Polynomial (1 Viewer)

Lukybear · Apr 21, 2010

Remainder of P(x) = x^3-3x^2 + 4x -2 when divided by (x+i)

Isnt P(-i) = -i + 3 -4i -2

since (-1 x i)^2 = 1 x i^2 = -1?

stampede · Apr 21, 2010

yeh, wheres the problem?

Lukybear · Apr 22, 2010

Could you do it for me? Divide P(x) by (x+i) Just to make sure.

omgmynameislong · Apr 22, 2010

i+3-4i-2
=1-3i

Lukybear · Apr 22, 2010

One more question.

If P(x) is a monic, degree 4, integer coefficent and constant term of 2. P(x) has a zero i, and a rational zero. The sum of the zeroes of P(x) is a postive real number. Find P(x), factorise fully with R.

After solving question, we get ab = 2, where a and b are roots

The solution given states roots are 1,2 since 1+2 >0. However, can a and b be other values, that satisfy P(x), i.e. 1/2 and 4?

Lukybear · Apr 22, 2010

If someone could help me to solve this question, I would be very much appreciated.

P(x) is a monic polynomial of degree 4 with integer coefficents and constant term 4. One Zero is sqrt 2, another zero is rational and the sum of the zeros is positive. Factorise P(x) fully over R

stampede · Apr 22, 2010

Lukybear said:
One more question.

If P(x) is a monic, degree 4, integer coefficent and constant term of 2. P(x) has a zero i, and a rational zero. The sum of the zeroes of P(x) is a postive real number. Find P(x), factorise fully with R.

After solving question, we get ab = 2, where a and b are roots

The solution given states roots are 1,2 since 1+2 >0. However, can a and b be other values, that satisfy P(x), i.e. 1/2 and 4?

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ontVertAlignInTxbx/> <w:Word11KerningPairs/> <w:CachedColBalance/> </w:Compatibility> <w:BrowserLevel>MicrosoftInternetExplorer4</w:BrowserLevel> <m:mathPr> <m:mathFont m:val="Cambria Math"/> <m:brkBin m:val="before"/> <m:brkBinSub m:val="--"/> <m:smallFrac m:val="off"/> <m:dispDef/> <m:lMargin m:val="0"/> <m:rMargin m:val="0"/> <m:defJc m:val="centerGroup"/> <m:wrapIndent m:val="1440"/> <m:intLim m:val="subSup"/> <m:naryLim m:val="undOvr"/> </m:mathPr></w:WordDocument> </xml><![endif]--><style>  </style> eqn in root form is:

x^4 - (α+β+γ+δ)x^3 + (αβ+αγ+αδ+βγ+βδ+γδ)x^2 - ......
where a, β, γ, δ are the roots.

sum of roots: α+β+γ+δ = α+β (since if i is a root, then -i is also a root, ie: α + β + i - i = α+β)

now, sum of roots = -b/a which has to be an integer since the q says so.
therefore α+β = -b/a = integer = k

therefore: α+β = k

if α and β are .5 and 4, they do not add up to an integer, k.
get it?

Lukybear · Apr 22, 2010

stampede said:
<link rel="File-List" href="file:///C:%5CUsers%5CMohamad%5CAppData%5CLocal%5CTemp%5Cmsohtmlclip1%5C01%5Cclip_filelist.xml"><link rel="Preview" href="file:///C:%5CUsers%5CMohamad%5CAppData%5CLocal%5CTemp%5Cmsohtmlclip1%5C01%5Cclip_preview.wmf"><link rel="themeData" href="file:///C:%5CUsers%5CMohamad%5CAppData%5CLocal%5CTemp%5Cmsohtmlclip1%5C01%5Cclip_themedata.thmx"><link rel="colorSchemeMapping" href="file:///C:%5CUsers%5CMohamad%5CAppData%5CLocal%5CTemp%5Cmsohtmlclip1%5C01%5Cclip_colorschememapping.xml"><style>  </style> eqn in root form is:

x^4 - (α+β+γ+δ)x^3 + (αβ+αγ+αδ+βγ+βδ+γδ)x^2 - ......
where a, β, γ, δ are the roots.

sum of roots: α+β+γ+δ = α+β (since if i is a root, then -i is also a root, ie: α + β + i - i = α+β)

now, sum of roots = -b/a which has to be an integer since the q says so.
therefore α+β = -b/a = integer = k

therefore: α+β = k

if α and β are .5 and 4, they do not add up to an integer, k.
get it?

But -b/a doesnt really have to be interger does it? It says a+b add to a no.>0 and only specify that b and a are integers. So b/a could be fraction

stampede · Apr 22, 2010

Lukybear said:
But -b/a doesnt really have to be interger does it? It says a+b add to a no.>0 and only specify that b and a are integers. So b/a could be fraction

from ur q:

If P(x) is a monic, degree 4, integer coefficent

also, a is 1 since the eqn is monic
therefore -b/a is just -b
so alpha+beta = -b which has to be an integer

untouchablecuz · Apr 22, 2010

http://community.boredofstudies.org/14/mathematics-extension-2/239096/polynomials.html#post4943090

Polynomial (1 Viewer)

Lukybear

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stampede

doin it tuff

Lukybear

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omgmynameislong

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Lukybear

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Lukybear

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stampede

doin it tuff

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stampede

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untouchablecuz

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