How do you divide #(6x^4+6x^3-12x^2-7x-7)/(x-2) #?

Answer 1

I got:

#= 6x^3 + 18x^2 + 24x + 41 + 75/(x-2)#

Synthetic division is a pretty simple method for doing it.

The coefficients of each term are used for this. It works similarly to standard long division but is quicker and/or more compact.

#6x^4 + 6x^3 - 12x^2 - 7x - 7#
#=># #6" "6" "-12" "-7" "-7#
The factor divided by is as if you were setting it equal to #0# and solved for it. If we have #x - 2#, the factor in the upper-left square is #2#, because #x - 2 = 0 => x = 2#.

Thus, we start here:

#color(white)([(color(black)(ul(2)|), color(black)(6), color(black)(6), color(black)(-12), color(black)(-7), color(black)(-7)),(color(black)(+),color(black)(ul" "),color(black)(ul" "),color(black)(ul" "),color(black)(ul" "),color(black)(ul" ")),(color(black)(),color(black)(),color(black)(),color(black)(),color(black)(),color(black)())])#

Next, the fundamental actions are:

Thus, we would receive:

#color(white)([(color(black)(ul(2)|), color(black)(6), color(black)(6), color(black)(-12), color(black)(-7), color(black)(-7)),(color(black)(+),color(black)(ul" "),color(black)(ul" "),color(black)(ul" "),color(black)(ul" "),color(black)(ul" ")),(color(black)(),color(black)(6),color(black)(),color(black)(),color(black)(),color(black)())])#
#color(white)([(color(black)(ul(2)|), color(black)(6), color(black)(6), color(black)(-12), color(black)(-7), color(black)(-7)),(color(black)(+),color(black)(ul" "),color(black)(ul"12"),color(black)(ul" "),color(black)(ul" "),color(black)(ul" ")),(color(black)(),color(black)(6),color(black)(18),color(black)(),color(black)(),color(black)())])#
#color(white)([(color(black)(ul(2)|), color(black)(6), color(black)(6), color(black)(-12), color(black)(-7), color(black)(-7)),(color(black)(+),color(black)(ul" "),color(black)(ul"12"),color(black)(ul"36"),color(black)(ul" "),color(black)(ul" ")),(color(black)(),color(black)(6),color(black)(18),color(black)(24),color(black)(),color(black)())])#
#color(white)([(color(black)(ul(2)|), color(black)(6), color(black)(6), color(black)(-12), color(black)(-7), color(black)(-7)),(color(black)(+),color(black)(ul" "),color(black)(ul"12"),color(black)(ul"36"),color(black)(ul"48"),color(black)(ul" ")),(color(black)(),color(black)(6),color(black)(18),color(black)(24),color(black)(41),color(black)())])#
#color(white)([(color(black)(ul(2)|), color(black)(6), color(black)(6), color(black)(-12), color(black)(-7), color(black)(-7)),(color(black)(+),color(black)(ul" "),color(black)(ul"12"),color(black)(ul"36"),color(black)(ul"48"),color(black)(ul"82")),(color(black)(),color(black)(6),color(black)(18),color(black)(24),color(black)(41),color(black)(75))])#

Simply reassign the coefficients to the polynomial solution. It is necessary for our answer to have decreased by one degree from the initial quartic.

Our response is a cubic, for this reason:

#= q(x) + r(x)#
#= color(blue)(stackrel(q(x), "quotient")overbrace(6x^3 + 18x^2 + 24x + 41) + stackrel(r(x), "remainder")overbrace(75/(x-2)))#
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Answer 2

To divide (6x^4+6x^3-12x^2-7x-7) by (x-2), you can use long division. Here are the steps:

  1. Divide the first term of the dividend (6x^4) by the first term of the divisor (x). The result is 6x^3.
  2. Multiply the entire divisor (x-2) by the result obtained in step 1 (6x^3). This gives you 6x^4-12x^3.
  3. Subtract the result obtained in step 2 from the original dividend (6x^4+6x^3-12x^2-7x-7). This gives you -6x^3-12x^2-7x-7.
  4. Bring down the next term from the dividend (-6x^3), and repeat steps 1-3 until you have no more terms to bring down.
  5. Repeat steps 1-4 until you have no more terms to bring down.

The final result of the division is 6x^3-12x^2-7x-7.

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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.

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.

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.

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.

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