help with a couple of polynomial q's (1 Viewer)

[Patato]

hi all

these questions came from the resources, there are solutions but they aren't worked and i cant seem to get out a few. so if anyone can have a crack and explain to me their solution that would be cool



1) Given that there is a constant c such that (x4 + y4) = (x² + cxy + y²)(x² - cxy + y²) identically in x and
y, find c.



part b is the only one im having trouble with here, i just figured you should have part a

2). Factorise completely the polynomial p(x) = x^3 - x² - 8x + 12, given that the equation p(x) = 0
has a repeated root.
b. The polynomial q(x) has the form q(x) = p(x)(x + a), with p(x) as in (a) and where the
constant a is chosen so that q(x)  0 for all real values of x. Find all possible values of a


show (x^2 - 3x + 1) has no common zeros with (2x^2 - 4x - 2). i can do this, but its a long method..i feel like im missing a simple solution.



thanks

 

[xV1P3R]

1. For the first one, you either expand (which might take a while) or take a look at coefficients.

So for example, coefficient of x²y² on the LHS is 0, while on the RHS is 1+1-c², then solve from there.

2. Is that > or < 0?

3. Find the roots to 1 and 2 and show that they're not equal? (or is that the long method)

 

[blackops23]

hi all

these questions came from the resources, there are solutions but they aren't worked and i cant seem to get out a few. so if anyone can have a crack and explain to me their solution that would be cool



1) Given that there is a constant c such that (x4 + y4) = (x² + cxy + y²)(x² - cxy + y²) identically in x and
y, find c.



part b is the only one im having trouble with here, i just figured you should have part a

2). Factorise completely the polynomial p(x) = x^3 - x² - 8x + 12, given that the equation p(x) = 0
has a repeated root.
b. The polynomial q(x) has the form q(x) = p(x)(x + a), with p(x) as in (a) and where the
constant a is chosen so that q(x)  0 for all real values of x. Find all possible values of a


show (x^2 - 3x + 1) has no common zeros with (2x^2 - 4x - 2). i can do this, but its a long method..i feel like im missing a simple solution.



thanks

Q1. x^4 + y^4 = (x^2 + y^2)^2 - 2(x^2)(y^2)
Difference of 2 squares....
therefore:
x^4 + y^4 = (x^2 + y^2 - xy*root(2)]*[x^2 + y^2 + xy*root(2)]
=(x^2 + y^2 - xyc)(x^2 + y^2 + xyc)

obviously we can see that c = root 2 = 1.414... wat eva


Q2. P(x) = (x-2)(x-2)(x+3)
Q(x)=(x-2)(x-2)(x+3)(x+a) = 0

are you sure you worded the question right? the value of a can be anything at all can't it? "Find possible values of a" -- question's answer seems like all real a, but is there something i'm missing from the question?

Also the last part about showing the roots aren't equal:
SMART PPL, PLEASE CHECK ME ON MY FOLLOWING SOLUTION BECAUSE IT SEEMS BS TO ME

FIRST EQUATION: x^2 - 3x + 1 =0 has some 2 roots
2ND EQUATION: 2x^2 - 4x - 2 =0
NOW DIVIDE THIS BY 2
x^2 - 2x - 1 = 0

x^2 - 3x + 1 (not equal to) x^2 - 2x - 1, as coefficients are different

therefore they have different roots

 

[xV1P3R]

@ blackops

c = +/- sqrt(2)

And I think your solution for the last part might look BS because the question is pretty BS lol!

 

[blackops23]

@ blackops

c = +/- sqrt(2)

And I think your solution for the last part might look BS because the question is pretty BS lol!

yeah thanks, also, yeah its pretty BS because i didn't even understand the question! is that whole bold part for Q2 ONE WHOLE QUESTION, OR are there two distinct separate parts...? lol