Question

does anyone know how to put x^3-4x^2+2x+3
in quadratic form

Answer 1

y = x-(4/3) ; x = y+(4/3) prolly not, but Decartes would do it

Answer 2

As I've mentioned, I basically know nothing about the quadratic form, but if i remember right--people like it more for 2nd degree polynomials, so... this factors to
(-3+x) (-1-x+x^2)
does that make it any easier?

Answer 3

yea but how did u factor the -3+x out

Answer 4

\[y^3+\frac{10}{3}y+\frac{25}{27}=0\]
will tells us the roots for our sub; then we just -4/3 for it :)

Answer 5

amistre64: how did you do this?

Answer 6

i was reading in hawkings book today the chapter on descartes; he had a reduction method to get rid of the second term

Answer 7

x = y + coeff/leading power

Answer 8

it just shifts the graph really; then the roots of the new graph are just how far off you moved it

Answer 9

5/3 - 4/3 = 1/3 as an original root

Answer 10

|its difficult understanding that the degree is 4 but theres only 3 zeros because the original equation was \[x^4-7x^3+14x^2-3x-9 \] so what i did was find p/q first so the only rational zero was 3 bc i used synthetic division and i was left with x^3-4x^2+2x+3 so now im suppose to use the quadratic equation but i dont know how to factor something out in order to make it a quadratic

Answer 11

lol, lets try it on that one :)

Answer 12

http://www.wolframalpha.com/input/?i=y%3D%28x%2B7%2F4%29%5E4%E2%88%927%28x%2B7%2F4%29%5E3%2B14%28x%2B7%2F4%29%5E2%E2%88%923%28x%2B7%2F4%29%E2%88%929

it got rid of the second term alright; now if only I knew what to do with it after that lol

Answer 13

like i know how to do it but i dont remember how to factor it out to leave it as a quadratic can u help me on that

Answer 14

you pretty much set up a pool of options as you best guesses

Answer 15

\[\pm\frac{factor.last.term}{factor.first.term}\]

Answer 16

since the leading term is a 1 that means that all our rational option are factors of 9

+- 1,3,9

Answer 17

okay

Answer 18

try out each term to see what zeros out

x4−7x3+14x2−3x−9
0 1 -6 8
------------------
1 )1 -6 8 5 ... not it



x4−7x3+14x2−3x−9
0 3 -12 6 9
---------------------
3 )1 -4 2 3 0 ... 3 IS a root , and the remaining digits are our coeffs


(x-3)(x^3 -4x^2 +2x +3)

then we try the same technique for this new cubic

Answer 19

okay im following

Answer 20

+- 1,3

x^3 -4x^2 +2x +3
0 3 -3 -3
--------------------
3) 1 -1 -1 0 .... that worked out nice, we got another zero


(x-3)(x-3)(x^2 -x -1) the remaining is a quadratic that can be done up with whatever manner you desire

Answer 21

synthetic division as it is called tends to be a bit smoother than longhand for this

Answer 22

oh okay so u got 3 twice

Answer 23

yes, which tells me that it touches the xaxis at x=3 and bend back again instead of actually going thru it

Answer 24

http://www.wolframalpha.com/input/?i=+x4%E2%88%927x3%2B14x2%E2%88%923x%E2%88%929

the graph shows that :)

Answer 25

oh okay thanks dude i understand now u really helped dude