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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.
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To complete the factorization and find the solutions for x, then (x^2 - 9)(x^2 - 4) = 0 must be factored once more.
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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.