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How do you divide #(2x^3-4x^2+16x-8) / (x-2)#?

Answer 1

Nathan Axel

#2x^2+16+24/(x-2)#

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

#"consider the numerator"#

#color(red)(2x^2)(x-2)cancel(color(magenta)(+4x^2))cancel(-4x^2)+16x-8#

#=color(red)(2x^2)(x-2)color(red)(+16)(x-2)color(magenta)(+32)-8#

#=color(red)(2x^2)(x-2)color(red)(+16)(x-2)+24#

#"quotient "=color(red)(2x^2+16)," remainder "=24#

#rArr(2x^3-4x^2+16x-8)/(x-2)=2x^2+16x+24/(x-2)#

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Answer 2

Eli Assouline

To divide (2x^3-4x^2+16x-8) by (x-2), you can use polynomial long division.

First, divide the highest degree term of the dividend (2x^3) by the highest degree term of the divisor (x). This gives you 2x^2.

Next, multiply the entire divisor (x-2) by the quotient you just found (2x^2), and subtract the result from the dividend (2x^3-4x^2+16x-8). This will give you a new polynomial.

Repeat the process by dividing the highest degree term of the new polynomial by the highest degree term of the divisor. Continue this process until you have divided all the terms.

The final quotient is 2x^2 + 4x + 24, and the remainder is 40.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.