How can I factor a quartic equation without a calculator for the AIME?

[p53ud0 dr34m5]

this was on the aime, and i was wondering how to factor it without a calculator.

x^4-4x^3+6x^2-4x=2005

 

[shmoe]

Do the coefficients 1, -4, 6, -4, look familiar? (think pascals triangle)

 

[p53ud0 dr34m5]

will this method yield both imaginary and real roots? I am not following your hint.

 

[shmoe]

It will give all roots. Furthur hint:

x^4-4x^3+6x^2-4x+1=(x-1)^4

 

[p53ud0 dr34m5]

ok, so:
i can add one to each side and have:
x^4-4x^3+6x^2-4x+1=2006
now, i can use the (x-1)^4
(x-1)^4=2006?

 

[saltydog]

ok, so:
i can add one to each side and have:
x^4-4x^3+6x^2-4x+1=2006
now, i can use the (x-1)^4
(x-1)^4=2006?

Yes. Next, use the formula for complex roots. You know:

\sqrt[4]{1}=+1,-1,+i,-i

So:

\sqrt[4]{2006}

is?

 

[p53ud0 dr34m5]

thanks for your help. i have some more questions about other concepts, but ill post a new thread. I am drained from getting up early for the sat. good thing it was easy.

 

[p53ud0 dr34m5]

oh and how do you find the 4th root of a number without your calculator?

 

[Data]

Well, you can get as close as you like to the positive real root by using Newton's Method:

We want the positive, real root of f(x)=x^4 - 2006. Since 6^4 = 1296 < 2006 < 2401 = 7^4 we make a first guess that the root is x_1 = \frac{13}{2} = 6.5. Newton's method tells us that a better guess is

x_2 = x_1 - {f(x_1) \over f^\prime (x_1)} = \frac{13}{2} - \frac{\left(\frac{13}{2}\right)^4- 2006}{4\cdot \left(\frac{13}{2}\right)^3} =\frac{13}{2} - \frac{\frac{28561}{16} - \frac{32096}{16}}{\frac{8788}{8}} = \frac{13}{2} + \frac{3535}{17576} = \frac{114244}{17576} + \frac{3535}{17576} = \frac{117779}{17576} \approx 6.701

By iterating the method (using 6.701 as a second guess, and applying the same process) you can get arbitrarily close, since our function is continuous and the method does converge for this function. The root is really 6.6924129...

 

[tongos]

specifically the question was this...
Let P be the product of the nonreal roots of x^4-4x^3+6x^2-4x=2005. Find the greatest integer that is less than or equal to P.

(1+fourthroot of 2006)(1-fourthroot of 2006)P= -2005
when you factor, you'll see that all roots have to multiply to give -2005

((2006^0.5)-1)P=2005

rationalizing the denominator, 2005(sqrt(2006)+1)/2005
sqrt(2006) is between 44 and 45. thus adding one will yield 45.(something). the greatest integer less than this is 45.

 

[tongos]

i personally like the solution to number 8,

2^(333x-2)+2^(111x+2)=2^(222x)+1
has three real roots
find the sum of the roots.

let 2^(111x)=r

then r^3+16r-8r-4=0
let a,b,c be the roots of the polynomial.

then 2^(111x)=a x= Ina/(111In2)

then 2^(111x)=b x=Inb/(111In2)

then 2^(111x)=c x=Inc/(111In2)

adding them and letting (Ina+Inb+Inc)/(111In2)=Inabc/(111In2)

by the third degree polynomial we know that the roots multiply to give 4.

thus (2In2)/(111In2)= 2/111 which is the sum of the roots. m/n, m+n=113

 

[p53ud0 dr34m5]

see, i could get the answer to the one about the imaginaries with a calculator, but i couldn't without one. i suck without a calculator. the other problem you showed was really easy. i got that one right for sure, because i got 113, but i went about a different way to get the 2/111. i feel really really really stupid that i didnt think of Newtons method to find the root. *sigh*

 

[tongos]

there was no need for Newton's method, that's why they say the greatest integer that is less than P. Though it did require to know that the sqrt(2006) is between 44 and 45.

 

[tongos]

i was surprised of the easiness of 1-12. 13,14, and 15 are killer problems though.

 

[p53ud0 dr34m5]

heh, i was off by 3 on 15. it sucked. I am not that good at math that requires so much thinking, because this is my first year actually doing any math competition. i was surprised that i even made above a 100 on the amc 12, haha. i think on 13 you have to use factorials, and i have no idea about 14. unfortunately, i thought a lot of th eproblems were hard. i only got 3 on the AIME right. i could have gotten at least a 6, but i can not add without a calculator. and i was off by no more than 5 on three that i missed. oh, well. tongos, if i have a question about a problem, could i ask you. do you have aim?