Factorization

To expand algebraic expressions, we need to use the distributive property.
The reverse of expansion, called factorization or factoring,
consists of using the distributive property in reverse and writing
the expression as product of simpler ones. For example, we can write

We say that $x-3$ and $x+1$ are factors of $x^{2}-2x-3$.

Why Is Factorization Important?

If we can put an equation in a factored form $P\cdot Q=0$, where $P$ and $Q$ are expressions in terms of the unknown variable, solving
the problem reduces to solving two independent and often simpler equations
$P=0$ and $Q=0$. (Recall from Section: Sets of Numbers that if $ab=0$ then $a=0$ or $b=0$). For
example, because $x^{2}-2x-3=(x-3)(x+1)$, instead of solving

$ x^{2}-2x-3=0, $

we can solve

$ x-3=0\quad,\quad x+1=0 $

which shows at once

$ x=3\quad\text{or}\quad x=-1. $

Similarly because

$ x^{3}-8x^{2}+17x-10=(x-1)(x-2)(x-5) $

we immediately conclude

$ x^{3}-8x^{2}+17x-10=0\Leftrightarrow x=1\quad\text{or}\quad x=2\quad\text{or}\quad x=5. $

Also evaluating $x^{3}-8x^{2}+17x-10$ reduces to only three subtractions
and two multiplications.

Factorization is not always possible, and even when it is possible, it does
not necessarily yield simpler factors. For example, $x^{5}-32$ can
be factored into $x-2$ and $x_{4}+2x^{3}+4x^{2}+8x+16$, but clearly
we would rather solve $x_{5}-32=x_{5}-2_{5}=0$ than solve the following
two equations

$ x-2=0,\quad x^{4}+2x^{3}+4x^{2}+8x+16=0. $

Another application of factorization can be simplification of rational
expressions. For example, $(x^{2}-1)/(x^{2}-2x-3)$ can be simplified
as

$ \frac{x^{2}-1}{x^{3}-2x-3}=\frac{(x-1)\cancel{(x+1)}}{(x-3)\cancel{(x+1)}}=\frac{x-1}{x-3}\qquad(x\neq-1) $

Now let's review some techniques of factorization.

Common Factors

When there is a factor common to every term of an expression, we can
simply factor it out by applying the distributive property in reverse.

Example

Factor each expression

(a) $12x^{3}-15x^{2}$

(b) $4x^{5}-8x^{4}-16x^{3}-20x^{2}$

(c) $(2x-5)(3x-7)+4(3x-7)$

Solution

(a) The greatest common factor of the terms $12x^{3}$ and $-15x_{2}$
is $3x^{2}$, so we have

$ 12x^{3}-15x_{2}=3x^{2}(4x-5) $

(b) The greatest common factor of all terms is $4x^{2}$, so we have

$ 4x^{5}-8x_{4}-16x^{3}-20x^{2}=4x^{2}\left(x^{3}-2x^{2}-4x-5\right). $

(c) The greatest common factor of $(2x-5)(3x-7)$ and $4(3x-7)$ is
$(3x-7)$. Thus

$\begin{aligned} (2x-5)(3x-7)+4(3x-7) & =(3x-7)\left[(2x-5)+4\right] & =(3x-7)(2x-1) \quad (after simplification) \end{aligned}$

{In
calculus sometimes we need to factor expressions with fractional or
negative exponents. In this case, we factor out the common factor
with the smallest exponent.}

Example

Factor each expression

(a) $x^{3/2}+4x^{1/2}-7x^{-1/2}$

(b) $(x-4)^{-3/5}+5(x-4)^{2/5}$

Solution

(a) The common factor is $x$ and its smallest exponent is $-1/2$.
Thus

$ x^{3/2}+4x^{1/2}-7x^{-1/2}=x^{-1/2}(x^{2}+4x-7) $

Note that $x^{-1/2}x^{2}=x^{(2-1/2)}=x^{3/2}$ and $x^{-1/2}x=x^{(1-1/2)}=x^{1/2}$.

(b) The common factor is $x-4$ and its smallest exponent is $-3/5$.
Thus

$\begin{aligned} (x-4)^{-3/5}+5(x-4)^{2/5} & =(x-4)^{-3/5}\left[1+(x-4)^{1}\right] & =(x-4)^{-3/2}(x-3). \end{aligned}$

Note $(x-4)^{-3/5}\cdot(x-4)=(x-4)^{(1-3/5)}=(x-4)^{2/5}$.

