \\&2(x+1)^2 - 7(x+1) + 3\end{align}

Intuition

Consider this expansion problem \begin{align}\definecolor{lavender}{RGB}{179,154,240} \definecolor{lightyellow}{RGB}{236, 231, 141} \definecolor{lightblue}{RGB}{142, 181, 227} (2x + 1)(3x - 5) &= {\color{lavender} 6 }x^2 \ {\color{lightblue}-} \ {\color{lightblue}10}x \ {\color{lightblue}+} \ {\color{lightblue}3}x \ {\color{lightyellow} -} \ {\color{lightyellow} 5} \\&= {\color{lavender}6}x^2 \ {\color{lightblue}-} \ {\color{lightblue}7}x \ {\color{lightyellow}-} \ {\color{lightyellow} 5}\end{align}

Here, ${\color{lavender} a = 6}, {\color{lightblue}b = -7}, {\color{lightyellow} c = -5}$.

Arrange the factors vertically and erase the $x$: \begin{align}(2 \ \ & \ + 1) \\ (3 \ \ & \ - 5) \end{align}

Insights:

First column product is
- \[2 \cdot 3 = 6 = {\color{lavender}a}\]
- \begin{align}(2 \ \ & \ + 1) \\ \downarrow \ & \ \ \ \ \\ (3 \ \ & \ - 5) \end{align}
Second column product is
- \[1 \cdot (-5) = -5 = {\color{lightyellow}c}\]
- \begin{align}(2 \ \ & \ + 1) \\ \ \ & \ \ \ \ \downarrow \\ (3 \ \ & \ - 5) \end{align}
Cross product and sum
- \begin{align}2 \cdot (-5) + 3 \cdot 1 &= {\color{lightblue}-10 + 3} \\&= -7 \\&= {\color{lightblue}b}\end{align}
- \begin{align}(2 \ \ & \ + 1) \\ \ & \hspace{-0.38em}{\LARGE\diagup} \hspace{-1.45em}{\LARGE\diagdown}\ \ \ \ \ \\ (3 \ \ & \ - 5) \end{align}

But we are trying to factor. Given $\color{lavender} a = 6$, you do not know beforehand that it must be $2\cdot 3$. It could be $1\cdot 6$!

Therefore you need to generate candidates for ${\color{lavender} a}$ and ${\color{lightyellow} c}$!

See this handout for more examples.

"new" X method

Given $ax^2 + bx + c$:

First, write the coefficient structure: \begin{align} ( \ \_ \ &\ \ \_ \ ) \\ (\ \_ \ &\ \ \_ \ ) \end{align} Then:

Generate candidates for $a$ (left column product, all possibilities of two numbers that multiply to $a$).
Generate candidates for $c$ (right column product, all possibilities of two numbers that multiply to $c$).
Choose the right candidates by checking the cross-product and sum against $b$.

Factor: \[x^2 + 7x + 12\]

Solution:

We first identify ${\color{lavender} a = 1}, {\color{lightblue}b = 7}, {\color{lightyellow}c = 12}$.

Make your coefficient structure
\begin{align} ( \ \_ \ &\ \ \_ \ ) \\ (\ \_ \ &\ \ \_ \ ) \end{align}

Candidates for ${\color{lavender} a = 1}$ (left column): \begin{array}{c} 1\\1 \end{array}
Candidates for ${\color{lightyellow}c = 12}$ (right column): \begin{array}{c|c|c|c|c|c} 1 & 12 & 2 & 6 & 3 & 4\\ 12 & 1 & 6 & 2 & 4 & 3 \end{array}
Now check candidates against ${\color{lightblue}b = 7}$ until cross-product is ${\color{lightblue}7}$: \begin{align}(1 \ \ & \ \ \ \ \ \ 1) \\ \ & \hspace{-0.38em}{\LARGE\diagup} \hspace{-1.45em}{\LARGE\diagdown}\ \ \ \ \ \\ (1 \ \ &\ \ \ \ 12) \end{align} $ 1 \cdot 12 + 1 \cdot 1 = 13$. Wrong choice, try another candidate. \begin{align}(1 \ \ & \ \ \ \ \ 4) \\ \ & \hspace{-0.38em}{\LARGE\diagup} \hspace{-1.45em}{\LARGE\diagdown}\ \ \ \ \ \\ (1 \ \ &\ \ \ \ \ 3) \end{align} $1\cdot 3 + 1 \cdot 4 = {\color{lightblue} 7}$. Matches with ${\color{lightblue} b}!$
Put $x$'s back into first column and signs in second column \begin{align}(1x \ \ & \ + \ 4) \\ \ & \ \ \ \ \ \\ (1x \ \ &\ + \ 3) \end{align} and write in one line: $\boxed{(x+4)(x+3)}$

If $a \neq 1$, you only need to check the "flipped" candidates only for left column or only for right column. If you check flipped candidates for both right/left column, you'll be checking repeated candidates. Two examples follow.

