![]() = (5 + x)(5 - x) Ĭlick here to find more information on quadratic equations. This is because a² - b² = (a + b)(a - b). If you are asked to factorise an expression which is one square number minus another, you can factorise it immediately. Unfortunately, the only other method of factorising is by trial and error. It is worth studying these examples further if you do not understand what is happening. Once you work out what is going on, this method makes factorising any expression easy. We need to split the 2x into two numbers which multiply to give -8. Now, make the last two expressions look like the expression in the bracket: The first two terms, 12y² and -18y both divide by 6y, so 'take out' this factor of 6y.Ħy(2y - 3) - 2y + 3 One systematic method, however, is as follows: There is no simple method of factorising a quadratic expression, but with a little practise it becomes easier. This video shows you how to solve a quadratic equation by factoring. So if you were asked to factorise x² + x, since x goes into both terms, you would write x(x + 1). The first step of factorising an expression is to 'take out' any common factors which the terms have. This is an important way of solving quadratic equations. 2 x( x + 3) = 2x² + 6x ).įor an expression of the form (a + b)(c + d), the expanded version is ac + ad + bc + bd, in other words everything in the first bracket should be multiplied by everything in the second.įactorising is the reverse of expanding brackets, so it is, for example, putting 2x² + x - 3 into the form (2x + 3)(x - 1). But if we look closely at the example we just did, try. Questionsįactor each of the following polynomials and solve what you can.This section shows you how to factorise and includes examples, sample questions and videos.īrackets should be expanded in the following ways:įor an expression of the form a(b + c), the expanded version is ab + ac, i.e., multiply the term outside the bracket by everything inside the bracket (e.g. This question is what factoring quadratic expressions is all about, and it can be pretty tricky. Checking for any others by using the discriminant reveals that all other solutions are complex or imaginary solutions. The two real solutions are x = 2 and x = -1. 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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