Polynomials (2 Viewers)

[Yellow]

suppose you're given a factor, say g(x), to a polynomial f(x). how exactly do you find the other factor if there is no remainder without having to perform long division?

 

[Lazarus]

example:

f(x) = x + 4x + x - 6

g(x) = x + 2

solution:

x + 4x + x - 6 = (x + 2)(x + bx + c)

Clearly, we need to begin with the x term. The RHS will then expand to give 2x, but we need 4x, so b = 2.

x + 4x + x - 6 = (x + 2)(x + 2x + c)

This will now expand to give 4x, but we only want 1x, so in order to lose 3x, c = -3.

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

You can see that this expands to give a constant term of -6.

The quadratic term is easily factorised.

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


Edit: Listen to quartic for polynomials of degree > 3.

 

[quartic]

The process is called inspection or intelligent inspection if you have a polynomial

P(x) = ax^4 + bx^3 + cx^2 + dx +e

with a factor (kx - r)^3

then

P(x) = (sx + l)(kx - r)^3

in this simple case s*k = a and l*(-r) = e

That is good if you are told that P(x) has a tripple zero (or if you are asked to show that) but in other cases what you basically do is balance the coefficients of each power of x and do this sequentially untill you have broken the polynomial into irreducable factors.

e.g. P(x) = ax^3 + bx^2 + cx + d

has a root at x = m

P(x) = (x-m)(ax^2 + [b + am]x + [c + {bm + am^2}])

if x = m is a legitimate root of P(x) then (-m)(c + [bm + am^2]) = d

then you are left with a quadratic that may or may not be factorised further.

That was a somewhat cumbersome explanation but I hope it helped.

Edit: bugger beaten to it by Lazarus and with a better explaination too.

 

[quartic]

Originally posted by Lazarus

Edit: Listen to quartic for polynomials of degree > 3.

I actually hate polynomials so my username really dosn't make much sense.

 

[Lazarus]

It sounds cool though. =P

 

[Dumbarse]

or u could just use long division...

 

[Lazarus]

He specifically asked for a method other than long division.


The other day, my calculus lecturer long divided a quintic by a quartic in under 10 seconds.

 

[quartic]

sure he didn't just memorise it all to impress people.

anyway despite the somewhat cumbersome explaination required inspection can be quite an intuitive prosess and can be much faster than long division.

 

[Dumbarse]

really??
i'm not too aware of this method, or maybe it just looks weird typed out, it always looks more complicated on the computer

 

[Yellow]

ah i get it now!

both explanations were good but i like quartic's! it's more structured i suppose. i need something like that to actually see what's going on but i'll be using Lazarus' method now coz it's faster.

 

[Dumbarse]

ohh ok, i decided to write it down, seems basic now written on paper, is it helpful u reckon?

 

[Dumbarse]

what happens when theres a remainder?. still work?

 

[Lazarus]

If there's a remainder, I don't think you'll be able to factorise the quadratic term (if you even get that far).

 

[Dumbarse]

yeh dont u have to look for a number that divides into the single coefficient, then test if it equal 0, then use long division to obtain other factor, and remaindrt??

 

[spice girl]

The funny thing is, Lazarus's first soln to a non-long-division way is actually longer than doing long division.

Long division's the way to go, like it or not. You might as well apply it to everything if u need the practice.

 

[Yellow]

long division? then you'll have to rewrite everything. there's also more chance of making a mistake.

but anyway, i've almost finished understanding this topic but there's this one question that i'm stuck with. mind if someone helped me out? it's a de moivre question. i don't know how to do the last bit.

cos(4x) = 8cos^4(x) - 8cos^2(x) + 1
Solve the equation 8y^4 - 8y^2 + 1 = 0 and deduce the exact values of cos(pi/8) and cos(5pi/8).

 

[Dumbarse]

yeh u solve it for de moivres theorem and also by binomial theorem,,, then match real parts, it should work out

for the ' deduce the exact values of cos(pi/8) and cos(5pi/8)."
cant u just, because youve just solved for cos(4x), put x=pi/32 to get cos (pi/8),, then put x=5pi/32 to get cos(5pi/8)
???

 

[Yellow]

yeh that's what i tried but how do you find the exact value of cos(pi/32)?

 

[Dumbarse]

so if u want the exact value of cos(pi/32)

put x= pi/128

 

[quartic]

Originally posted by quartic


P(x) = (sx + l)(kx - r)^3

in this simple case s*k = a and l*(-r) = e

Hmmmm I made a booboo. I believe what I meant is that in the above case s*k^3 = a and l*(-r)^3 = e

I don't believe no one picked me up on that one.