Polynomial Questions (2 Viewers)

[Iruka]

a^2(b+c)+b^2(c+a)+c^2(a+b) + kabc

=(a+b)(ab + c^2) + a^2c + b^2 c + kabc

=(a+b)(ab+c^2) + c(a^2 + b^2 + kab)

once we get to this point, we see that if k=2, then that last term becomes a perfect square, so

(a+b)(ab+c^2) + c(a^2 + b^2 + kab) = (a+b)(ab+c^2) + c(a+b)^2

= (a+b)(ab+c^2 + c(a+b))

=(a+b)[c(a+c) + b(a+c)]

=(a+b)(a+c)(c+b)

It's a symmetric polynomial!!

 

[lolokay]

more questions...

Find the value of k if (a+b) is a factor of a^2(b+c)+b^2(c+a)+c^2(a+b) + kabc and with this value of k factorize the expression.

I found k already but I dont know how to factorize the expression

start with (a+b)(c^2+?)
now to get a^2(b+c) there must be an ab and ac in the (?), to multiply with the a
(a+b)(c^2+ab+ac+?)
When multiplied by b, this gives ab^2 + abc
Now all that's left is cb^2 and (k-1)abc
So add in a bc, to multiply with the b
(a+b)(c^2+ab+ac+bc)
Which also gives + abc when the bc is multiplied by the a, so you have 2abc - and since (a+b) is not a factor of kabc, k must be 2.

To completely factorise, you just need to factorise (c^2+ab+ac+bc), which can be dones easily with the grouping method to become (a+c)(b+c)
Giving (a+b)(a+c)(b+c)

or you could've just guessed how to factorise it, since its so symmetrical and skipped all this

 

[3unitz]

sub in b = - a
a^2(c - a) + a^2(c + a) - ka^2c = 0 (factor theorem)
c - a + c + a - kc = 0
2c - kc = 0
k = 2

expanding we get
ba^2 + ca^2 + cb^2 + ab^2 + ac^2 + bc^2 + 2abc
= b(a^2 + c^2 + 2ac) + b^2(c + a) + ac(a + c)
= b(a + c)^2 + b^2(c + a) + ac(a + c)
= (a + c)(ab + bc + b^2 + ac)
= (a + c)[a(b + c) + b(c + b)]
= (a + c)(a + b)(b + c)

 

[Fortian09]

damn right u love maths

Whats the concept behind this though?



Another question i dont know
Prove that (b-c) is a factor of a³(b-c)+b³(c-a)+c³(a-b) and write down two other factors of this expression. Hence factorize the expression completely.



Harder questions...

I dun get these questions either.

The roots of x³ - 9root2x² + 46x - 30root2 = 0 are in arithmetic progression. solve the equation.

The roots of the equation 8x³-14x²-21x+27=0 are in G.P., solve the equation.

 

[3unitz]

damn right u love maths

Whats the concept behind this though?

Another question i dont know
Prove that (b-c) is a factor of a³(b-c)+b³(c-a)+c³(a-b) and write down two other factors of this expression. Hence factorize the expression completely.

if we know (b-c) is a factor the thing can be put in the form:
a^3(b-c) + b^3(c-a) + c^3(a-b) = (b - c)(.........)(..........)etc

subing in c=b will give RHS = (c - c)(.........)(..........)etc = 0
LHS = a^3(c - c) + b^3(b - a) + b^3(a - b)
= b^3(b - a) + b^3(a-b)
= b - a + a - b
= 0
= RHS
therefore (b-c) is a factor

ba^3 - ca^3 + cb^3 - ab^3 + c^3a - bc^3
= -a(b^3 - c^3) + a^3(b - c) + cb(b^2 - c^2)
= -a(b - c)(b^2 + bc + a^2) + a^3(b - c) + cb(b - c)(b + c)
= (b - c)[-a(b^2 + bc + a^2) + a^3 + cb(b+c)]
= (b - c)[-ab^2 - abc - a^3 + a^3 + cb(b + c)]
= (b - c)[-ab(b + c) + cb(b + c)]
= (b - c)(b + c)(cb - ab)
= b(b - c)(b + c)(c - a)

Harder questions...

I dun get these questions either.

The roots of x³ - 9root2x² + 46x - 30root2 = 0 are in arithmetic progression. solve the equation.

