Help..my algrbrator is frozen, I can't divide?

Question

(14y + 8y2 + y3 + 12) ÷ (6 + y)

Answer 1

Well Dear Schnitzel;

First rewrite it like that;
y^3 + 8y^2 + 14y + 12 ÷ y +6

▪Step1
( y +6 ) ( y^2) = y^3 + 6y^2 ; {let R1 = y^2}
y^3 + 8y^2 + 14y + 12 - ( y^3 + 6y^2 ) =
y^3 + 8y^2 + 14y + 12 - y^3 - 6y^2 = 2y^2 + 14y + 12 {remainder}

▪Step 2
2y^2 + 14y + 12 ÷ y +6
(y+6)(2y) = 2y^2 + 12y ; { let R2 =2y }
2y^2 + 14y + 12 - ( 2y^2 + 12y) =
2y^2 + 14y + 12 - 2y^2 - 12y= 2y+12 {remainder}

▪Step 3
2y+12 ÷ y +6
(y+6)(2) = 2y +12 ; { let R3 = 2 }
2y+12 - ( 2y +12) =
2y+12 - 2y - 12 = 0 {remainder}

▪Step 4
Sum R1 , R2 , R3
R1 + R2 + R3
y^2 + 2y +2
(14y + 8y2 + y3 + 12) ÷ (6 + y) = R1 + R2 + R3
So;
(14y + 8y2 + y3 + 12) ÷ (6 + y) = y^2 + 2y +2
And Remainder is Zero'

Good Luck dear.

Answer 2

Divide the leading terms: y^3/y=y^2, so y^2 is the first term in the quotient.

(y^3 + 8y^2 + 14y + 12) - y^2(y+6) = 2y^2 + 14y + 12.

Now compute (2y^2+14y+12) / (6+y). Divide the leading terms, 2y^2/y = 2y, so 2y is the second term in the quotient.

(2y^2+14y+12) - 2y(y+6) = 2y+12.

Now compute (2y+12)/(6+y), but this is just 2. This is the third and last term in your quotient, which is y^2 + 2y + 2.

Answer 3

First rearange in descending order of operations:
(y^3 + 8y^2 + 14y + 12) / (y + 6)
Then divide synthetically (long division gets you the same results, but is so much harder to type):

-6 | 1...8 ....14....12
..........-6 ...-12 ..-12
------------------------
....1.. .2.... 2.......0

Therefore:
(y^3 + 8y^2 + 14y + 12) / (y + 6) = (y^2 + 2y + 2)

Answer 4

Use synthetic division.
Write the polynomial in descending order: y^3 + 8y^2 + 14y + 12
Write down only the coefficients.
y+6 = 0 when y = -6

-6|__1__8___14__12
_______-6__-12__-12
____1__2____2__0
Answer: y^2 + 2y + 2

Answer 5

rewrite the equation:
.. y3+8y2+14y+12 : y+6 =y2+2y+2
.- y3+6y2
------------------------------
........2y2+14y+12
.......-2y2+12y
-----------------------------
.................2y+12
................-2y+12
----------------------------
........................0