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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)
A pool is surrounded through a deck that has the similar width all the way around. The total area of the deck only is 400 square feet. The dimensions of the pool are 18 feet throug
8l550ml - 1/4l =
Slope of Tangent Line : It is the next major interpretation of the derivative. The slope of the tangent line to f ( x ) at x = a is f ′ ( a ) . Then the tangent line is given by,
simplify the expression 3/5/64
int*sin^-1 x dx=?
Example of Partial Fraction Decomposition Evaluate the following integral. ∫ (3x+11 / x 2 -x-6) (dx) Solution: The 1 st step is to factor the denominator so far as
greens function for x''''=0, x(1)=0, x''(0)+x''(1)=0 is G(t,s)= {1-s for t or equal to s
How to find a function
In the earlier section we looked at first order differential equations. In this section we will move on to second order differential equations. Just as we did in the previous secti
Find out the Taylor Series for f (x) = e x about x = 0. Solution In fact this is one of the easier Taylor Series that we'll be asked to calculate. To find out the Taylor
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