![]() ![]() We know that any number multiplied by 0 gets 0. We have two factors when multiplied together gets 0. We find that the two terms have x in common. We can factorize quadratic equations by looking for values that are common. If the coefficient of x 2 is greater than 1 then you may want to consider using the Quadratic formula. This is still manageable if the coefficient of x 2 is 1. In other cases, you will have to try out different possibilities to get the right factors for quadratic equations. In some cases, recognizing some common patterns in the equation will help you to factorize the quadratic equation.įor example, the quadratic equation could be a Perfect Square Trinomial (Square of a Sum or Square of a Difference) or Difference of Two Squares. Sometimes, the first step is to factor out the greatest common factor before applying other factoring techniques. Quadratics are algebraic expressions that include the term, x2, in the general form. Steps Step 1) Determine the product of ac (the coefficients in a quadratic equation) Step 2) Determine what factors of ac sum to b Step 3) ungroup the mi. The simplest way to factoring quadratic equations would be to find common factors. Solving Quadratic Equations by Factorising. Solving Quadratic Equations using the Quadratic Formula The two numbers here would be -8 and 1.Factoring Quadratic Equations (Square of a sum, Square of a difference, Difference of 2 squaresįactoring Quadratic Equations where the coefficient of x 2 is greater than 1įactoring Quadratic Equations by Completing the Square Notice how we have even numbers in 4, -14 and -8. The two numbers which satisfy these conditions are 6 and 2 (since \(6 \times 2 = 12\) and \(6 2 = 8\)). We seek two numbers which multiply to \(3 \times 4 = 12\) and add up to \(b = 8\). ![]() Consider the first terms as one pair and the last two terms as another pair.Ĭommon factor from the first two terms and common factor from the last two terms.Ĭommon factor one more time to achieve the factored form. Notice how there are now four terms instead of three terms. Using the numbers \(j\) and \(k\) decompose \(bx\) into \(jx kx\) or \(kx jx\). Here are the steps of factoring a quadratic equation in the form of \(y = ax^2 bx c\) through decomposition.ĭetermine two numbers \(j\) and \(k\) such that \(jk = ac\) and \(j k = b\). The final answer would still be the same but the steps would be slightly different. We can also break down the \(13x\) into \(x 12x\) instead of \(12x x\). To summarize the example, here are the steps in full. To check that \(y = (x 3)(4x 1)\) is indeed the factored form of \(y = 4x^2 13x 3\), we use the FOIL method when multiplying binomials. Factorisation into double brackets is the reverse process of expanding. Factoring out the \((x 3)\) gives us the factored form. To factorise a quadratic expression in the form x2 bx c we need double brackets. Notice that we now have a common factor of \((x 3)\). The first common factoring is on the first two terms and the second common factoring would be applied on the third and fourth terms. The equation is now \(y = 4x^2 12x x 3\).įrom \(y = 4x^2 12x x 3\) we do common factoring twice. Using the numbers 12 and 1 we can decompose the \(13x\) into \(12x\) and \(x\) which matches the 12 and 1. Unlike the factoring method when \(a = 1\), we add another step before the final factored form. The two numbers which fit that criteria are 12 and 1 since \(12 \times 1 = 12\) and \(12 1 = 13\). We need two numbers \(j\) and \(k\) which are factors of \(4 \times 3 = 12\) and satisfy \(j \times k = 12\) and \(j k = 13\). Suppose that we are given \(y = 4x^2 13x 3\). As an example, we can break down a number like 10 into 5 and 5, 3 and 7 or even 6 and 4. Before we mention the decomposition factoring method, it is important to explore the math trick of decomposition. ![]()
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