Factor: \[2x^2 - 5x + 2\]

We first identify ${\color{lavender} a = 2}, {\color{lightblue}b = -5}, {\color{lightyellow}c = 2}$.

Make your coefficient structure
\begin{align} ( \ \_ \ &\ \ \_ \ ) \\ (\ \_ \ &\ \ \_ \ ) \end{align}

Candidates for ${\color{lavender} a = 2}$ (left column): \begin{array}{c|c} 2 & 1 \\ 1 & 2 \end{array} Keep the flipped candidates.
Candidates for ${\color{lightyellow}c = 2}$ (right column): \begin{array}{c|c|c|c} {\color{grey}2} & -2 & {\color{grey}1} &{\color{grey}-2} \\ {\color{grey}1} & -1 & {\color{grey}2} & {\color{grey}-1} \end{array}
- Because ${\color{lightblue}b = -5}$ is negative, you'll need to account for the product ${\color{lightyellow}c = 2} = (-2)(-1)$ in order for the cross-product to be negative.
- Ignore the right column's flipped candidates (the last two greyed out candidates). Only the left column candidates or right candidates need to be flipped (in our case, left is flipped alread).
- The first candidate is greyed out because ${\color{lightblue}b}$ must be negative. Only the second candidate could generate a negative ${\color{lightblue}b}$.
Now check candidates against ${\color{lightblue}b = -5}$ until cross-product is ${\color{lightblue}5}$: \begin{align}(2 \ \ & \ \ -2) \\ \ & \hspace{-0.38em}{\LARGE\diagup} \hspace{-1.45em}{\LARGE\diagdown}\ \ \ \ \ \\ (1 \ \ &\ \ -1) \end{align} $ 2 \cdot (-1) + 1 \cdot (-2) = 4$. Wrong choice, try another candidate. \begin{align}(1 \ \ & \ \ -2) \\ \ & \hspace{-0.38em}{\LARGE\diagup} \hspace{-1.45em}{\LARGE\diagdown}\ \ \ \ \ \\ (2 \ \ &\ \ -1) \end{align} $1 \cdot (-1) + 2\cdot (-2) = {\color{lightblue}-5}$. Matches with ${\color{lightblue} b}$!
Put $x$'s back into first column and signs in second column \begin{align}(1x \ \ & \ - \ 2) \\ \ & \ \ \ \ \ \\ (2x \ \ &\ - \ 1) \end{align} and write in one line: $\boxed{(x-2)(2x-1)}$

Factor: \[6x^2 + 7x - 5\]

We first identify ${\color{lavender} a = 6}, {\color{lightblue}b = 7}, {\color{lightyellow}c = -5}$.

Make your coefficient structure
\begin{align} ( \ \_ \ &\ \ \_ \ ) \\ (\ \_ \ &\ \ \_ \ ) \end{align}

Candidates for ${\color{lavender} a = 6}$ (left column): \begin{array}{c|c|c|c} 6 & 3 & 2 & 1 \\ 1 & 2 & 3 & 6 \end{array} Keep the flipped candidates.
Candidates for ${\color{lightyellow}c = -5}$ (right column): \begin{array}{c|c|c|c} -5 & -1 & {\color{grey} 5} & {\color{grey} 1} \\ 1 & 5 & {\color{grey} -1} & {\color{grey} -5} \end{array}
Here, the last two candidates are greyed out because we don't have to check them. They are flipped from the first two.
Advice: If ${\color{lightyellow}c}$ is negative (in our case), then you should look at ${\color{lightblue}b}$ before checking candidates. Here, ${\color{lightblue}b = 7}$. So we should not pick the candidate \begin{align}-5 \\ 1\end{align} first when checking because $-5$ is far from our required result of ${\color{lightblue}b = 7}$. That leaves one good candidate for the right column.
Now check candidates against ${\color{lightblue}b = 7}$ until cross-product is ${\color{lightblue}7}$: \begin{align}(6 \ \ & \ \ -1) \\ \ & \hspace{-0.38em}{\LARGE\diagup} \hspace{-1.45em}{\LARGE\diagdown}\ \ \ \ \ \\ (1 \ \ &\ \hphantom{+} \ 5) \end{align} $ 6 \cdot 5 + 1 \cdot (-1) = 29$. Wrong choice, try another candidate. \begin{align}(3 \ \ & \ \ -1) \\ \ & \hspace{-0.38em}{\LARGE\diagup} \hspace{-1.45em}{\LARGE\diagdown}\ \ \ \ \ \\ (2 \ \ &\ \hphantom{+} \ 5) \end{align} $3 \cdot 5 + 2\cdot (-1) = 13$. Getting close to 7. Try again: \begin{align}(2 \ \ & \ \ -1) \\ \ & \hspace{-0.38em}{\LARGE\diagup} \hspace{-1.45em}{\LARGE\diagdown}\ \ \ \ \ \\ (3 \ \ &\ \hphantom{+} \ 5) \end{align} $2 \cdot 5 + 3\cdot (-1) = {\color{lightblue} 7}$. Matches with ${\color{lightblue} b}$!
Put $x$'s back into first column and signs in second column \begin{align}(2x \ \ & \ - \ 1) \\ \ & \ \ \ \ \ \\ (3x \ \ &\ + \ 5) \end{align} and write in one line: $\boxed{(2x-1)(3x+5)}$