The roots of the equation 8x³-14x²-21x+27=0 are in G.P., solve the equation.

have to go to work,

but for a) let roots be: a - d, a, a + d
1. use sum of roots to find the value of a,
2. then use product of roots to get d

b) let roots be: a/r , a, ar
1. use product of roots to find the value of a,
2. use sum of roots to get a quadratic for r, and solve.

 

[bored of sc]

The roots of x³ - 9root2x² + 46x - 30root2 = 0 are in arithmetic progression. solve the equation.

Let roots be: a, a + n, a + 2n

3a + 3n = 9root2 -------- (1)

3a² + 6an + 2n² = 46 ----------- (2)

a³ + 3a²n + 2an² = 30root2 -------- (3)

After that the algebra gets insane.

 

[tacogym27101990]

Let roots be: , @ + n,@ ,@+ 2n


After that the algebra gets insane.

let the roots be a-n, a , a+n alot less messy

because when you do the sum of the roots you'll just get @=-b/3a
so you have one root straight away
the just do the product of the roots, sub in what ever @ is and find n

 

[Fortian09]

can someone plz solve the arithmetic progression question plz?

 

[bored of sc]

can someone plz solve the arithmetic progression question plz?

x³ -9root2 x² + 46x - 30root2 = 0

a = 1, b = -9root2, c = 46, d = -30root2

Let roots be @ - n, @ and @ + n.

Sum roots:
@ + (@ - n) + (@ + n) = -b/a = 9root2
3@ = 9root2
@ = 3root2 ------------ (1)

Product roots:
(@ - n)(@ + n)@ = -d/a = 30root2
@³ - @n² = 30root2 --------- (2)

Sub (1) into (2)

(3root2)³ - (3root2)n² = 30root2
54root2 - 3root2n² = 30root2
3root2n² = 24root2
n² = 8
n = 2root2 (root8 simplified)

Therefore, the roots are 3root2, 3root2 - 2 root2, 3root2 + 2root2.

Therefore x = 3root2, root2 and 5root2

 

[Fortian09]

The roots of x³+(2-P)x²-(Q+3)x=8 are in G.P. Find P and Q

 

[bored of sc]

The roots of x³+(2-P)x²-(Q+3)x=8 are in G.P. Find P and Q

Let roots be @/r, @ and @r.

a = 1, b = (2 - P), c = -(Q-3), d = -8

Sum roots (one at a time):

@ + @/r + @r = -b/a = P-2
@ + @/r + @r = P-2 ---------------(1)

Sum of roots (two at a time):

@² /r + @² + @²r = c/a = -(Q-3)
@²/r + @² + @²r = -(Q-3) ------------(2)

Product of the roots:

@ (@/r) (@r) = -d/a = 8 ------------(3)

From (3):

@³ = 8
@ = 2

Sub @ = 2 into both (1) and (2).

 

[tommykins]

回复: Re: Polynomial Questions

let roots be a/r, a, ar

x3+(2-P)x2-(Q+3)x=8

rearrange

x3+(2-P)x2-(Q+3)x - 8 = 0

a/r*a*ar = a^3 = 8. product of roots
a = 2

a/r + a + ar = -b/a solve there, subbing a = 2, solve for P.
a^2/r + a^2 + a^2r = c/a solve again for Q

 

[bored of sc]

Re: 回复: Re: Polynomial Questions

let roots be a/r, a, ar

x3+(2-P)x2-(Q+3)x=8

rearrange

x3+(2-P)x2-(Q+3)x - 8 = 0

a/r*a*ar = a^3 = 8. product of roots
a = 2

a/r + a + ar = -b/a solve there, subbing a = 2, solve for P.
a^2/r + a^2 + a^2r = c/a solve again for Q

But you don't know what r is...

 

[Fortian09]

Re: 回复: Re: Polynomial Questions

Hmm BOSC I tried your way but I couldnt get the P and Q values...

 

[tommykins]

Re: 回复: Re: Polynomial Questions

But you don't know what r is...

What...I'm doing it your way basically..

I've only been able to fnid a relationship between P and Q, not their values.

EDIT- got the idea of subbing in x = 2 in the original eqn as @ = 2, but you end up with 2p + Q = 1

and obviously 2p+q = 7 and 2p+q = 1 can't be solved