Special Factorization Formulas

If there is no common factor, to factor algebraic expressions we can
sometimes use the following formulas that we reviewed in Section: Special Product
and reverse the process.

Example

Factor each expression

(a) $9x^{2}-49$

(b) $x^{4}-16$

(c) $x_{2}+10x+25$

(d) $x_{3}-27$

(e) $8x_{3}+64$

Solution

(a) Rewriting as difference of squares

$ 9x_{2}-49=(3x)^{2}-7^{2} $

Let $A=3x$ and $B=7$ in $A_{2}-B_{2}=(A-B)(A+B)$. Then

$\begin{aligned} (3x)^{2}-7^{2} & =A^{2}-B^{2} & =(A-B)(A+B) & =(3x-7)(3x+7). \end{aligned}$

(b) Let $A=x^{2}$ and $B=4$, then

$\begin{aligned} x^{4}-16 & =A^{2}-B^{2} & =(A-B)(A+B) & =(x_{2}-4)(x_{2}+4) \end{aligned}$

We note that we can rewrite $x^{2}-4$ as $x_{2}-2^{2}$, so we can
use the Difference of Squares formula and factor it as

$ x^{2}-2^{2}=(x-2)(x+2) $

Thus

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

(c) Let $A=x$ and $B=5$. Because the middle term $10x=2AB$, the
polynomial is a perfect square. By the Perfect Square formula we have

$ x^2+10x+25=(x+5)^2. $

(d) Let $A=x$ and $B=3$, then

$\begin{aligned} x^3-27 & =A^3-B^3 & =(A-B)(A^2+AB+B^2) & =(x-3)(x[2]+3x+9). \end{aligned}$

(e) Here the terms $8x^3$ and 64 have the common factor 8, so first
of all we factor it out.

$ 8x^3+64=8(x^3+8). $

Then we can use the Sum of Cubes formula with $A=x$ and $B=2$:

$\begin{aligned} x^3+8 & =A^3+B^3 & =(A+B)(A^2-AB+B^2) & =(x+2)(x[2]-2x+4). \end{aligned}$

Therefore

$\begin{aligned} 8x^3+64 & =8(x^3+8) & =8(x+2)(x[2]-2x+2). \end{aligned}$

Example

Factor $9x^2-12xy+4y^2$.

Solution

Notice that $9x^2=(3x)^2$ and $4y^2=(2y)^2$. Since

$12xy=2 (3x)(2y)$

we have a perfect square. Given that the sign of the middle term is negative, we factor as

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

Example

Factor $a^2+b^2+c^2+2 a b+2 a c+2 b c$.

Solution

We can reduce this to the form of a trinomial square by grouping the terms thus:

$ \begin{aligned} a^2+2 a b+b^2+2 a c+2 b c+c^2 & =(a+b)^2+2(a+b) c+c^2 & =(a+b+c)^2 \end{aligned} $

Example

Factor $x^2-y^2-z^2+2 y z$.

Solution

$ \begin{aligned} x^2-y^2-z^2+2 y z & =x^2-\left(y^2-2 y z+z^2\right) & =x^2-(y-z)^2 &\qquad\qquad\qquad(\text{Now use }A^2-B^2=(A-B)(A+B)\ ) & =(x+y-z)(x-y+z@) \end{aligned} $

Example

Factor $x^4+x^2 y^2+y^4$.

Solution

$ \begin{aligned} x^2+x^2 y^2+y^2 & =x^4+2 x^2 y[2]+y^4-x^2 y^2 & =\left(x^2+y^2\right)[2]-x^2 y^2 & =\left(x^2+y^2+x y\right)\left(x^2+y^2-x y\right) \end{aligned} $

Factorization of Polynomials of the Form $x^2+bx c$ by Inspection

It is sometimes possible to factor \(x^2 + bx + c\) by inspection when \(b\) and \(c\) are integers.

Since

$ (x+p)(x+q) = x^{2} + (p+q)x + pq, $

we can factor polynomials of the form \(x^2 + bx + c\) if we can find numbers \(p\) and \(q\) such that:

$ p + q = b \quad \text{and} \quad pq = c. $

While such numbers are rarely rational, when they are, they must be integers, so we do not consider fractions like \(1/2\) or \(3/7\) during inspection.

