How do you divide $\frac{2 x^{2} + 8 x + 16}{x - 4}$ $(2x^2+8x+16) / (x-4)$ ?

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How do you divide $\frac{2 x^{2} + 8 x + 16}{x - 4}$ $(2x^2+8x+16) / (x-4)$ ?

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Answer 1 · 2018-06-23T19:23:03

$2 x + 16 + \frac{80}{x + 4}$ $2x+16+80/(x+4)$

Explanation:

$one way is to use the divisor as a factor in the numerator$ $"one way is to use the divisor as a factor in the numerator"$

$consider the numerator$ $"consider the numerator"$

$2 x (x - 4) + 8 x + 8 x + 16$ $color(red)(2x)(x-4)color(magenta)(+8x)+8x+16$

$= 2 x (x - 4) + 16 (x - 4) + 64 + 16$ $=color(red)(2x)(x-4)color(red)(+16)(x-4)color(magenta)(+64)+16$

$= 2 x (x - 4) + 16 (x - 4) + 80$ $=color(red)(2x)(x-4)color(red)(+16)(x-4)+80$

$quotient = 2 x + 16, remainder = 80$ $"quotient "=color(red)(2x+16)," remainder "=80$

$\frac{2 x^{2} + 8 x + 16}{x - 4} = 2 x + 16 + \frac{80}{x - 4}$ $(2x^2+8x+16)/(x-4)=2x+16+80/(x-4)$

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How do you divide $\frac{2 x^{2} + 8 x + 16}{x - 4}$ $(2x^2+8x+16) / (x-4)$ ?

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Explanation:

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How do you divide (2x^2+8x+16) / (x-4) 2x2+8x+16x−4(2x^2+8x+16) / (x-4) ?

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Explanation:

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How do you divide $\frac{2 x^{2} + 8 x + 16}{x - 4}$ $(2x^2+8x+16) / (x-4)$ ?