How do you write the prime factorization of #-75ab^2#?

Answer 1

#-1, 3, 5, 5, a, b, b#

The smaller prime numbers are 2, 3, 5, 7, 11. Try dividing the given number by a small prime number and keep dividing until it can't be divided any more. Prime numbers are only evenly divisible by 1 or by themselves.

Since two cannot divide seventy-five evenly, try dividing by three. You get twenty-five, which is also impossible to divide by three. Attempt to divide twenty-five by the next smallest prime number, five. This gives you twenty-five, which is divisible by five.

Similarly, the alphanumerics #a# and #b^2# each stand for numerals. Their primes are #a# and #b#. Finally, the original number was a negative, which means it must have been multiplied by #-1#.

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Answer 2

Gabriel Agustin

#-3*5^2ab^2#

#75=5 xx 5 xx 3#

then #3 xx5^2 ab^2#

#=- 3*5^2ab^2#

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Answer 3

Zoe Adams

The prime factorization of -75ab^2 is -1 * 3 * 5^2 * a * b^2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.