In this section, we choose $p$ and $q$ by trial and error.

Example

Factor $x^2+18 x+42$.

Solution

We seek two integers, $p$ and $q$, whose product is 42 and sum 18. As both $p q$ and $p+q$ are positive, both $p$ and $q$ must be positive. Hence among the positive integers whose product is 42—namely, 42 and 1, 21 and 2, 14 and 3, 7 and 6—we seek a pair whose sum is 18, and find 7 and 6.

Hence

$ x^2+18 x+42=(x+7)(x+6). $

Example

Factor $x^2-13 x+22$.

Solution

Here both $p$ and $q$ must be negative; for their product is positive and their sum negative. Hence, testing as before the pairs of negative integers whose product is 22, we find $-11$ and $-2$; for $-11-2=-13$.

Hence

$ x^2-13 x+22=(x-11)(x-2). $

Example

Factor $x^2-9 x-22$.

Solution

Here, since $p q$ is negative, $p$ and $q$ must have opposite signs; and since $p+q$ is negative, the one which is numerically greater must be negative. Hence we set $-22=-22 \times 1=-11 \times 2$, and, testing as before, find $p=-11$ and $q=2$; for $-11+2=-9$.

Hence

$ x^2-9 x-22=(x-11)(x-2). $

Example

Factor each expression

(a) $x^2+3x+2$

(b) $x^2-2x-15$

Solution

(a) We need to find $p$ and $q$ such that $p+q=3$ and $pq=2$.
By trial and error we find that they are 2 and 1. Thus the factorization
is

$ x^2+3x+2=(x+2)(x+1). $

(b) We need to choose $p$ and $q$ such that $p+q=-2$ and $pq=-15$.
By trial and error we find that they are $-5$ and $3$. Thus

$ x^2-2x-15=(x-5)(x+3). $

Factorization of Polynomials of the Form $ax^2+bx+c$ by Inspection

It is sometimes possible to factor $ax^2+bx+c$ by trial and error, when $a, b$, and $c$ are integers.

By multiplying and dividing by $a$, we may reduce $a x^2+b x+c$ to the form $\left[(a x)^2+b(a x)+a c\right] / a$, and then factor the bracketed expression with respect to $a x$ by the method just explained.

Example

Factor \( 6x^2 + 7x - 5 \).

Solution

Let's rewrite the expression:

$ 6x^2 + 7x - 5 = \frac{(6x)^2 + 7(6x) - 5 \times 6}{6} $

Let \( u = 6x \):

$ (6x)^2 + 7(6x) - 30 = u^2 + 7u - 30 $

and then factor the result. We need to find \( p \) and \( q \) such that \( p + q = 7 \) and \( pq = -30 \). Testing factors of $-30$:

$ -30 = 30 \times (-1) = 10 \times (-3) = 6 \times (-5) $

The pair $10$ and $-3$ satisfy our conditions. Thus

$ u^2 + 7u - 30 = (u + 10)(u - 3). $

Now substitute \( 6x \) for \( u \):

$ (6x) + 7(6x) - 30 = (6x + 10)(6x - 3) $

Finally:

$ \begin{aligned} 6x^2 + 7x - 5 &= \frac{(6x)^2 + 7(6x) - 30}{6} = \frac{(6x + 10)(6x - 3)}{6} &= (3x + 5)(2x - 1) \end{aligned} $

Example

Factor $2x^2+7x+3$.

Solution

$ \begin{aligned} 2x^2+7x+3 & =\frac{(2x)^2+7(2x)+6}{2} & =\frac{(2x+6)(2x+1)}{2}=(x+3)(2x+1) \end{aligned} $

Example

Factor $ab x^2+(a^2+b^2) x+a b$.

Solution

$ \begin{aligned} ab x^2+(a^2+b^2) x+Ab & =\frac{(abx)^2+(a^2+b^2)(abx)+a^2"b^2}{ab} & =\frac{(abx+a^2)(abx+b^2)}{ab} & =(bx+a)(ax+b). \end{aligned} $

Example

Factor $16x^2+72x-63$.

