Expanding and Factoring for a Purpose, foundations for calculus lesson

The same expression can be written in several ways, and each way is good at something different. The factored form $(x - 2)(x + 2)$ shows at a glance where the expression is zero and where it changes sign. The expanded form $x^2 - 4$ is easier to add to or subtract from other polynomials, term by term. Calculus asks you to switch between these forms all the time, so the useful skill is choosing the form that fits the job and checking that nothing changed along the way.

Key terms

Term

Definition The terms of an expression are the parts that are added together at the top level. A subtraction counts as adding the opposite, so in $a - b$ the terms are $a$ and $-b$ . A part inside parentheses belongs to a single term until the parentheses are removed.

In plain words Terms are the pieces separated by plus and minus signs that are not hidden inside parentheses. Each term carries its own sign.

Example The expression $3x^2 - 5x + 2$ has three terms, $3x^2$ , $-5x$ and $2$ . The expression $3(x + 2)$ has only one term, a product, even though $x + 2$ has two terms of its own inside.

Factor

Definition The factors of an expression are the parts that are multiplied together at the top level. For polynomials, a polynomial $q$ is a factor of a polynomial $P$ when $P = q \cdot r$ for some polynomial $r$ .

In plain words Factors are joined by multiplication, terms are joined by addition. Cancelling in a fraction works only with a factor shared by the whole numerator and the whole denominator, and only where that factor is not zero. It never works with terms.

Example In $3x(x + 2)$ the factors are $3$ , $x$ and $x + 2$ . In $x^2 + 2x$ , by contrast, $x$ is a factor of each term but the expression as written is a sum. Writing it as $x(x + 2)$ makes $x$ a factor of the whole expression. This is why

\frac{x^2 + 2x}{x} = x + 2, \qquad x \ne 0,

works, while cancelling the $x$ in $\frac{x + 2}{x}$ would be wrong, because there $x$ is a term of the numerator, not a factor.

Distributive property

Definition For all real numbers $a$ , $b$ and $c$ ,

a(b + c) = ab + ac.

Read left to right it multiplies out a product. Read right to left it pulls out a common factor.

In plain words Multiplying a sum means multiplying each of its terms. Going backwards, a factor that every term shares can be taken outside parentheses.

Example Left to right, $4x(x - 3) = 4x^2 - 12x$ . Right to left, $6x^2 + 9x = 3x(2x + 3)$ . Check the second one by distributing again, since $3x \cdot 2x + 3x \cdot 3 = 6x^2 + 9x$ .

Expansion

Definition To expand a product is to rewrite it as a sum of terms by applying the distributive property until no product of sums remains, then combining like terms. For two sums, every term of the first is multiplied by every term of the second.

In plain words Multiply everything by everything, then collect what matches.

Example Expanding $(x + 3)(x - 5)$ gives four products before combining,

\begin{aligned} &(x + 3)(x - 5) \\[0.4em] &\quad = x^2 - 5x + 3x - 15 \\[0.4em] &\quad = x^2 - 2x - 15. \end{aligned}

A quick test at $x = 1$ gives $(4)(-4) = -16$ on the left and $1 - 2 - 15 = -16$ on the right. A test like this can catch a mistake, but agreement at one number does not prove two expressions are equal.

Square of a sum

Definition For all real numbers $a$ and $b$ ,

(a + b)^2 = a^2 + 2ab + b^2,

and, replacing $b$ by $-b$ ,

(a - b)^2 = a^2 - 2ab + b^2.

The middle term $2ab$ comes from two separate products, $ab$ and $ba$ , when $(a + b)(a + b)$ is expanded. A square of side $a + b$ with $a > 0$ and $b > 0$ splits into a square of area $a^2$ , a square of area $b^2$ and two rectangles of area $ab$ each. That picture needs nonnegative lengths, but the identity itself holds for every real $a$ and $b$ .

In plain words Squaring a sum is not squaring each piece. You also get twice the product of the two pieces, the cross term.

Example Expanding gives $(x + 5)^2 = x^2 + 10x + 25$ . The tempting wrong answer $x^2 + 25$ fails already at $x = 1$ , where $(1 + 5)^2 = 36$ but $1 + 25 = 26$ . Calculus uses this identity in difference quotients, for instance

(3 + h)^2 = 9 + 6h + h^2.

