x^8 + x^2/x=?
(Simplify your answer.)
x^8 + x^2/x=?
(Simplify your answer.)
Hello, Jamal!
The first thing I look for in simplifying (or solving) a rational expression or equation is an opportunity to factor common terms. In the numerator it's x^2, yielding x^2(x^6 +1). One of the x's of the x^2 in the numerator will cancel with the x in the denominator, leaving x(x^6+1). Now it gets interesting... the (x^6+1) part of the numerator can be factored as the sum of cubes; the general formula is a^3+b^3 = (a+b)(a^2-ab+1).
When I look at the formula, I like to put it in plain language: In the first ( ), take the cube root of each term. In the second ( ), what I put in is based on what's in the first ( ): square the first term, put them together "as is" for the middle, then square the second term. The signs are automatic! In the sum of cubes the three signs will be, without question, "+" then "-" then "+." (If it's a difference you follow the same term pattern but the sign sequence is "-" then "+" then "+.")
So here we go! (x^6+1) = (x^2+1)(x^4-1x^2+1). Note: you might be able to factor the first ( ), but the second will never be factorable. If you're using "i" for imaginary numbers, you can factor the first ( ) as (x+i)(x-i) and you can use the quadratic formula to simplify the second set of ( ). If you're just using real numbers, you're done. Answer: x(x^2+1)(x^4-x^2+1).
The tricky part, I think, is seeing the sum of cubes. What helped me see it was remembering that a cube root is taken by dividing the exponent by 3, so that the cube root of x^6 is doable as x^2.
Hope this helped!
So, in (x^8 + x^2)/x, we could say that the x in the denominator is in actuality, x^1 (the 1 is usually not written for simplicity sake). Since we know that, we could rewrite the equation as: [(x^8)/x^1 +(x^2)/x^1]. Even though this may look weird, it's actually super easy form here. When the bases are equal, you can simply subtract the exponents when dividing. In this case, 8-1 and 2-1, respectively. This gives you can answer of: x^7 +x. Which, if you really wanted to, you could rewrite as x(x^6+1).