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Factor Expressions Involving Large Powers, Radicals, and Trig Functions
You can use substitution to factor expressions involving large powers, radicals, and trig functions
Large powers : If all the powers of your variables are multiplies of the same number, that's a good time to use substitution.x12 - 7x6 - 8 Since the degree of x in each term is a multiple of 6, you can rewrite the expression as a function of x6 :(x6)2 - 7x6 - 8and solve using substitution. Make a new variable A to represent x6 , and you'll haveA2 - 7A - 8,Which you can factor like this:(A - 8)(A + 1).Then you can substitute back the x6 .(x6 + 8)(x6 + 1).You're not done! Don't forget that you might not be done factoring. The first factor happens to be a difference of cubes and the second factor is a sum of cubes:
x6 - 8 = (x2)3 - 23= (x2 - 2)(x4 + 2x2 + 4) x6 + 1 = (x2)3 + 13 = (x2 + 1)(x4 -x2 + 1)So your final factorization is:(x2 - 2)(x4 + 2x2 + 4) = (x2 + 1)(x4 - x2 + 1)
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Solve : 4x2+2x+3=0 Ans) x^2 + (1/2)x = -(3/4) (x+1/4)^2 = 1/16 - 3/4 = -11/16 implies x = (-1+i(11)^(1/2))/4 and its conjugate.
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