Chapter 10: Quadratics

# 10.5 Solving Quadratic Equations Using Substitution

Factoring trinomials in which the leading term is not 1 is only slightly more difficult than when the leading coefficient is 1. The method used to factor the trinomial is unchanged.

Example 10.5.1

Solve for in .

First start by converting this trinomial into a form that is more common. Here, it would be a lot easier when factoring There is a standard strategy to achieve this through substitution.

First, let . Now substitute for every , the equation is transformed into .

factors into .

Once the equation is factored, replace the substitutions with the original variables, which means that, since , then becomes .

To complete the factorization and find the solutions for , then must be factored once more. This is done using the difference of squares equation: .

Factoring thus leaves .

Solving each of these terms yields the solutions .

This same strategy can be followed to solve similar large-powered trinomials and binomials.

Example 10.5.2

Factor the binomial .

Here, it would be a lot easier if the expression for factoring was .

First, let , which leaves the factor of .

easily factors out to .

Now that the substituted values are factored out, replace the with the original . This turns into .

The factored and terms can be recognized as the difference of cubes.

These are factored using and .

And so, factors out to and factors out to .

Combining all of these terms yields:

The two real solutions are and . Checking for any others by using the discriminant reveals that all other solutions are complex or imaginary solutions.

# Questions

Factor each of the following polynomials and solve what you can.

<a class=”internal” href=”/intermediatealgebraberg/back-matter/answer-key-10-5/”>Answer Key 10.5