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OpenStudy (anonymous):

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

OpenStudy (anonymous):

idk

OpenStudy (amistre64):

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

OpenStudy (anonymous):

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

OpenStudy (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

OpenStudy (amistre64):

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

OpenStudy (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

OpenStudy (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

OpenStudy (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 :)

OpenStudy (amistre64):

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

OpenStudy (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 :)

OpenStudy (amistre64):

:) youre welcome

OpenStudy (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

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (anonymous):

4=x

OpenStudy (anonymous):

-4

OpenStudy (anonymous):

my bad wrong sign

OpenStudy (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?

OpenStudy (amistre64):

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

OpenStudy (amistre64):

and the x intercept

OpenStudy (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

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (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....

OpenStudy (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

OpenStudy (anonymous):

ok im still following so were good lol

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (anonymous):

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

OpenStudy (amistre64):

:) theyll clear out by suppertime :)

OpenStudy (anonymous):

countin on it

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