Solution

$ \begin{aligned} 16 x^2+72 x-63 & =(4x)^2+18(4x)-63 & =(4x+21)(4x-3) \end{aligned} $

Factorization of $ax^2+bx c$ by Completing the Square

While the preceding methods apply in particular cases only, the following is perfectly general.

Since

$ \left(x+\frac{p}{2}\right)^2=x^2+p x+\frac{p^2}{4} $

we can make $x^2+p x$ a perfect square by adding $\frac{p^2}{4}$.

This process is called completing the square of $x^2+p x$®

1. We will not affect the value of $x^2+p x+q$ if we both add and subtract $p^2 / 4$. By this means we can transform it into a difference of squares:
   $
   \begin{aligned}
   x^2+p x+q & =x^2+p x+\frac{p^2}{4}-\frac{p^2}{4}+q

& =\left(x+\frac{p}{2}\right)^2-\frac{p^2-4 q}{4}

&=\left(x+\frac{p}{2}+\frac{\sqrt{p^2-4 q}}{2}\right)\left(x+\frac{p}{2}-\frac{\sqrt{p^2-4 q}}{2}\right)

& =\left(x+\frac{p+\sqrt{p^2-4 q}}{2}\right)\left(x+\frac{p-\sqrt{p^2-4 q}}{2}\right) &&(i)
\end{aligned}

$ 2. Since $

a x^2+b x+c=a\left(x^2+\frac{b}{a} x+\frac{c}{a}\right),

$ we may obtain the factors by substituting $b / a$ for $p$ and $c / a$ for $q$ in Equation (i). Simplifying, we get: $

a x^2+b x+c=a\left(x+\frac{b+\sqrt{b^2-4 a c}}{2 a}\right)\left(x+\frac{b-\sqrt{b^2-4 a c}}{2 a}\right).

$ > **Example** > > Factor $x^2-6 x+2$. > > > **Solution** > > > > $

\begin{aligned}
x^2-6 x+2 & =x^2-6 x+3^Š-3^2+2

& =(x-3)^Š-7

& =(x-3+\sqrt{7})(x-3-\sqrt{7})
\end{aligned}
$

Example

Factor $3 x^2-5 x+1$.

Solution

$ \begin{aligned} 3 x^2-5 x+1 & =3\left[x^2-\frac{5}{3} x+\frac{1}{3}\right] & =3\left[x^2-\frac{5}{3} x+\left(\frac{5}{6}\right)^2-\left(\frac{5}{6}\right)^2+\frac{1}{3}\right] & =3\left[\left(x-\frac{5}{6}\right)^2-\frac{13}{36}\right] & =3\left(x-\frac{5}{6}+\frac{\sqrt{13}}{6}\right)\left(x-\frac{5}{6}-\frac{\sqrt{13}}{6}\right) \end{aligned} $

Example

Factor $x^2+8 x+20$.

Solution

$ \begin{aligned} x^2+8 x+20 & =x^2+8 x+4^2-4^2+20 & =(x*4)^2+4 & =(x+4)^2-4 (\sqrt{-1})^2 &=(x+4)^2-(2\sqrt{-1})^2 & =(x+4+2\sqrt{-1})(x+4-2\sqrt{-1}) \end{aligned} $

Here we first obtain a sum of squares, and then transform it into a difference by replacing 4 by $-4(\sqrt{-1})^2$; this introduces complex numbers.

Factorization by Grouping Terms

Sometimes there is no common factor (other than $\pm1$) to all
terms of a polynomial. However, the polynomial can sometimes be factored
if we suitably group the terms that have common factors. This strategy
may work for polynomials with at least four terms.

For example, we can factor $x^{3}+3x^{2}+4x+12$ by grouping:

$\begin{aligned} x^{3}+3x^{2}+4x+12 & =(xe{3}+3x^{2})+(4x+12) & =x_{2}(x+3)+4(x+3) & =(x+3)(x^{2}+4) \end{aligned}$

Example

Factor $2x^{3}+x^{2}-18x-9$.

Solution

$\begin{aligned} 2x^{3}+x_{2}-18x-9 & =(2x^{3}+x^{2})-(18x+2) & =x^{2}(2x+1)-9(2x+1) & =(2x+1)(x^{2}-9) & =(2x+1)(h-3)(x+3) \end{aligned}$