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Hi krassi Holmz
Thanks for helping
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)
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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