Question

How do you divide `(2x^2+10x+12) /(x+3)` using polynomial long division?

Answer 1

`(2x^2+10x+12)div(x+3) = (2x+4)` with remainder `(0)`

Explanation:

`{: (,,2x,+4,), (,,"------","------","------"), (x+3,")",2x^2,+10x,+12), (,,2x^2,+6x,), (,,"------","------","------"), (,,,4x,+12), (,,,4x,+12), (,,,"------","------"), (,,,,0) :}`

Detailed explanation (ignore if the above made adequate sense)

Starting from:
`{: (,,"------","------","------"), (x+3,")",2x^2,+10x,+12) :}`
considering only the first term of the divisor (`x`) and of the dividend (`2x^2`)
we see that `2x^2 div x = 2x`; so we write `2x` above `2x^2`
`{: (,,2x,,), (,,"------","------","------"), (x+3,")",2x^2,+10x,+12) :}`

now we multiply the divisor (`x+3`) by the term we just wrote above the line (i.e. `2x`) to get `2x^2+6x` which we write under the dividend
`{: (,,2x,+4,), (,,"------","------","------"), (x+3,")",2x^2,+10x,+12), (,,2x^2,+6x,) :}`

Subtract from the dividend
`{: (,,2x,+4,), (,,"------","------","------"), (x+3,")",2x^2,+10x,+12), (,,2x^2,+6x,), (,,"------","------","------"), (,,,4x,+12) :}`

Again using only the first term of the divisor (`x`) and the first term of our remainder (`4x`)
we see that (`4x div x = 4`; so we write `4` as the next term above the top line.
`{: (,,2x,+4,), (,,"------","------","------"), (x+3,")",2x^2,+10x,+12), (,,2x^2,+6x,), (,,"------","------","------"), (,,,4x,+12) :}`

Multiply the divisor (`x+3`) by the new quotient term (`+4`) to get `4x+12`; then subtract to get a final remainder of `0`
`{: (,,2x,+4,), (,,"------","------","------"), (x+3,")",2x^2,+10x,+12), (,,2x^2,+6x,), (,,"------","------","------"), (,,,4x,+12), (,,,4x,+12), (,,,"------","------"), (,,,,0) :}`