• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

Polynomials (1 Viewer)

Yellow

Member
Joined
Aug 15, 2002
Messages
82
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

Retired
Joined
Jul 6, 2002
Messages
5,965
Location
CBD
Gender
Male
HSC
2001
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

Member
Joined
Aug 6, 2002
Messages
82
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.
 
Last edited:

quartic

Member
Joined
Aug 6, 2002
Messages
82
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

Retired
Joined
Jul 6, 2002
Messages
5,965
Location
CBD
Gender
Male
HSC
2001
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. :eek:
 

quartic

Member
Joined
Aug 6, 2002
Messages
82
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.
 

Yellow

Member
Joined
Aug 15, 2002
Messages
82
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.
 

Lazarus

Retired
Joined
Jul 6, 2002
Messages
5,965
Location
CBD
Gender
Male
HSC
2001
If there's a remainder, I don't think you'll be able to factorise the quadratic term (if you even get that far).
 

spice girl

magic mirror
Joined
Aug 10, 2002
Messages
785
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

Member
Joined
Aug 15, 2002
Messages
82
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).
 
Last edited:

Dumbarse

Member
Joined
Aug 9, 2002
Messages
423
Location
BOS moderator & operator
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)
???
 

quartic

Member
Joined
Aug 6, 2002
Messages
82
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. :rolleyes:
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top