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 [latex]x[/latex] in [latex]x^4 - 13x^2 + 36 = 0[/latex].
First start by converting this trinomial into a form that is more common. Here, it would be a lot easier when factoring [latex]x^2 - 13x + 36 = 0.[/latex] There is a standard strategy to achieve this through substitution.
First, let [latex]u = x^2[/latex]. Now substitute [latex]u[/latex] for every [latex]x^2[/latex], the equation is transformed into [latex]u^2-13u+36=0[/latex].
[latex]u^2 - 13u + 36 = 0[/latex] factors into [latex](u - 9)(u - 4) = 0[/latex].
Once the equation is factored, replace the substitutions with the original variables, which means that, since [latex]u = x^2[/latex], then [latex](u - 9)(u - 4) = 0[/latex] becomes [latex](x^2 - 9)(x^2 - 4) = 0[/latex].
To complete the factorization and find the solutions for [latex]x[/latex], then [latex](x^2 - 9)(x^2 - 4) = 0[/latex] must be factored once more. This is done using the difference of squares equation: [latex]a^2 - b^2 = (a + b)(a - b)[/latex].
Factoring [latex](x^2 - 9)(x^2 - 4) = 0[/latex] thus leaves [latex](x - 3)(x + 3)(x - 2)(x + 2) = 0[/latex].
Solving each of these terms yields the solutions [latex]x = \pm 3, \pm 2[/latex].
This same strategy can be followed to solve similar large-powered trinomials and binomials.
Example 10.5.2
Factor the binomial [latex]x^6 - 7x^3 - 8 = 0[/latex].
Here, it would be a lot easier if the expression for factoring was [latex]x^2 - 7x - 8 = 0[/latex].
First, let [latex]u = x^3[/latex], which leaves the factor of [latex]u^2 - 7u - 8 = 0[/latex].
[latex]u^2 - 7u - 8 = 0[/latex] easily factors out to [latex](u - 8)(u + 1) = 0[/latex].
Now that the substituted values are factored out, replace the [latex]u[/latex] with the original [latex]x^3[/latex]. This turns [latex](u - 8)(u + 1) = 0[/latex] into [latex](x^3 - 8)(x^3 + 1) = 0[/latex].
The factored [latex](x^3 - 8)[/latex] and [latex](x^3 + 1)[/latex] terms can be recognized as the difference of cubes.
These are factored using [latex]a^3 - b^3 = (a - b)(a^2 + ab + b^2)[/latex] and [latex]a^3 + b^3 = (a + b)(a^2 - ab + b^2)[/latex].
And so, [latex](x^3 - 8)[/latex] factors out to [latex](x - 2)(x^2 + 2x + 4)[/latex] and [latex](x^3 + 1)[/latex] factors out to [latex](x + 1)(x^2 - x + 1)[/latex].
Combining all of these terms yields:
\[(x – 2)(x^2 + 2x + 4)(x + 1)(x^2 – x + 1) = 0\]
The two real solutions are [latex]x = 2[/latex] and [latex]x = -1[/latex]. 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.
- [latex]x^4-5x^2+4=0[/latex]
- [latex]y^4-9y^2+20=0[/latex]
- [latex]m^4-7m^2-8=0[/latex]
- [latex]y^4-29y^2+100=0[/latex]
- [latex]a^4-50a^2+49=0[/latex]
- [latex]b^4-10b^2+9=0[/latex]
- [latex]x^4+64=20x^2[/latex]
- [latex]6z^6-z^3=12[/latex]
- [latex]z^6-216=19z^3[/latex]
- [latex]x^6-35x^3+216=0[/latex]
