How do you write $(–3r + 4r^2 – 3) – (4r^2 + 6r – 2)$ in standard form?

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Answer 1 · 2017-02-20T14:01:28

$-9r - 1$

First remove the individual terms from parenthesis being careful to manage the signs of the individual terms correctly:

$-3r + 4r^2 - 3 - 4r^2 - 6r + 2$

Next, group like items:

$4r^2 - 4r^2 - 3r - 6r - 3 + 2$

Now, combine like items:

$(4 - 4)r^2 + (-3 - 6)r - 1$

$0r^2 - 9r - 1$

$-9r - 1$

How do you write (–3r + 4r^2 – 3) – (4r^2 + 6r – 2) (–3r + 4r^2 – 3) – (4r^2 + 6r – 2) in standard form?