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Splitting the Middle Term

(Factoring Quadratics)

We now learn how to split the middle term.
Splitting the middle term is a method for factoring quadratic equations.

By the end of this section we'll know how to write quadratics in factored form.

For example, we'll know how to show that: \[2x^2+7x+3 = \begin{pmatrix} 2x + 1 \end{pmatrix} \begin{pmatrix} x + 3 \end{pmatrix}\] We start by watching a tutorial to learn a five-step method for factornig quadratics.

Tutorial

In the following tutorial we learn how to factor quadratics by splitting the middle term. Watch it now.

Exercise 1

Write each of the following quadratics in factored form:

  1. \(2x^2 + 7x + 3\)
  2. \(6x^2+x-2\)
  3. \(3x^2 - 13x + 4\)
  4. \(3x^2-19x + 20\)
  5. \(4x^2+9x+5\)

Note: this exercise can be downloaded as a worksheet to practice with: Worksheet 1

Answers Without Working

  1. \(2x^2+7x + 3 = \begin{pmatrix} 2x + 1 \end{pmatrix} \begin{pmatrix} x + 3 \end{pmatrix}\)
  2. \(6x^2 + x - 2 = \begin{pmatrix} 2x - 1 \end{pmatrix} \begin{pmatrix} 3x + 2 \end{pmatrix}\)
  3. \(3x^2 - 13x + 4 = \begin{pmatrix}x - 4 \end{pmatrix} \begin{pmatrix} 3x - 1 \end{pmatrix}\)
  4. \(3x^2 - 19 x + 20 = \begin{pmatrix} x - 5 \end{pmatrix} \begin{pmatrix} 3x - 4 \end{pmatrix}\)
  5. \(4x^2 + 9x + 5 = \begin{pmatrix} 4x + 5 \end{pmatrix} \begin{pmatrix} x + 1 \end{pmatrix}\)

Solving Quadratic Equations

Now that we know how to factor quadratics, by splitting the middle term, we learn how to solve quadratic equations by factoring.

The method/technique rests on the null factor law, explained here.

Null Factor Law

Given two quantities \(A\) and \(B\) such that: \[A\times B = 0\] then one of the following must be true:

  • \(A=0\)
  • \(B=0\)
  • \(A=B=0\)
This law can be used for solving quadratic equations.

Null Factor Law for Quadratics

Once a quadratic, \(ax^2+bx+c\), has been written in factored form, we can use the null factor law to solve the equation: \[ax^2+bx+c = 0\]

Example

In factored form, the quadratic \(2x^2+7x+3\) can be written: \[\begin{pmatrix}2x+1 \end{pmatrix} \begin{pmatrix}x+3 \end{pmatrix}\] Consequently we can write: \[2x^2+7x+3 = 0\]

has same solutions as: \[\begin{pmatrix}2x+1 \end{pmatrix} \begin{pmatrix}x+3 \end{pmatrix} = 0\]
Now, using the null factor law we can state that one of the following must be true:
  • \(2x+1 = 0\), or
  • \(x+3 = 0\), or
  • \(2x+1 = 0\) and \(x+3 = 0\)
Consequently, the solutions to the quadratic equation are the solutions to \(2x+1 = 0\) and \(x+3 = 0\).

Starting with \(2x+1 = 0\): \[\begin{aligned} & 2x+1 = 0 \\ & 2x = -1 \\ & x = -\frac{1}{2} \end{aligned}\] We now solve \(x+3 = 0\): \[\begin{aligned} & x+3 = 0 \\ & x = -3 \end{aligned}\] The equation \(2x^2+7x+3 = 0\) has two solutions:

\(x = -\frac{1}{2}\) and \(x = -3\)

Tutorial

In the following tutorial we learn how to solve quadratic equations that have been written in their factored form, by splitting the middle term, using the null factor law.

Exercise 2

Solve each of the following quadratic equations by factoring (splitting the middle term):

  1. \(4x^2 + 7x + 3 = 0\)
  2. \(x^2+3x - 4 = 0\)
  3. \(6x^2 - 7x +2 = 0\)
  4. \(6x^2 + 7x - 3 = 0\)
  5. \(-2x^2 + x + 3 = 0\)

Solution Without Working

  1. In factored form, we write: \(4x^2 + 7x + 3 = (4x+3)(x+1)\) and the solutions to \(4x^2 + 7x + 3 = 0\) are:

    \(x = -1\) and \(x = -\frac{3}{4}\)

  2. In factored form, we write: \(x^2+3x - 4 = (x+4)(x-1)\) and the solutions to \(x^2+3x - 4 = 0\) are:

    \(x = -4\) and \(x = 1\)

  3. In factored form, we write: \(6x^2 - 7x +2 = (3x-2)(2x-1)\) and the solutions to \(6x^2 - 7x +2 = 0\) are:

    \(x = \frac{2}{3}\) and \(x = \frac{1}{2}\)

  4. In factored form, we write: \(6x^2 + 7x - 3 = (2x+3)(3x-1)\) and the solutions to \(6x^2 + 7x - 3 = 0\) are:

    \(x = -\frac{3}{2}\) and \(x = \frac{1}{3}\)

  5. In factored form, we write: \(-2x^2 + x + 3 = (3-2x)(x+1)\) and the solutions to \(-2x^2 + x + 3 = 0\) are:

    \(x = \frac{3}{2}\) and \(x = -1\)

Solution With Working


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