Using long division solve #(12x^2+9x-7)-:3x#?

Question

Eli Assouline · Accepted Answer

#4x+3-7/(3x)# #(12x^2+9x-7) color(blue)(-:3x)# #" "12x^2+9x-7# #color(magenta)(4x)color(blue)((3x))->color(white)(.) ul(12x^2" ")larr" subtract"# #" "0+9x-7# #color(magenta)(3)color(blue)((3x))->" "ul(color(white)(.)9x" ")larr" subtract"# #" "0color(magenta)(-7""larr" remainder")# ;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(magenta)(4x+3-7/(color(blue)(3x)))#

Savannah Anderson · Answer

undefined To solve (12x^2+9x-7) ÷ 3x using long division, follow these steps:

1. Divide the first term of the dividend (12x^2) by the divisor (3x). The result is 4x.
2. Multiply the divisor (3x) by the result obtained in step 1 (4x), giving you 12x^2.
3. Subtract the product obtained in step 2 (12x^2) from the first term of the dividend (12x^2). The result is 0.
4. Bring down the next term from the dividend, which is 9x.
5. Divide the term obtained in step 4 (9x) by the divisor (3x). The result is 3.
6. Multiply the divisor (3x) by the result obtained in step 5 (3), giving you 9x.
7. Subtract the product obtained in step 6 (9x) from the term brought down (9x). The result is 0.
8. Bring down the last term from the dividend, which is -7.
9. Divide the term obtained in step 8 (-7) by the divisor (3x). The result is -7/3x.
10. Multiply the divisor (3x) by the result obtained in step 9 (-7/3x), giving you -7.
11. Subtract the product obtained in step 10 (-7) from the term brought down (-7). The result is 0.

The final result of the division is 4x + 3 - (7/3x), or simply written as 4x + 3 - 7/3x.