Factoring becomes simpler when the leading coefficient is equal to 1.The difference of perfect squares technique is useful for factoring quadratic equations.Factoring allows us to rewrite quadratic equations as a product of two binomials. ![]() The choice of method depends on the complexity of the equation and individual preferences. However, in cases where factoring is not possible or when a more systematic approach is needed, the quadratic formula provides an effective alternative. In conclusion, factoring is a useful method to solve quadratic equations, particularly when the equation can be easily factored. ![]() Plugging these values into the quadratic formula, we get:īy using the quadratic formula, we obtain the same solutions as we obtained through factoring. For this equation, a = 8, b = 2, and c = -15. Let's consider the equation 8x² + 2x - 15 = 0. The quadratic formula states that for an equation in the form ax² + bx + c = 0, the solutions for x can be found using the formula: In such situations, the quadratic formula provides an alternative method to solve quadratic equations. In some cases, factoring may not be feasible or straightforward. ![]() Setting each factor equal to zero gives 4x + 12 = 0, which gives x = -3, and 2x - 10 = 0, resulting in x = 5/2. Thus, the factored form of the equation is (4x + 12)(2x - 10). After exploring different combinations, we find that the numbers are 12 and -10. To factor this expression, we need to find two numbers that multiply to -120 (product of the leading coefficient and constant term) but add up to 2 (coefficient of the linear term). Let's consider the trinomial 8x² + 2x - 15. If the leading coefficient of the quadratic equation is not equal to 1, factoring can involve some additional steps. Example 5: Factoring Trinomials with Leading Coefficient not equal to 1 Setting each factor equal to zero gives x - 5 = 0, giving x = 5, and x + 3 = 0, resulting in x = -3. Therefore, the factored form of the equation is (x - 5)(x + 3). To factor this expression, we need to find two numbers that multiply to -15 but add up to -2. Let's consider the trinomial x² - 2x - 15. When the leading coefficient of the quadratic equation is 1, factoring becomes simpler. Example 4: Factoring Trinomials with Leading Coefficient of 1 Setting each factor equal to zero yields 3x + 8 = 0, giving x = -8/3, and 3x - 8 = 0, resulting in x = 8/3. Thus, the factored form of the equation is (3x + 8)(3x - 8). The square root of 9 is 3, and the square root of 64 is 8. Example 3: 9x² - 64 = 0įor the equation 9x² - 64 = 0, we can directly apply the difference of perfect squares technique. Setting each factor equal to zero gives x + 5 = 0, resulting in x = -5, and x - 5 = 0, giving x = 5. So, the factored form of the equation is (x + 5)(x - 5). The square root of x² is x, and the square root of 25 is 5. Now we can Apply the difference of perfect squares technique. Dividing each term by 3, we obtain x² - 25 = 0. However, we can first find the greatest common factor (GCF) of the equation, which is 3. We cannot directly use the difference of perfect squares technique for this equation. In this example, we have the equation 3x² - 75 = 0.
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