How will you factor quadratic trynomial?

Answer 1

If the leading coefficient is 1 (or you don't see any number in front of x^2) then find 2 factors f1,f2 which multiply to get the constant term, c, and add to get the coefficient of x, or b,. Then the factors would be (x+f1)(x+f2). Always watch for negatives. Example: x^2 + 2x - 8 B= 2 C= -8 We want 2 numbers that multiply to -8 and add to 2. Since they multiply to a negative number, one must be positive and one must be negative. Since the sum is positive, the larger must also be positive. Analyzing signs can help with determining the factors. With our example, the factors would be 4, -2 so the factor of x^2 + 2x - 8 is (x + 4)(x - 2) To factor quadratic trinomials with a coefficient of the quadratic term , other than 1, I would try the AC method. Here is an example Factor 2x^2 - 3x - 2 With the AC method, find the product of the a, the coefficient of the quadratic term, and c, the constant Here AC = 2*-2 = -4 Then find your b, or coefficient of x B = -3 AC = -4 B = -3 To factor, you need to find a pair of factors that multiply to get AC but adds up to B Our factors would be -4 and 1 since -4*1 = -4 and -4 + 1 = -3 Then rewrite your factors (Ax + f1)(Ax + f2) where f1, f2 are the factors you just found. For our example, (2x - 4)(2x + 1). Finally, factor out any common factor from each binomial (here we can factor 2 out of 2x-4 to get (x - 2)(2x +1)) If this doesn't work, resort to the Quadratic Formula. The factors are then (x - (-b + sqrt(b^2-4ac))/2a)(x - (-b - sqrt(b^2 - 4ac))/2a)