How do you divide #(2x^3 - 6x^2 + 8) / (x^2 - 4)#?

Question

Ethan Adkins · Accepted Answer

#(2x^3-6x^2+8)/(x^2-4)# #color(white)(........)color(white)(.)2x-6# #x^2-4|overline(2x^3-6x^2 color(white)(......)+8)# #color(white)(............)ul(2x^3 color(white)(.........)-8x# #color(white)(................)-6x^2+8x+8# #color(white)(..................)ul(-6x^2 color(white)(.......)+24)# #color(white)(..............................)8x-16# #(2x^3-6x^2+8) / (x^2-4) = 2x-6 # and remainder of # (8x-16)/(x^2-4)#

Avery Alexander · Answer

undefined To divide (2x^3 - 6x^2 + 8) by (x^2 - 4), we can use polynomial long division.

First, we divide the highest degree term of the numerator (2x^3) by the highest degree term of the denominator (x^2). This gives us 2x as the quotient.

Next, we multiply the entire denominator (x^2 - 4) by the quotient (2x), which gives us 2x(x^2 - 4) = 2x^3 - 8x.

We then subtract this result from the numerator (2x^3 - 6x^2 + 8) to get the remainder: (2x^3 - 6x^2 + 8) - (2x^3 - 8x) = -6x^2 + 8x + 8.

Now, we repeat the process with the remainder (-6x^2 + 8x + 8) and the denominator (x^2 - 4).

We divide the highest degree term of the remainder (-6x^2) by the highest degree term of the denominator (x^2), which gives us -6x as the quotient.

We multiply the entire denominator (x^2 - 4) by the quotient (-6x), which gives us -6x(x^2 - 4) = -6x^3 + 24x.

We subtract this result from the remainder (-6x^2 + 8x + 8) to get the new remainder: (-6x^2 + 8x + 8) - (-6x^3 + 24x) = 6x^3 - 14x + 8.

Since the degree of the new remainder (6x^3 - 14x + 8) is less than the degree of the denominator (x^2 - 4), we have reached the end of the division.

Therefore, the quotient is 2x - 6x and the remainder is 6x^3 - 14x + 8.