2). What is the answer from the following?
(6y^4 + 15y^3 + 28y + 6) ÷ (y + 3)
Synthetic Division:
-3)....6....15.....0....28....6
........6.....-3....9.....1...|..3
Quotient: 6x^3 - 3x^2 + 9x + 1
Remainder: 3
============================
3).Find all the numbers for which the rational expression
below is NOT defined:
x^2-16/x^2-7x+12
---
Solve: x^2-7x+12 = 0
(x-6)(x+3) = 0
x = 6 or x = -3
Not defined for x = 6 or x = -3
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4). simplify
1 1
---------------- + -----------
a^2 + 7a + 12 a^2 + a - 6
]--------------------------------------------------------------------------------
1 1
--------------------------- + ------------------------------
a^2 + 2a - 8 a^2 + 5a + 4
===============================================================================
{1/[(a+3)(a+4)] + 1/[a+3)(a-2)]} / {1/[(a+4)(a-2)] + 1/[(a+4)(a+1)]
-----------------------------------------------------------------------
[(a-2)/lcm + (a+4)/lcm] / [(a+1)/lcm + (a-2)/lcm]
----
[(2a+2)/lcm] / [(2a-1)/lcm]
----
Invert the denominator and multiply:
2(a+2)/[(a+3)(a+4)(a-2)] * [(a+4)(a-2)(a+1)]/(2a-1)
-------------------------------------------------------
Cancel where you can to end up with:
[2(a+2)(a+1)]/[(2a-1)(a+3)]
=============================
Cheers,
Stan H.
You can put this solution on YOUR website! #2 one way is to start by setting it up like long division.
the first step
y+3 |into| 6y4 + 15y3 goes 6y3 times
6y^3(y+3)=6y^4+18y^3
so we subtract from 6y4 + 15y3 and get -3y^3 remember that 15-18=-3
how many times does y go into -3y^3 answer -3y^2
we don't have any y^2 in our term
so multiply out (y+3)(-3y^2) and get -3y^3-9y^2
so next we subtract 9y^2 from zero and get 9y^2
how many times does y go into 9y^2
9y of course
(y+3)*9y=9y^2+27y
but we have 28y
so we get y as the remainder
now we have y+6
y+3 goes into y+6 once with a remainder of 3
so we have (6y^3-3y^2+9y+1)(y+3) +1