1.14: Factoring a Monomial from a Polynomial and GCF

We have learned how to multiply polynomials. For example using the distributive property on \(2 x(x-1)\) we get \(2 x^{2}-2 x\) or using \(\mathrm{FOlL}\) on \((x+1)(2 x-5)\) results in \(2 x^{2}-3 x-5 .\) Factoring is an inverse procedure of multiplying polynomials. If we write \(2 x^{2}-2 x=2 x(x-1)\) or \(2 x^{2}-3 x-5=(x+1)(2 x-5),\) this is called factoring. In general, factoring is a procedure to write a polynomial as a product of two or more polynomials. In the first example \(2 x\) and \((x-1)\) are called factors of \(2 x^{2}-2 x .\) In the second example \((x+1)\) and \((2 x-5)\) are factors of \(2 x^{2}-3 x-5\).

Factoring is extremely useful when we try to solve polynomial equations and simplify algebraic fractions. In the following three chapters, we will learn several methods of factoring.

Factoring the Greatest Common Factor (GCF)

In this section we will explain how to factor the greatest common factor (GCF) from a polynomial. The GCF is the largest polynomial that divides every term of the polynomial. For example, the polynomial \(9 x^{3}+6 x^{2}+12 x^{4}\) is the addition of three terms \(9 x^{3}, 6 x^{2}\) and \(12 x^{4}\). From factorization, we have:

\[9 x^{3}=3 \cdot 3 \cdot x \cdot x \cdot x=\left(3 x^{2}\right)(3 x)\nonumber\]

\[6 x^{2}=2 \cdot 3 \cdot x \cdot x=\left(3 x^{2}\right)(2)\nonumber\]

\[12 x^{4}=2 \cdot 2 \cdot 3 \cdot x \cdot x \cdot x \cdot x=\left(3 x^{2}\right)\left(4 x^{2}\right)\nonumber\]

Therefore, the monomial \(3 x^{2}\) is the \(G C F\) of \(9 x^{3}, 6 x^{2}\) and \(12 x^{4}\). Once the GCF is identified, we can use the distributive property to factor out the GCF as follows:

\[\begin{align*}
9 x^{3}+6 x^{2}+12 x^{4} &=3 x^{2}(3 x)+3 x^{2}(2)+3 x^{2}\left(4 x^{2}\right) \\
&=3 x^{2}\left(3 x+2+4 x^{2}\right)
\end{align*}\nonumber\]

Example \(\PageIndex{1}\)

Factor out the GCF from the given polynomial.

a) \(6 a^{4}-8 a\)

The GCF is \(2 a\)

\[6 a^{4}-8 a=2 a\left(3 a^{3}\right)-2 a(4)=2 a\left(3 a^{3}-4\right)\nonumber\]

b) \(15 a b^{3} c-10 a^{2} b c+25 b^{2} c^{3}\)

The GCF is \(5 b c\)

\[\begin{align*}
15 a b^{3} c-10 a^{2} b c+25 b^{2} c^{3} &=5 b c\left(3 a b^{2}\right)-5 b c\left(2 a^{2}\right)+5 b c\left(5 b c^{2}\right) \\
&=5 b c\left(3 a b^{2}-2 a^{2}+5 b c^{2}\right)
\end{align*}\nonumber\]

c) \(3 x^{2}-6 x+12\)

The GCF is 3.

\[3 x^{2}-6 x+12=3\left(x^{2}\right)-3(2 x)-3(4)=3\left(x^{2}-2 x-4\right)\nonumber\]

In the problems we have done so far, the first term of the polynomial has been positive. If the first term of the polynomial is negative, it is better to factor the opposite of the GCF. The procedure is exactly the same, but we need to pay attention to the sign of each term.

