![]() Questionsįactor each of the following polynomials and solve what you can. These two values are the solution to the original quadratic equation. Checking for any others by using the discriminant reveals that all other solutions are complex or imaginary solutions. Now I can solve each factor by setting each one equal to zero and solving the resulting linear equations: x + 2 0 or x + 3 0. The two real solutions are x = 2 and x = -1. Improve your math knowledge with free questions in 'Solve a quadratic equation by factoring' and thousands of other math skills. The factored (x^3 - 8) and (x^3 + 1) terms can be recognized as the difference of cubes. Now that the substituted values are factored out, replace the u with the original x^3. Here, it would be a lot easier if the expression for factoring was x^2 - 7x - 8 = 0.įirst, let u = x^3, which leaves the factor of u^2 - 7u - 8 = 0. ![]() This same strategy can be followed to solve similar large-powered trinomials and binomials.įactor the binomial x^6 - 7x^3 - 8 = 0. Solving each of these terms yields the solutions x = \pm 3, \pm 2. This is done using the difference of squares equation: a^2 - b^2 = (a + b)(a - b).įactoring (x^2 - 9)(x^2 - 4) = 0 thus leaves (x - 3)(x + 3)(x - 2)(x + 2) = 0. To complete the factorization and find the solutions for x, then (x^2 - 9)(x^2 - 4) = 0 must be factored once more. ![]() 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. Now substitute u for every x^2, the equation is transformed into u^2-13u+36=0. There is a standard strategy to achieve this through substitution.įirst, let u = x^2. Here, it would be a lot easier when factoring x^2 - 13x + 36 = 0. ![]() Solve for x in x^4 - 13x^2 + 36 = 0.įirst start by converting this trinomial into a form that is more common. ![]()
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