$factoring quadratic equations problems$

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We now set each factor equal to zero and solve for x. Thus, when factored, the original equation becomes (2 x + 1)( x – 4) = 0. We will next factor out a 2 x from the first two terms. We will then group the first two terms and the last two terms. Those two numbers which multiply to give –8 and add to give –7 are –8 and 1. We must now think of two numbers that will multiply to give us –8, but will add to give us –7 (the coefficient in front of the x term). (We could use the quadratic formula, but it's easier to factor when we can.)īecause the coefficient in front of the is not equal to 1, we need to multiply this coefficient by the constant, which is –4. In other words, to find the roots of a function, we must set the function equal to zero and solve for the possible values of x. Whenever a function passes through a point on the x-axis, the value of the function is zero. The roots of a function are the x intercepts of the function. When we set each of these pieces equal to zero, we get x – 2 = 0 and x + 3 = 0. Then, because we need the positive value to be larger than the negative value, we can determine that the correct factorization is (x – 2)(x + 3). Since only a combination of +/–2 and +/–3 can add to 1, we can eliminate our first two options. However, because we need our middle value to be 1, the combination of the outer and inner values (O and I in FOIL), must add up to positive 1. The factors of 6 are 1, 2, 3, and 6, so our last values must multiply together to make 6. We also know that one factor must be negative and one must be positive if our last value (the L in FOIL) is to be negative. Because we have x 2, we know that both factors must include x. However, there is a lot of room for computational error using this method.Īnother option is to factor the equation using the reverse FOIL method. In this problem, a = 1, b = 1, and c = –6. One option is to use the quadratic formula: There are two ways to solve this problem.

Factoring quadratic equations problems