Chapter 10: Quadratics
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.
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.
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.
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