How do you divide #(4x^2+8x+6) / (x+2) #?

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How do you divide #(4x^2+8x+6) / (x+2) #?

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Answer 1 · 2018-02-06T20:02:08

#4x, R?#

Explanation:

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to solve, you would use polynomial long division. starting with #x# into #4x^2# yields #4x# but when you multiply that by 2 you get #8x# and that subtracts to zero. further dividing just gets more zeros.

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How do you divide #(4x^2+8x+6) / (x+2) #?

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