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[Sirius Black]

hi all,
I am a yr 11 student but I am doing yr12 4u maths so it is hard for me

Here is a question
How to factoriseP(x)= x^3-3x^2+4 and P(x)=x^4+x^3-3x^3-5x-2 and find the multiple root

And has anyone goy any general idea about solving polynomials. You know like P(x)=x^4-2x^3+2x-1 has a multiple zero ,find the zero.
In this question how come they can get P(1)=0 directly in the first step??

Thank you:apig:

 

[Xayma]

Remainder Theorem

for the first one
P(-1)=0

Therefore (x+1) is a factor. Do division of it.

For the second one P(2)=0
Therefore divide it by (x-2)

 

[CM_Tutor]

Originally posted by Sirius Black
Factorise P(x)= x^3-3x^2+4

To do this, you first need to find a solution of P(x) = 0. The usual way to do this is to try the factors of the constant. Here the constant is 4, so we'd try 1, 2, 4, -1, -2, and -4.

P(1) = 1³ -3(1)² + 4 = 1 - 3 + 4 = 2 <> 0 ..... No good, keep going.
P(2) = 2³ -3(2)² + 4 = 8 - 12 + 8 = 4 <> 0 ..... No good, keep going.
P(4) = 4³ -3(4)² + 4 = 64 - 48 + 4 = 20 <> 0 ..... No good, keep going.
P(-1) = (-1)³ -3(-1)² + 4 = -1 - 3 + 4 = 0 ..... We found one.

So, x = -1 is a solution of P(x) = 0, and hence (x + 1) is a factor of P(x). We could now do long division to find the quadratic factor, and then factorise that by the usual quadratic methods, or we could find the other factors using root theory, or we could use root theory to find the quadratic - I favour the third of these, as it is shortest.

Consider P(x) = 0 with roots a, b and -1.
Sum of roots: a + b - 1 = -(-3) / 1 and hence a + b = 4
Product of roots: a * b * (-1) = -(4) / 1 and hence a * b = 4
Since these are the roots of the quadratic factor, it must be x² - 4x + 4

So, P(x) = (x + 1)(x² - 4x + 4) = (x + 1)(x - 2)²

factorise ... P(x)=x^4+x^3-3x^3-5x-2 and find the multiple root

Firstly, does this mean P(x) = x⁴ + x³ - 3x² - 5x - 2? I'm going to assume that it does.
Now, by trying factors of -2, we quickly see that P(-1) = 0 and so (x + 1) is a factor of P(x).

We also know that P(x) has a multiple root, which means that P(x) = 0 and P'(x) = 0 share a root.
P'(x) = 4x³ + 3x² - 6x - 5
Notice that P'(-1) = 0, and so x = -1 is that multiple root.
Thus, (x + 1)² is a factor of P(x).

We have now the same choices as before, and I will use the same approach.
Consider P(x) = 0 with roots a, b, -1 and -1.
Sum of roots: a + b - 1 - 1 = -(1) / 1 and hence a + b = 1
Product of roots: a * b * (-1) * (-1) = -2 / 1 and hence a * b = -2
Since these are the roots of the quadratic factor, it must be x² - x - 2

So, P(x) = (x + 1)²(x² - x - 2) = (x + 1)²(x - 2)(x + 1) = (x + 1)³(x - 2)

And has anyone goy any general idea about solving polynomials. You know like P(x)=x^4-2x^3+2x-1 has a multiple zero ,find the zero.
In this question how come they can get P(1)=0 directly in the first step??

Thank you:apig:

I'm going to leave you to think about this one.

 

[Sirius Black]

Thank you

what is the formula for the sum/product of roots of ploynomials?

By the way, how to type the superscrip or subscript?

 

[Xayma]

Ok sub script is < sub > < /sub > without the spaces. Super is the same with sup.

Ok for a polynomical, the first coefficent (that of the highest power) is "a", the second is "b" (even if it is 0) third is "c" etc.

The sum of roots one at a time (alpha+beta+gamma etc)=-b/a
The sum of roots two at a time (alpha*beta+alpha*gamma+beta*gamma)= c/a
Three at a time=-d/a
Four at a time=e/a
Etc with the sign alternating.

 

[Grey Council]

I swear, cm_tutor explains better than my teacher. seriously. my teacher just tells us its easy, and to figure it out ourselves.

btw, which school do you go to sirius black? the heck, never heard of a school accelerating 4u maths in year 11. you must be good, to have accelerated.

 

[Heinz]

Originally posted by Grey Council
I swear, cm_tutor explains better than my teacher. seriously. my teacher just tells us its easy, and to figure it out ourselves.

if onle cm_tutor was qualified to teach.

btw, which school do you go to sirius black? the heck, never heard of a school accelerating 4u maths in year 11. you must be good, to have accelerated.

youll find that alot of selective schools accelerate in 4u maths. ruse is one surprising exception.

 

[Grey Council]

hurm, sydney boys doesn't accelerate either. thats why i'm surprised.

i didn't know heaps of schools accelerate. thats quite ironic, really. ruse doesn't.

 

[Xayma]

Originally posted by Grey Council
i didn't know heaps of schools accelerate. thats quite ironic, really. ruse doesn't. but then, the top ruse maths class finishes off 4u, 3u and 2u maths by like term 1. they don't need to accelerate. :|

Or blink or breath.