Polynomial Questions (2 Viewers)

[Fortian09]

whats the reason though?

 

[lyounamu]

whats the reason though?

The original equation you provided was not divisible by (x-1) as the Remainder theorem suggested. When we actually divided the original equation by (x-1), we learnt that there was a remainder of -7. Therefore, if you wish to make this divisible, you had to add 7 to the original equation.

 

[Aerath]

whats the reason though?

The question asked you to make it divisible. Look at my answer in purple pen on the previous page (unless it was too small). =\

 

[Fortian09]

More questions

What property is common to all equations of the form
x ³ + mx + k = 0?

 

[lyounamu]

More questions

What property is common to all equations of the form
x ³ + mx + k = 0?

Wow, what is that meant to ask?

 

[tommykins]

Wow, what is that meant to ask?

All I can say is that its discriminant = m^2 - 4k

It's x^3, not x^2.

 

[Fortian09]

i dont get this question either...

 

[independantz]

More questions

What property is common to all equations of the form
x ³ + mx + k = 0?

I'll take a stab and say the sum of roots equals zero lol.

 

[lyounamu]

It's x^3, not x^2.

I am blind.

 

[tommykins]

I'll take a stab and say the sum of roots equals zero lol.

Correct.

 

[Fortian09]

So the sum of roots is the correct answer?

 

[vds700]

So the sum of roots is the correct answer?

yeah it should be. Also the product of the roots = -k.

This from a past trial?

 

[Fortian09]

Why is it the sum and product of roots equal 0?

 

[Slidey]

The sum of roots is zero because there is no coefficient for the degree 2 term, and the product is -k because the coefficient of the degree zero term is k (and the degree is odd).

 

[bored of sc]

More questions

What property is common to all equations of the form
x ³ + mx + k = 0?

Would saying the co-efficient of the x² being 0 be correct?

 

[Slidey]

Would saying the co-efficient of the x² being 0 be correct?

Interestingly, you can convert any cubic polynomial into the depressed form y^3+A^y=B by making the substitution x=y-b/(3a) in ax^3+bx^2+cx+d=0. Indeed, this is how the cubic formula was found.

 

[bored of sc]

Interestingly, you can convert any cubic polynomial into the depressed form y^3+A^y=B by making the substitution x=y-b/(3a) in ax^3+bx^2+cx+d=0. Indeed, this is how the cubic formula was found.

Wow.

 

[tacogym27101990]

In AT Progression Series let alpha = a beta = a + d gamma = a +2d and so on.

id use a-d, a and a+d, itd work out nicer for the sum of the roots

 

[Fortian09]

I have more problems...
hmm i didnt expect this to be turned into a hot thread...
but oh wellz
ummm
people should ask polynomial questions on this coz i feel bad when its just me asking...

so yeah

how would i completely factorize the expression a²(b+c) + b² + c²(a+b) + 2abc when (a+b) is a factor?

 

[Fortian09]

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