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How can $(2^{100}-2^{98})(2^{99}-2^{97})$ be written in terms of its prime factors?

I tried to expand it: $2^{199}-4^{197}+2^{195}$
What do I do next?

The answer choices are:

A. $2^{100}\cdot3 \cdot 5$

B. $2^{195}\cdot 3^{2}$

C. $2^{199}\cdot 5^{2}$

D. $2^{394}$

E. $2^{195}\cdot 7 \cdot 5$

  • 4
    Don't expand, factor. – 2017-01-11
  • 0
    $2^{100}-2^{98} = 2^{98} (2^2 - 1) = 3 \cdot2^{98}$ and $2^{99} - 2^{97} = 3 \cdot2^{97}$ – 2017-01-11
  • 1
    Linear algebra? This needs elementary algebra such as knowing that $2^x +2^x$ does not equal 4^x$ – 2017-01-11

2 Answers 2

1

Just factor the expression to get the answer:

$(2^{100}-2^{98})\cdot(2^{99}-2^{97}) = [2^{98}(2^2-1)][2^{97}(2^2-1)] = 2^{98+97}\cdot 3^2 = 2^{195}\cdot 3^2$

0

$2^{a+2}-2^a = 2^{a}(2^2-1)\\ 2^{98}\cdot3\cdot2^{97}\cdot 3\\ 2^{97+98}3^2$