How do you divide #(-2x^3+7x^2+9x+15)/(3x-1) #?

Question

How do you divide #(-2x^3+7x^2+9x+15)/(3x-1)   #?

Kennedy Adams · Accepted Answer

#=>-2/3x^2+19/2x+100/27+505/(81x-27)#

#" "color(white)(..)-2x^3+7x^2+9x+15# #" "color(magenta)(-2/3x^2)(3x-1) ->color(white)(.)ul( -2x^3+2/3x^2)" "larr" Subtract"# #" "color(white)(.)0+19/3x^2+9x+15# #" "color(magenta)(+19/9x)(3x-1)-> " "color(white)(.)ul(+19/3x^2-19/9x)larr" Subtract"# #" "color(white)(..)0+100/9x+15# #" "color(magenta)(+100/27)(3x-1)->" " ul(+100/9x-100/27)"Subt."# #" remainder "rarr 0 color(white)(....) color(magenta)(+505/27)#

' ~ ~

#(-2x^3+7x^2+9x+15)/(3x-1) = color(magenta)(-2/3x^2+19/9x+100/27+[505/27-:(3x-1)]#

#=>-2/3x^2+19/2x+100/27+505/(81x-27)#

Avery Alexander · Answer

undefined To divide (-2x^3+7x^2+9x+15) by (3x-1), you can use long division. Here are the steps:

1. Divide the first term of the numerator (-2x^3) by the first term of the denominator (3x). The result is -2/3x^2.

2. Multiply the entire denominator (3x-1) by the result obtained in step 1 (-2/3x^2). This gives you (-2/3x^2)(3x-1) = -2x + 2/3x^2.

3. Subtract the result obtained in step 2 from the numerator (-2x^3+7x^2+9x+15) by changing the signs. This gives you (-2x^3+7x^2+9x+15) - (-2x + 2/3x^2) = -2x^3 + 7x^2 + 9x + 15 + 2x - 2/3x^2.

4. Combine like terms in the resulting expression. In this case, you have -2x^3 + 2/3x^2 + 7x^2 + 2x + 9x + 15.

5. Repeat steps 1-4 with the new expression obtained in step 4 (-2x^3 + 2/3x^2 + 7x^2 + 2x + 9x + 15) until you have no more terms to divide.

By following these steps, you can continue the long division process until you have divided all the terms and obtained a quotient.