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 x in x^4 - 13x^2 + 36 = 0.

First start by converting this trinomial into a form that is more common. Here, it would be a lot easier when factoring x^2 - 13x + 36 = 0. There is a standard strategy to achieve this through substitution.

First, let u = x^2. Now substitute u for every x^2, the equation is transformed into u^2-13u+36=0.

u^2 - 13u + 36 = 0 factors into (u - 9)(u - 4) = 0.

Once the equation is factored, replace the substitutions with the original variables, which means that, since u = x^2, then (u - 9)(u - 4) = 0 becomes (x^2 - 9)(x^2 - 4) = 0.

To complete the factorization and find the solutions for x, then (x^2 - 9)(x^2 - 4) = 0 must be factored once more. This is done using the difference of squares equation: a^2 - b^2 = (a + b)(a - b).

Factoring (x^2 - 9)(x^2 - 4) = 0 thus leaves  (x - 3)(x + 3)(x - 2)(x + 2) = 0.

Solving each of these terms yields the solutions x = \pm 3, \pm 2.

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

Example 10.5.2

Factor the binomial x^6 - 7x^3 - 8 = 0.

Here, it would be a lot easier if the expression for factoring was x^2 - 7x - 8 = 0.

First, let u = x^3, which leaves the factor of u^2 - 7u - 8 = 0.

u^2 - 7u - 8 = 0 easily factors out to (u - 8)(u + 1) = 0.

Now that  the substituted values are factored out, replace the u with the original x^3. This turns (u - 8)(u + 1) = 0 into (x^3 - 8)(x^3 + 1) = 0.

The factored (x^3 - 8) and (x^3 + 1) terms can be recognized as the difference of cubes.

These are factored using a^3 - b^3 = (a - b)(a^2 + ab + b^2) and a^3 + b^3 = (a + b)(a^2 - ab + b^2).

And so, (x^3 - 8) factors out to (x - 2)(x^2 + 2x + 4) and (x^3 + 1) factors out to (x + 1)(x^2 - x + 1).

Combining all of these terms yields:

    \[(x - 2)(x^2 + 2x + 4)(x + 1)(x^2 - x + 1) = 0\]

The two real solutions are x = 2 and x = -1. 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.

  1. x^4-5x^2+4=0
  2. y^4-9y^2+20=0
  3. m^4-7m^2-8=0
  4. y^4-29y^2+100=0
  5. a^4-50a^2+49=0
  6. b^4-10b^2+9=0
  7. x^4+64=20x^2
  8. 6z^6-z^3=12
  9. z^6-216=19z^3
  10. x^6-35x^3+216=0

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

License

Icon for the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License

Intermediate Algebra by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

Share This Book