I highly recommend this method. It handles all cases of $ax^2 + bx + c$, even when $b$ or $c$ is negative. It is also quick to check cross-products, so you can find the factors quickly.

Method 3: Special Factoring Formulas

The three special product formulas reversed give the special factoring formulas.

$A^2 - B^2 = (A-B)(A+B)$
$A^2 + 2AB + B^2 = (A+B)^2$
$A^2 - 2AB + B^2 = (A-B)^2$

Factor:

$4x^2 - 25$

We have two terms, so we will use $A^2 - B^2$. Our task is to determine $A$ and $B$.
Line the terms up, giving $A^2 = 4x^2$, and $B^2 = 25$.
We need to write $4x^2$ as $(\text{factor})^2$. But currently, only the $x$ is squared. By the exponent laws, we can write \[A^2 = 4x^2 = 2^2x^2 = (2x)^2\] and now it is one factor squared, giving $A = 2x$.
Now $B^2 = 25 = 5^2$ so $B = 5$. Using the factoring formula: \[4x^2 - 25 = \boxed{(2x - 5)(2x + 5)}\]
$x^2 + 6x + 9$

Three terms with middle term positive, try the second formula. Our task is to determine $A$ and $B$, and check against $2AB$.
Line the terms up, giving $A^2 = x^2$, and $B^2 = 9 = 3^2$. Therefore $A = x$ and $B = 3$.
Now check if $2AB = 6x$: \[2AB = 2\cdot x \cdot 3 = 6x \checkmark\] We're good to use the formula: \[x^2 + 6x + 9 = \boxed{(x + 3)^2}\]

Method 4: Grouping

This method works on four terms. First order terms in descending powers.

We group first two/last two terms together, then use GCF on both groups.

Factor:

$x^3 + x^2 + 4x + 4$

\begin{align} x^3 + x^2 + 4x + 4 &= (x^3 + x^2) + (4x + 4) && \text{Group first/last two terms} \\ &= x^2(x+1) + 4(x+1) && \text{GCF factoring on both groups} \\&= \boxed{(x+1)(x^2 + 4)} && \text{GCF factoring on global terms} \end{align}
$x^3 - 2x^2 - 9x + 18$

\begin{align} x^3 - 2x^2 - 9x + 18 &= (x^3 - 2x^2) + (- 9x + 18) && \text{Group first/last two terms} \\ &= x^2(x-2) -9(x-2) && \text{GCF factoring on both groups (note the $-9$)} \\&= (x-2)(\underbrace{x^2}_{A^2} - \underbrace{9}_{B^2}) && \text{GCF factoring on global terms} \\&= \boxed{(x-2)(x-3)(x+3)} && \text{Special Factoring Formula 1} \end{align}

The last example shows even though we did have all factors, we sometimes can keep factoring a factor. We may need to use multiple methods.

Multiple Methods

Remember, before even starting a factoring problem, identify the terms and how many there are.

Doing so helps you to choose a correct strategy. Here's what I use (in order from left to right):

Number of Terms	Factoring Methods to Try (in order)
2 terms	GCF, $A^2 - B^2$
3 terms	GCF, $\text{"new" X method}, A^2 + 2AB + B^2, A^2 - 2AB + B^2$
4 terms	GCF, grouping
$\geq$ 5 terms	GCF

GCF factoring is the most ubiquitous factoring method; always try it first!

Factor:

$2x^4 - 8x^2$
$3x^2(x-1) - 16x(x-1) + 5(x-1)$