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 \(12x + x\). Then, we do all the math to simplify the expression. To use the Quadratic Formula, we substitute the values of a, b, and c into the expression on the right side of the formula. To summarize the example, here are the steps in full. The solutions to a quadratic equation of the form ax2 + bx + c 0, a 0 are given by the formula: x b ± b2 4ac 2a. 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. Factoring out the \((x + 3)\) gives us the factored form. 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. We can use the methods for solving quadratic equations that we learned in this section to solve for the missing side.Before we mention the decomposition factoring method, it is important to explore the math trick of decomposition. Because each of the terms is squared in the theorem, when we are solving for a side of a triangle, we have a quadratic equation. We use the Pythagorean Theorem to solve for the length of one side of a triangle when we have the lengths of the other two. It has immeasurable uses in architecture, engineering, the sciences, geometry, trigonometry, and algebra, and in everyday applications. ![]() It is based on a right triangle, and states the relationship among the lengths of the sides as \(a^2+b^2=c^2\), where \(a\) and \(b\) refer to the legs of a right triangle adjacent to the \(90°\) angle, and \(c\) refers to the hypotenuse. One of the most famous formulas in mathematics is the Pythagorean Theorem.
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