when i divide (3x^2+7x-18)/(x-3)
i get 3x+16- over 30 with three squares left for me to fill it
does x-3 go in those squares
2007-07-26 16:18:04 · 2 answers · asked by yoshi_gurl08 1 in Science & Mathematics ➔ Mathematics
Synthetic division:
3.....3 ..7 ..-18
............9 ....48
-----------------
........3 16 30
So the answer to (3x^2+7x-18)/(x-3) is:
3x+16 remainder 30
In other words,
(x-3)(3x+16)+30/(x-3)=3x^2+7x-18
-//-//-
2007-07-27 12:55:01 · answer #1 · answered by Popo B 3 · 0⤊ 0⤋
In (3x^2+7x-18) ignore the terms in just "x" for the moment
now what do you multiply (x-3) by to get (3x^2+ something -18)
that is (3x + 6) since 3x*x = 3x^2 and 6*(-3) = -18
Bring back the "x" term and (3x + 6)*(x - 3) gives 3x^2 - 3x -18
The difference between your original expression and this is +10x
so (3x^2+7x-18) = (3x +6)*(x - 3) + 10x
and (3x^2+7x-18)/(x-3) = 3x + 6 + 10x/(x - 3)
2007-07-26 16:32:02 · answer #2 · answered by Captain Mephisto 7 · 0⤊ 0⤋