Solve using long division( x^4-4x^2-1)/x-4 I know what the answer is but dont understand how to get it

- anonymous

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- schrodinger

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- anonymous

idk

- amistre64

do you recall how to do regular long division with numbers?

- anonymous

http://www.youtube.com/watch?v=l6_ghhd7kwQ
he is really helpful

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## More answers

- amistre64

recall that when we get a "solution" that we "subtract" to get a remainder... like:
1 <---- 3 time 1 = 3; so we -3 from 4
--------
3 | 40
-3 <---- subtract 3 to get
------
10 <---- we get a 1 and bring down the '0' to start again

- amistre64

its the same steps with dividing polynomials.... just remember to subtract...

- amistre64

x^3 <--(x^3)(x-4) = (x^4 -4x^3)
----------------
x-4 | x^4 -4x^2 -1
-x^4 +4x^3
------------
4x^3 -4x^2 -1 <- bring down what doesnt cancel
out and start over

- amistre64

x^3 +4x^2 <--(why 4x^2 ?)
----------------
x-4 | x^4 -4x^2 -1
-x^4 +4x^3
------------
4x^3 -4x^2 -1
-4x^3 +16x^2
--------------
12x^2 -1

- amistre64

x^3 +4x^2 +12x +48
----------------------
x-4 | x^4 -4x^2 -1
-x^4 +4x^3
------------
4x^3 -4x^2 -1
-4x^3 +16x^2
--------------
12x^2 -1
-12x^2 +48x
--------------
48x -1
-48x +192
----------
191 <-- remainder :)

- amistre64

the total answer is:
191
x^3 +4x^2 +12x +48 + ----
x-4

- anonymous

lol i was sittin here i figured it out too i was butsting my brain trying to remember its been so long since long division i was forgetting about the place holders thank you again :)

- amistre64

:) youre welcome

- anonymous

okay your a smart cookie help me with this one i got the answer but i dont understand what it means by finding the zeros of f? the problem was once again synthetic division to divide x^3 +8x^2-3x-90 by X=5 i got X^2+3x-18 as the answer which is said to be correct but i dont understand the zeros question. it gave -6,-5.3 as the answers where did it come from

- amistre64

finding the zeros means that you want to find the x intercepts of the graph; which only happen when y=0

- amistre64

suppose we are given the equation:
y = 5x+20; what would the zero be for this:
0 = 5x + 20 when x = ?

- anonymous

4=x

- anonymous

-4

- anonymous

my bad wrong sign

- amistre64

also, when the synthetic division has a remainder of "0"; you know that it is a factor :)
5 | 1 +8 -3 -90
0 +5 +65 +310
---------------
1 +13 +62 ...... <-- not a zero... did I typo it?

- amistre64

yes, the zero for that linear equation is -4 :) also known as a root

- amistre64

and the x intercept

- amistre64

-5 | 1 +8 -3 -90
0 -5 -15 +90
---------------
1 +3 -18 0 <-- better;
so when x=-5 we get a root, a zero

- amistre64

(x+5) is a factor then..
(x+5)(x^2 +3x -18)

- amistre64

when we factor out the quadractic we get...
x^2 +3x -18 = (x+6)(x-3)

- amistre64

all our factors of the cubic are then:
(x+5)(x+6)(x-3) = 0 ; remember when you multiply by '0' you get zero
5(0) = 0
143(0)=0
3(0)(16) = 0
in other words, there are 3 solutions here....

- amistre64

(0)(x+6)(x-3) = 0 when x=-5
(x+5)(0)(x-3) = 0 when x= -6
(x+5)(x+6)(0) = 0 when x=3

- anonymous

ok im still following so were good lol

- amistre64

our zeros, our roots, our y=0, our x intercepts are when :)
x= -6,-5,3

- amistre64

http://www.wolframalpha.com/input/?i=x%5E3+%2B8x%5E2-3x-90+from+-7+to+4

- amistre64

you can see from that that the graph cross the x axis at -6, -5 and 3

- anonymous

thank u sooooo much lol im havin a hard day full of brain farts

- amistre64

:) theyll clear out by suppertime :)

- anonymous

countin on it

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