Factorization

Definition To factor an expression is to rewrite it as a product. A polynomial with real coefficients is factored completely over the reals when it is written as a constant times nonconstant polynomials with real coefficients, none of which can be written as a product of two real polynomials of lower degree. Over the reals such unbreakable factors are the linear ones and the quadratics with no real zero. A quadratic factor $px^2 + qx + r$ with $p \ne 0$ can be split further over the reals exactly when it has a real zero, that is, when $q^2 - 4pr \ge 0$ .

In plain words Factoring is expansion run backwards. Keep going until no piece can be broken down further using real numbers.

Example Start with $2x^3 - 8x$ . Pull out the common factor $2x$ , then split the difference of squares that remains,

\begin{aligned} 2x^3 - 8x &= 2x(x^2 - 4) \\[0.4em] &= 2x(x - 2)(x + 2). \end{aligned}

Now the zeros $x = 0$ , $x = 2$ and $x = -2$ can be read straight off the factors.

Difference of squares

Definition For all real numbers $a$ and $b$ ,

a^2 - b^2 = (a - b)(a + b).

A sum of squares $x^2 + c^2$ with $c \ne 0$ does not factor over the reals, because it is at least $c^2 > 0$ for every real $x$ and so has no real zero.

In plain words Something squared minus something else squared always splits into a minus version times a plus version. A plus between two squares does not split this way.

Example Factor $x^4 - 16$ completely over the reals. It is a difference of squares with $a = x^2$ and $b = 4$ , and one of the new factors is again a difference of squares,

\begin{aligned} x^4 - 16 &= (x^2 - 4)(x^2 + 4) \\[0.4em] &= (x - 2)(x + 2)(x^2 + 4). \end{aligned}

The factor $x^2 + 4$ is at least $4$ for every real $x$ , so it stays as it is. Stopping at $(x^2 - 4)(x^2 + 4)$ would leave the job unfinished.

Equivalent expressions

Definition Two expressions in $x$ are equivalent on a set of real numbers when they are defined and have the same value for every $x$ in that set. Unless a set is named, it is the set of all $x$ where both expressions are defined. The polynomial identities on this page are equivalences for every real $x$ . Checking a rewrite by reversing it, such as expanding a factored form, confirms equivalence. Agreement at a few sample values does not, though a single disagreement proves two expressions are not equivalent.

In plain words Equivalent means the same value every single time, not just sometimes. The safest check is to undo your step and see the original come back.

Example The factored form $(x - 3)(x + 3)$ expands back to $x^2 - 9$ , so the two are equivalent for every real $x$ . On the other hand $(x + 3)^2$ and $x^2 + 9$ are not equivalent, because at $x = 1$ they give $16$ and $10$ . Watch the domain as well. The expressions $\frac{x^2 - 9}{x - 3}$ and $x + 3$ agree for every $x \ne 3$ , but only the second is defined at $x = 3$ , so they are equivalent on the set $x \ne 3$ and not on all real numbers.

Check yourself

Expand $(2x - 3)^2$ .

Show the answer

With $a = 2x$ and $b = 3$ in the square of a difference,

(2x - 3)^2 = 4x^2 - 12x + 9.

The cross term $-12x$ is twice the product $2x \cdot 3$ , with a minus sign.

Factor $x^4 - 81$ completely over the reals.

Show the answer

It is a difference of squares twice over,

\begin{aligned} x^4 - 81 &= (x^2 - 9)(x^2 + 9) \\[0.4em] &= (x - 3)(x + 3)(x^2 + 9). \end{aligned}

The factor $x^2 + 9$ is at least $9$ for every real $x$ , so it has no real zero and does not factor further.

For $h \ne 0$ , simplify

\frac{(x + h)^2 - x^2}{h}.

Show the answer

Expanding the square gives $x^2 + 2xh + h^2$ , so the numerator is $2xh + h^2 = h(2x + h)$ . Since $h$ is now a factor and $h \ne 0$ , it cancels,

\begin{gathered} \frac{(x + h)^2 - x^2}{h} = 2x + h, \\[0.4em] h \ne 0. \end{gathered}

Expanding and Factoring for a Purpose

Key terms

Term

Factor

Distributive property

Expansion

Square of a sum

Factorization

Difference of squares

Equivalent expressions

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus