If there is more than one denominator, multiply by the LCM of the denominators. Click on "Solve Similar" button to see more examples.ģ*x^2/3+3*5x-3*18=3*0 Multiply each term on both sides of the equation by 3. Let’s see how our equation solver solves this and similar problems. Since both factors are the same, there is only one solution.Ĥ(x-3)(x+2)=0 The constant factor 4 can never be 0 and does not affect the solution. X^2+9x-22=0 One side of the equation must be 0. Solve the following quadratic equations by factoring. Using techniques other than factoring to solve quadratic equations is discussed in Chapter 10. Not all quadratics can be factored using integer coefficients. Each of these solutions is a solution of the quadratic equation. Putting each factor equal to 0 gives two first-degree equations that can easily be solved. Equations of the formįactoring the quadratic expression, when possible, gives two factors of first degree. Polynomials of second degree are called quadratics. The reason is that a product is 0 only if at least one of the factors is 0. Thus, to solve an equation involving a product of polynomials equal to 0, we can let each factor in turn equal 0 to find all possible solutions. Since we have a product that equals 0, we allow one of the factors to be 0. This procedure does not help because x^2-5x+6=0 is not any easier to solve than the original equation. Now consider an equation involving a product of two polynomials such as But did you think that x - 2 had to be 0? This is true because 5 * 0 = 0, and 0 is the only number multiplied by 5 that will give a product of 0. How would you solve the equation 5(x-2)=0? Would you proceed in either of the following ways?īoth ways are correct and yield the solution x = 2. 5.4 Solving Quadratic Equations by Factoring
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