IIT JEE , AIEEE Forum - Mathematics , Algebra

rushali

If x²-x+1=0 is a factor of ax³+bx²+cx+d=0 then

d - a = b - 2c

a - d = b + 2c

a + d = b - 2c

none of these

with solution plzzzzzzzz

allamraju

The roots of the eqn. are -w,-w².Substituting them in the eqn.,we get,

-a+bw²-cw+d=0 and -a+bw-cw²+d=0.Adding both the eqns,we get,

-2a-b+c+2d=0[since w+w²=-1].So,I think the ans is none of these.

rushali

hey the ans is b

rishabh29288

well i think the answer is b).since x^2-x+1 ia factor of the given cubic eqn and hence the cubic eqn has two complex roots the other root has to be real.let that be -k
then (x+k)(x^2-x+1)=ax^3+bx^2+cx+d
compare the coeff to get the answer

animal

yes rishbhs method is correct but i cant find any mistake in allamraju's method also so i m confused rite now i also dis the same way allam did .

so can u plz tell the ans.

rushali

hey the roots of x^2+x+1 are -w n -w^2 ............... and not of x^2-x+1 ............. ya ?

animal

no the roots of x²+x+1=0 r w ,w ² and that of x²-x+1 r -w ,-w²

hope u got it.

thedumbheadwithnobrain

allamraju's approach is correct but here it wont be helpful as it does not solve the given conditions,,,but by putting (x+k) ${x}^{2}-x+1=a{x}^{3}+b{x}^{2}+cx+d$ we get

a=1

k-1=b

-k+1=c

k=d

and match the options

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If x2-x+1=0 is a factor of ax3+bx2+cx+d=0 then

The roots of the eqn. are -w,-w2.Substituting them in the eqn.,we get,

-a+bw2-cw+d=0 and -a+bw-cw2+d=0.Adding both the eqns,we get,

-2a-b+c+2d=0[since w+w2=-1].So,I think the ans is none of these.

If x²-x+1=0 is a factor of ax³+bx²+cx+d=0 then

The roots of the eqn. are -w,-w².Substituting them in the eqn.,we get,

-a+bw²-cw+d=0 and -a+bw-cw²+d=0.Adding both the eqns,we get,

-2a-b+c+2d=0[since w+w²=-1].So,I think the ans is none of these.