How do you write the prime factorization of $- 75 a b^{2}$ $-75ab^2$ ?

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How do you write the prime factorization of $- 75 a b^{2}$ $-75ab^2$ ?

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Answer 1 · 2017-01-04T14:50:43

$- 1, 3, 5, 5, a, b, b$ $-1, 3, 5, 5, a, b, b$

Explanation:

Prime numbers are only divisible evenly by 1 or by themselves. The smaller prime numbers are 2, 3, 5, 7, 11. Try dividing the given number by a small prime number and continue dividing until it can no longer be divided.

75 cannot be divided evenly by 2, so try dividing by 3. You get 25. This cannot be divided by 3, so try the next smallest prime number, 5. 25 is divisible by 5, with the result of 5, which is also divisible by 5.

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

Answer 2 · 2017-01-04T14:55:30

$- 3 \cdot 5^{2} a b^{2}$ $-3*5^2ab^2$

Explanation:

$75 = 5 \times 5 \times 3$ $75=5 xx 5 xx 3$

then $3 \times 5^{2} a b^{2}$ $3 xx5^2 ab^2$

$= - 3 \cdot 5^{2} a b^{2}$ $=- 3*5^2ab^2$

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How do you write the prime factorization of $- 75 a b^{2}$ $-75ab^2$ ?

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How do you write the prime factorization of $- 75 a b^{2}$ $-75ab^2$ ?