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

Question

Eva Adamson · Accepted Answer

#(2x³+4x²+2x+2)/(x²+4x-2)=2x-4+(22x-6)/(x²+4x-2)# #(2x³+4x²+2x+2)/(x²+4x-2)# #=2(x³+2x²+x+1)/(x²+4x-2)# #=2(x³+4x²-2x-2x²+3x+1)/(x²+4x-2)# #=2x(cancel((x²+4x-2)/(x²+4x-2)))^(=1)+2(-2x²+3x+1)/(x²+4x-2)# #=2x+2(-2x²-8x+4+11x-3)/(x²+4x-2)# #=2x+4(cancel((-x²-4x+2)/(x²+4x-2)))^(=-1)+(22x-6)/(x²+4x-2)# #=2x-4+(22x-6)/(x²+4x-2)# \0/ here's our answer !

Julia Adams · Answer

undefined To divide (2x^3+4x^2+2x+2) by (x^2+4x-2), you can use long division or synthetic division. Here is the step-by-step process using long division:

1. Divide the highest degree term of the numerator (2x^3) by the highest degree term of the denominator (x^2). This gives you 2x.

2. Multiply the entire denominator (x^2+4x-2) by the quotient obtained in step 1 (2x). This gives you 2x(x^2+4x-2) = 2x^3 + 8x^2 - 4x.

3. Subtract the result obtained in step 2 from the numerator (2x^3+4x^2+2x+2) to get the new numerator: (4x^2+2x+2) - (2x^3 + 8x^2 - 4x) = -2x^3 - 4x^2 + 6x + 2.

4. Bring down the next term from the numerator (-2x^3 - 4x^2 + 6x + 2), which is 6x.

5. Divide the new numerator (6x) by the highest degree term of the denominator (x^2). This gives you 6.

6. Multiply the entire denominator (x^2+4x-2) by the quotient obtained in step 5 (6). This gives you 6(x^2+4x-2) = 6x^2 + 24x - 12.

7. Subtract the result obtained in step 6 from the new numerator (-2x^3 - 4x^2 + 6x + 2) to get the new numerator: (6x + 2) - (6x^2 + 24x - 12) = -6x^2 - 18x + 14.

8. Bring down the next term from the numerator (-6x^2 - 18x + 14), which is 14.

9. Divide the new numerator (14) by the highest degree term of the denominator (x^2). This gives you 0.

10. Since the degree of the new numerator (14) is less than the degree of the denominator (x^2+4x-2), the division process is complete.

Therefore, the result of dividing (2x^3+4x^2+2x+2) by (x^2+4x-2) is 2x + 6 with a remainder of -6x^2 - 18x + 14.