Questionsįactor each of the following polynomials and solve what you can. Describe any patterns you notice between the square of the linear binomial and the resulting quadratic trinomial. Checking for any others by using the discriminant reveals that all other solutions are complex or imaginary solutions. Expand each expression below as a product of two linear binomials and then multiply to write the product in standard form. The two real solutions are x = 2 and x = -1. (ax2 + bx + c 0) Factor the quadratic expression. Set the equation equal to zero, that is, get all the nonzero terms on one side of the equal sign and 0 on the other. (x - 2)(x^2 + 2x + 4)(x + 1)(x^2 - x + 1) = 0 To solve quadratic equations by factoring, we must make use of the zero-factor property. The factored (x^3 - 8) and (x^3 + 1) terms can be recognized as the difference of cubes. 2) How can the vertex of a parabola be used in solving real world problems 3) Explain why the condition of a 0 is imposed in the definition of the quadratic function. Now that the substituted values are factored out, replace the u with the original x^3. 1) Explain the advantage of writing a quadratic function in standard form. 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. Further Maths GCSE Revision Revision Cards Books Factorising Quadratics Practice Questions. 5-a-day GCSE 9-1 5-a-day Primary 5-a-day Further Maths More. Next: Ratio Simplifying Textbook Exercise. Welcome Videos and Worksheets Primary 5-a-day. 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. Previous: Quadratics: Solving using Completing the Square Textbook Exercise. 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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