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Finding the Product and simplifying fractions that contain exponents?

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1 Answer

First, if possible, factor each trinomial separately into the product of two binomials:

     a2 + 17a + 70 = (a + 7)(a + 10)

     a2 + 3a - 10 = (a + 5)(a - 2)

     a2 + 2a - 8 = (a + 4)(a - 2)

     a2 + 20a + 100 = (a + 10)(a + 10)

Replace each trinomial in the problem with its factored version:

     [(a2 + 17a + 70)/(a2 + 3a - 10)]·[(a2 + 2a - 8)/(a2 + 20a + 100)]

  =  [(a + 7)(a + 10)/(a + 5)(a - 2)]·[(a + 4)(a - 2)/(a + 10)(a + 10)]

When multiplying fractions, we multiply the terms in the numerators of both fractions and the terms in the denominator of both fractions. In doing so, we arrive at the following:

  =  [(a + 7)(a + 10)·(a + 4)(a - 2)]/[(a + 5)(a - 2)·(a + 10)(a + 10)]

Now, any like binomials in the numerator and the denominator will cancel out. That is,

  =  [(a + 7)(a + 10)(a + 4)(a - 2)]/[(a + 5)(a - 2)(a + 10)(a + 10)]

This leaves us with the following:

  =  [(a + 7)(a + 4)]/[(a + 5)(a + 10)]

Since no other terms cancel out, this is as far as the product of the fractions can reduce. We can leave it in this form or find the product of the binomials in the numerator and the denominator.

     (a + 7)(a + 4) = a·a + 4·a + 7·a + 7·4 = a2 + 11a + 28

     (a + 5)(a + 10) = a·a + 10·a + 5·a + 5·10 = a2 + 15a + 50

Therefore,

     [(a + 7)(a + 4)]/[(a + 5)(a + 10)] = [(a2 + 11a + 28)]/[(a2 + 15a + 50)]

Thus, the final solution is:

     (a2 + 11a + 28)/(a2 + 15a + 50)