by remainder therom find the remainder when x^3-6x^2+2x-4 is divided by 1-3x/2
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Let p(x) =x^3 - 6x^2+2x-4
And g(x) =1-3x/2
By remainder theorem put g(x) =0
1-3x/2=0
3x/2=1
x=2/3
Hence remainder is
P(2/3)=(2/3)^3 - 6(2/3)^2 +2(2/3)-4
=8/27-6(4/9)+4/3-4
=8/27-24/9+4/3-4
=(8-72+36-108)/27
=-136/27
And g(x) =1-3x/2
By remainder theorem put g(x) =0
1-3x/2=0
3x/2=1
x=2/3
Hence remainder is
P(2/3)=(2/3)^3 - 6(2/3)^2 +2(2/3)-4
=8/27-6(4/9)+4/3-4
=8/27-24/9+4/3-4
=(8-72+36-108)/27
=-136/27
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