Example \(\PageIndex{2}\)

Factor out the GCF from the given polynomial:

a) \(-8 x^{2} y+12 x y^{2}-16 x y\)

The GCF is \(4 x y .\) since the first term of the polynomial is negative, we factor out \(-4 x y\) instead.

\[\begin{align*}
-8 x^{2} y+12 x y^{2}-16 x y &=(-4 x y)(2 x)+(-4 x y)(-3 y)+(-4 x y)(4) \\
&=-4 x y(2 x-3 y+4)
\end{align*}\nonumber\]

Note: Inside the parentheses, the sign has been changed for every term in the remaining factor.

b) \(-9 b^{5}-15 b^{4}+21 b^{3}+27 b^{2}\)

since the first term is negative, we factor out the opposite of the \(\mathrm{GCF}:-3 b^{2}\)

\[\begin{align*}
&-9 b^{5}-15 b^{4}+21 b^{3}+27 b^{2} \\
=&\left(-3 b^{2}\right)\left(3 b^{3}\right)+\left(-3 b^{2}\right)\left(5 b^{2}\right)+\left(-3 b^{2}\right)(-7 b)+\left(-3 b^{2}\right)(-9) \\
=&-3 b^{2}\left(3 b^{3}+5 b^{2}-7 b-9\right)
\end{align*}\nonumber\]

Again, notice how inside the parentheses, the sign of each of the remaining terms has been changed.

Factoring Out the Greatest Common Factor (GCF)

• Step 1: Identify the GCF of each term of the polynomial.
• Step 2: Write each term of the polynomial as a product of the GCF and remaining factor. If the first term of the polynomial is negative, we use the opposite of the GCF as the common factor.
• Step 3: Use the distributive property to factor out the GCF.

Factoring by Grouping

In many expressions, the greatest common factor may be a binomial. For example, in the expression \(x^{2}(2 x+1)+5(2 x+1)\), the binomial \((2 x+1)\) is a common factor of both terms. So we can factor out this binomial factor as following:

\[x^{2}(2 x+1)+5(2 x+1)=(2 x+1)\left(x^{2}+5\right)\nonumber\]

This idea is very useful in factoring a four-term polynomial using the grouping method. For example, \(6 a^{2}-10 a+3 a b-5 b\) is a polynomial with four terms. The GCF of all four terms is \(1 .\) However, if we group the first two terms and the last two terms, we get \(\left(6 a^{2}-10 a\right)+(3 a b-5 b)\). Notice that in the first group, \(2 a\) is a common factor and in the second group, \(b\) is a common factor, so we can factor them out: \(2 a(3 a-5)+b(3 a-5)\). The two terms now have a common binomial factor \(3 a-5\). After factoring it out, we get:

\[2 a(3 a-5)+b(3 a-5)=(3 a-5)(2 a+b) .\]

Factoring by grouping is a very useful method when we factor certain trinomials in a later session.

Example \(\PageIndex{3}\)

Factor the given polynomials by the grouping method.

a) \(x^{3}+3 x^{2}+2 x+6\)

\[\begin{align*}
x^{3}+3 x^{2}+2 x+6 &=\left(x^{3}+3 x^{2}\right)+(2 x+6) \\
&=x^{2}(x+3)+2(x+3) \\
&=(x+3)\left(x^{2}+2\right)
\end{align*}\nonumber\]

b) \(6 x^{2}-3 x-2 x y+y\)

\[\begin{align*}
6 x^{2}-3 x-2 x y+y &=\left(6 x^{2}-3 x\right)+(-2 x y+y) \\
&=3 x(2 x-1)+(-y)(2 x-1) \\
&=(2 x-1)(3 x-y)
\end{align*}\nonumber\]

c) \(10 a x+15 a y-8 b x-12 b y\)

\[\begin{align*}
10 a x+15 a y-8 b x-12 b y &=(10 a x+15 a y)+(-8 b x-12 b y) \\
&=5 a(2 x+3 y)+(-4 b)(2 x+3 y) \\
&=(2 x+3 y)(5 a-4 b)
\end{align*}\nonumber\]

Note: In the last two problems of the example, the third term in the polynomial is negative. When grouping, we need to add an extra addition sign between two groups, and in factoring the second group, factoring the opposite of the GCF is helpful.

Exit Problem

Factor completely: \(12 a^{5} b^{4} c^{5}-36 a^{6} b^{3} c-24 a b^{2}\)

Factor by grouping: \(35 x y+21 t y-15 x z-9 t z\)