Factoring Methods

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Methods of Factoring

We will discuss some systematic methods of factoring algebraic expressions.

Method of Common Factors

Consider a simple example: \(3x+9\)

By factorizing each term we get, \((3\times x)+(3 \times 3)\)

By distributive law, \(3x+9=(3\times x)+(3 \times 3)=3(x+3)\).

Method of Regrouping

Regrouping allows us to rearrange the terms of the expression that leads to factorization. 

Consider an expression: \(2ab+2b+7a+7\)

Notice that there is no single factor common to all the terms.

Let us write \(2ab+2b\) and \(7a+7\) in the factor form separately.

\[\begin{align}2ab+2b&=2b(a+1)\\7a+7&=7(a+1)\end{align}\]

So, \(2ab+2b+7a+7=2b(a+1)+7(a+1)\).

Now, we have a common factor \((a+1)\) on the right-hand side.

The given expression is factorized as \(2ab+2b+7a+7=2b(a+1)+7(a+1)=(2b+7)(a+1)\).

Method Using Algebraic Identities

Remember a few identities that are used for factorization.

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

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

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

Let's factorize \(4z^{2}-12z+9\)

Observe that \(4z^{2}=(2z)^{2}\), \(12z=2\times 3\times 2z\), and \(9=3^{2}\)

So, we can write \(4z^{2}-12z+9=(2z)^{2}-2\times 2z\times 3+3^{2}=(2z-3)^2\)

Solving Quadratic Equations By Factoring

We’ll do a few examples on solving quadratic equations by factorization.

Example 1: \[4x-12x^2=0\]

Given any quadratic equation, first check for the common factors.

In this example, check for the common factors among \(4x\) and \(12x^2\)

We can observe that \(4x\) is a common factor. Let’s take that common factor from the quadratic equation.

\[\begin{align}4x-12x^2&=0\\4x(1-3x)&=0\end{align}\]

Thus, by factoring the quadratic equation, we get \(4x(1-3x)=0\).

In other words, the quadratic equation can be factored as \((4x)(1-3x)=0\).

Example 2: \[x^2+5x+6=0\]

We’ll factor this in two binomials, namely \((x+a)\) and \((x+b)\).

Expand \((x+a)(x+b)=0\).

\[\begin{align}(x+a)(x+b)&=0\\x(x+b)+a(x+b)&=0\\x^2+bx+ax+ab&=0\\x^2+(a+b)x+ab&=0\end{align}\]  

Let’s try to find \(a\) and \(b\) such that \((a+b)\) is mapped to 5 and \(ab\) is mapped to 6. 

  

The factors of \(6\) are \(1\), \(2\), \(3\) and \(6\). Find the pairs \(a\) and \(b\) from \(1\), \(2\), \(3\) and \(6\) such that \(a+b=5\) and \(ab=6\)

This is the right pair: \(2\) and \(3\), because, \(2+3 = 5\) and \(2 \times 3=6\)

Thus, the quadratic equation can be factored as \((x+2)(x+3)=0\).

Calculator For Factoring

Here is a calculator for factoring different expressions.