How do you find the sum or difference of $(3n^3-5n+n^2)-(-8n^2+3n^3)$ ?

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Answer 1 · 2017-03-31T12:02:42

See the entire solution process below:

First, remove all of the terms from parenthesis. Be careful to handle the signs of each individual term correctly:

$3n^3 - 5n + n^2 + 8n^2 - 3n^3$

Next, group like terms:

$3n^3 - 3n^3 + n^2 + 8n^2 - 5n$

Now, combine like terms:

$3n^3 - 3n^3 + 1n^2 + 8n^2 - 5n$

$(3 - 3)n^3 + (1 + 8)n^2 - 5n$

$0n^3 + 9n^2 - 5n$

$9n^2 - 5n$

Answer 2 · 2017-03-31T12:33:02

$n(9n-5)$

$(3n^3-5n+n^2)-(8n^2+3n^3)$

$:.=3n^3-5n+n^2+8n^2-3n^3$

$:.=9n^2-5n$

$:.=n(9n-5)$

How do you find the sum or difference of (3n^3-5n+n^2)-(-8n^2+3n^3)(3n^3-5n+n^2)-(-8n^2+3n^3)?