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Factor x^6-y^9 and simplify 5t-10all over t-5

Sagot :

1.)   x^ - y^9 is a product of the sum and difference of binomial

           Factors:  (x² - y³) (x² + y³)

2.)  5t - 10   (i believe this is in simplified form)
         t-5
    Factor  5t - 10   = 5 (t-2)
   
    5(t-2)
      t-5

 

Bellalesseidn
Factor x⁶ - y⁹

   Since we're given a binomial, we have two choices: either use DOTS or the SDOTC method. We won't use DOTS because x⁶ isn't a perfect square whereas the terms should be a perfect square in order for us to use the DOTS method. On the other hand, we're going to use the SDOTC method because the terms are suitable for the requirements needed when finding the SDOTC.

(x² - y³)(x⁴ - x²y³ + y⁹) ---- factored form of x⁶ - y⁹ using the difference of two cubes

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Simplify 5t-10 all over t - 5.

[tex] \frac{5t-10}{t-5} [/tex] = [tex] \frac{5(t - 2)}{t - 5} [/tex]

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