Pre-Calculus question, answer fast please!!?

factorize (1-sin^2(x)) = (1+sin(x)).(1-sin(x)) divide by (1+sin(x)) = 1-sin(x))

as quickly as you have x^2 + 2x - 8, it quite is quite undemanding: discover 2 numbers whose product is -8 and whose sum is two those numbers are -2 and four. for this reason: x^2 + 2x - 8 = (x - 2) (x + 4) besides the shown fact that it quite is larger now no longer difficulty-loose on an identical time as optimum properly time-honored coefficient isn't a million we are able to apply the same technique. yet this time you're able to desire to locate 2 numbers whose product is = to synthetic from optimum properly time-honored coefficient and consistent coefficient (3 * -fifty six) = -168 and whose sum is -13 as basic those numbers are 8 and -21 Now replace -13x with -21x + 8x and ingredient in pairs: 3x^2 - 21x + 8x - fifty six = 3x (x - 7) + 8 (x - 7) = (x - 7) (3x + 8) ------------------------------ yet yet another technique is to end the sq.. Now with this social accumulating we are able to get nasty fractions, so quite of factoring out optimum properly time-honored coefficient, bypass consistent to superb area, and multiply the two aspects with 4 * optimum properly time-honored coefficient = 4*3 = 12 3x^2 - 13x = fifty six 12 (3x^2 - 13x) = 12 * fifty six 36x^2 - 12*13x = 672 Now upload sq. of unique coefficient of x (13^2 = 169) (6x)^2 - 2*6*13x + 13^2 = 672 + 169 observe that left area is now a perfect sq., so it might desire to extremely be factored: (6x - 13)^2 = 841 (6x - 13)^2 - 841 = 0 Now 841 = 29^2 so we've distinction of squares: (6x - 13 - 29) (6x - 13 + 29) = 0 (6x - 40 2) (6x + sixteen) = 0 6 (x - 7) * 2 (3x + 8) = 0 using fact all individuals began out multiplying the two aspects by making use of 12, divide by making use of 12 now: (x - 7) (3x + 8) = 0 M?thm?m