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#2 Re: Help Me ! » Simplifying Monomials Help (Algebra 2) » 2005-12-30 10:37:08

Hi


1)       1/(4xy^3)^2   *  125x^-9y^12
       
           = (125y^6)/16x^11



2)   Let us first divide (y^5-3y²-20)  by   (y-2)

The quotient will be y^4 + 2y^3 + 4y^2  +5y+10

which means that

(y-2)(y^4 + 2y^3 + 4y^2 +5y+10) = y^5-3y²-20

Therefore,   

    y-2  /  y^5-3y²-20
   
=y-2/ (y-2)(y^4 + 2y^3 + 4y^2 +5y+10)

(y-2) in the numerator and denominator gets cancelled and the answer will be

= 1/(y^4 + 2y^3 + 4y^2 +5y+10)

#3 Re: Help Me ! » Factor Theorem? » 2005-12-30 09:47:30

Hi Katy

According to the factor theorem,
For a polynomial P(x), x - a is a factor if P(a) = 0.

therefore substituting x=1 in the polynomial
P(x) =(x^3+3x^2+13x-15)

P(1) = 1+3+13-15
        =2

again substituting x=2 in P(x)

P(2)= 8+12+26-15
       =31

P(3)=27+27+39-15
       =93

As the remainder is not zero for x=1,2,3,4,.............so on

Hence the given polynomial P(x) =(x^3+3x^2+13x-15)
has no factor

I think the polynomial u have given is unfactorizable

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