How do you find the quotient of #(p^3-4p^2+9)div(p-1)# using long division?

Question

Kennedy Adams · Accepted Answer

#p^2-3p+3 # and remainder#12/(p-1)# #(p^3-4p^2+9)-:(p-1)# #color(white)(..........)color(white)(..........)p^2-3p+3# #p-1|overline(p^3-4p^2+0+9)# #color(white)(............)ul(p^3-p^2)# #color(white)(...............)-3p^2+0# #color(white)(................)ul(-3p^2-3p)# #color(white)(..................................)3p+9# #color(white)(..................................)ul(3p-3)# #color(white)(..........................................)12# #(p^3-4p^2+9) / (p-1) = p^2-3p+3 # and remainder#12/(p-1)#

Jameson Allen · Answer

undefined To find the quotient of (p^3-4p^2+9) divided by (p-1) using long division, follow these steps:

1. Divide the first term of the dividend (p^3) by the first term of the divisor (p). The result is p^2.

2. Multiply the divisor (p-1) by the quotient obtained in step 1 (p^2). The result is p^3-p^2.

3. Subtract the product obtained in step 2 (p^3-p^2) from the dividend (p^3-4p^2+9). This gives you -3p^2+9.

4. Bring down the next term from the dividend, which is -3p^2. Now you have -3p^2+9 as the new dividend.

5. Divide the first term of the new dividend (-3p^2) by the first term of the divisor (p). The result is -3p.

6. Multiply the divisor (p-1) by the quotient obtained in step 5 (-3p). The result is -3p^2+3p.

7. Subtract the product obtained in step 6 (-3p^2+3p) from the new dividend (-3p^2+9). This gives you 6p+9.

8. Bring down the next term from the dividend, which is 6p. Now you have 6p+9 as the new dividend.

9. Divide the first term of the new dividend (6p) by the first term of the divisor (p). The result is 6.

10. Multiply the divisor (p-1) by the quotient obtained in step 9 (6). The result is 6p-6.

11. Subtract the product obtained in step 10 (6p-6) from the new dividend (6p+9). This gives you 15.

12. There are no more terms left in the dividend, so the division is complete. The quotient is p^2-3p+6, and the remainder is 15.