what id the fasted way to solve a quartic (1 Viewer)

[123ash]

I know Like in a quartic eqn. such as Ax^4 + Bx^3 + Cx^2 + Dx + E

we usually sub in values up to +-4 and find a root, then do long division twice, I am sick of this method.

Does anyone has any other quicker method.

 

[nonsenseTM]

what?

 

[nonsenseTM]

if a ... e are integers , we sub the factors of e , after we get two factors ,we can use sum and product of roots

 

[123ash]

so if a=0 and e= 106,
we sub +- 2

but if we have e = 204,
we sub +-2, +-3, +-4, +-6

 

[123ash]

sorry A shud be 1,

but if A is a number, we do e/A

 

[buchanan]

You guys are describing using the factor theorem, which is OK if the roots are rational and coefficients are integers. BUT if they are not rational it might be harder.

In this case, you could use the quartic formula:

or use Wolfram|Alpha

 

[123ash]

You guys are describing using the factor theorem, which is OK if the roots are rational and coefficients are integers. BUT if they are not rational it might be harder.

In this case, you could use the quartic formula:

or use Wolfram|Alpha

where did the y come from??

 

[shaon0]

I know Like in a quartic eqn. such as Ax^4 + Bx^3 + Cx^2 + Dx + E

we usually sub in values up to +-4 and find a root, then do long division twice, I am sick of this method.

Does anyone has any other quicker method.

They won't ask you a general case in HSC. They'll ask you to maybe solve an easy case using rational root theorem, factor theorem and later maybe conjugate root theorem. OR they might give a symmetric quartic poly to be solved ie where A=E,B=D eg: x^4+4x^3+6x^2+4x+1.

 

[cutemouse]

You could try factorising out x^2 and then completing the square etc.

 

[buchanan]

where did the y come from??

To solve a general quartic which can't be solved by other means such as the factor theorem, one needs to solve a resolvent cubic first.