How do you divide #5x^2 - 6x^3 + 1 + 7x# by #3x - 4#?

1 Answer
Lotusbluete
Jan 29, 2016

#(-6x^3 + 5x^2 + 7x + 1) -: (3x - 4) = -2x^2 - x +1#

with remainder #5#.

Explanation:

First of all, order the terms by the power of #x#. In your case, this means

#- 6 x^3 + 5x^2 + 7x + 1#

for the first term.

Now let me walk you through the polynomial long division.

You are basically doing the following operations:

  • Divide the dividend's term with the highest power by the divisor's term with the highest power. In your case, that's #(- 6 x^3) -: (3x) = -2 x^2#

  • Multiply the result, in your case #-2x^2#, with the divisor: #(-2x^2) * (3x - 4) = -6x^3 + 8x^2#

  • Subtract the result from the last step from your divident: #(-6x^3 + 5x^2 + 7x + 1) - (- 6 x^3 + 8x^2) = - 3x^2 + 7x + 1#

  • Now, you can repeat all those steps with the term #-3x^2 + 7x + 1# as a new divident.... etc.

In total, your division should look like this:

#color(white)(xx) (-6x^3 + 5x^2 + 7x + 1) -: (3x - 4) = -2x^2 - x +1#
#- (- 6 x^3 + 8 x^2)#
# color(white)(xx) color(white)(xxxxxxxxx) / #
# color(white)(xxxxxxx) - 3 x^2 + 7x #
# color(white)(xxxxx) -(- 3 x^2 + 4x) #
# color(white)(xxxxxxx) color(white)(xxxxxxxxx) / #
# color(white)(xxxxxxxxxxxxxx) 3x +1#
# color(white)(xxxxxxxxxxxii) -(3x-4)#
# color(white)(xxxxxxxxxxxxx) color(white)(xxxxxxxxx) / #
# color(white)(xxxxxxxxxxxxxxxxxx) 5#

This means that

#(-6x^3 + 5x^2 + 7x + 1) -: (3x - 4) = -2x^2 - x +1#

with remainder #5#.

Or, if you prefer a different notation,

#(-6x^3 + 5x^2 + 7x + 1) -: (3x - 4) = -2x^2 - x +1 + 5 / (3x - 4)#

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