Question

Let F(X)=2x^3-3/2x^2-1/8x+3/16 and x=y/a then the least positive integral value of a for which coefficients of f(y)=0 become integral is

Answer 1

\[f(x)=2x^3-\frac{ 3 }{ 2 }x^2-\frac{ 1 }{ 8 }x+\frac{ 3 }{ 16} and x=\frac{ y }{ a }\]

Answer 2

x = y/a

y = xa

\(f(y) = 2(xa)^3 - 3/2 (xa)^2 - 1/8 x + 3/16 \)

Answer 3

don't u think there should be a even after

Answer 4

1/8x

Answer 5

I am not sure
f(y) = 0 in the question is throwing me off -- if you set f(y) to 0, you can multiply anything both sides and always have integral coefficients :o

Answer 6

f(y) needs to be in terms of y
\[f(y) = \frac{2}{a^{3}}y^{3}-\frac{3}{2a^{2}}y^{2}-\frac{1}{8a}y+\frac{3}{16} = 0\]

Answer 7

y=xa then why r u not putting value as eqn multiplied by a and leaving that constant term

Answer 8

??
oh i got it, multiply equation by a^3
\[2y^{3}-\frac{3a}{2} y^{2}-\frac{a^{2}}{8}y+\frac{3a^{3}}{16}=0\]
find integer whose cube is multiple of 16, square is multiple of 8, and even
a = 4

Answer 9

we can as well multiply the whole equation by 16 and get rid of fractions right ?

Answer 10

yes but they want a value for "a" to get rid of fractions

Answer 11

asking for integral coefficients of f(y) = 0 makes no sense to me still

Answer 12

question should be asking for integral coefficients of f(y) i think

Answer 13

@samigupta8 , do you see my posts above?

Answer 14

why u multiplied whole equation by a^3

Answer 15

so that "a" is in numerator ... impossible to make 2/a^3 an integer

Answer 16

okk...

Answer 17

sorry not impossible but a=1 wouldnt work for other coefficients
its just a way to manipulate equation to get integer coefficients