How do you divide $\frac{4 x^{2} + 8 x + 6}{x + 2}$ $(4x^2+8x+6) / (x+2)$ ?

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How do you divide $\frac{4 x^{2} + 8 x + 6}{x + 2}$ $(4x^2+8x+6) / (x+2)$ ?

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Answer 1 · 2018-02-06T20:02:08

$4 x, R ?$ $4x, R?$

Explanation:

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

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How do you divide $\frac{4 x^{2} + 8 x + 6}{x + 2}$ $(4x^2+8x+6) / (x+2)$ ?

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How do you divide (4x^2+8x+6) / (x+2) 4x2+8x+6x+2(4x^2+8x+6) / (x+2) ?

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How do you divide $\frac{4 x^{2} + 8 x + 6}{x + 2}$ $(4x^2+8x+6) / (x+2)$ ?