Please help me to solve this quartic equation: (6x+7)^2 (3x+4) (x+1) = 6

[Farzin]

Please help me to solve this quartic equation.
(6x+7)^2(3x+4)(x+1)=6


I managed to come this far:
108x^4+504x^3+879x^2+679x+190=0
and also:
3x^2(6x+7)^2+7x(6x+7)^2+4(6x+7)^2=6

 

[tkhunny]

Please help me to solve this quartic equation.
(6x+7)^2(3x+4)(x+1)=6


I managed to come this far:
108x^4+504x^3+879x^2+679x+190=0
and also:
3x^2(6x+7)^2+7x(6x+7)^2+4(6x+7)^2=6

It's not clear what is wanted. If a polynomial has rational solutions, where you started is where you want to end up.

Do you mean "expand". You did that right after "this far:".

So, really, what is wanted? Can you provide the entire problem statement or do you just no know that you are done?

 

[Farzin]

It's not clear what is wanted. If a polynomial has rational solutions, where you started is where you want to end up.

Do you mean "expand". You did that right after "this far:".

So, really, what is wanted? Can you provide the entire problem statement or do you just no know that you are done?

I just want to find the roots. I know there are 2 rationals and a pair of irrationals.

 

[Dr.Peterson]

I just want to find the roots. I know there are 2 rationals and a pair of irrationals.

I would use the rational root theorem to search for rational roots, possibly using a rough sketch of the graph (or, if allowed, technology) to estimate where to look for roots. Once you have divided out the two rational factors, you can use the quadratic formula if necessary.

When I did it, I found the two rational roots (which are simple fractions) and two non-real roots. I could be wrong on the latter.

 

[Farzin]

I would use the rational root theorem to search for rational roots, possibly using a rough sketch of the graph (or, if allowed, technology) to estimate where to look for roots. Once you have divided out the two rational factors, you can use the quadratic formula if necessary.

When I did it, I found the two rational roots (which are simple fractions) and two non-real roots. I could be wrong on the latter.

Thank you very much. Yes, I did use the rational root theorem and I found -2/3 and -5/3, but it was tedious because I had to check many possibilities. I wonder if there is any other method to solve it more algebraically.

 

[Steven G]

108x^4+504x^3+879x^2+679x+190=0

1st I would find the gcf of 108,504,879, 679 and 190 to get smaller coefficients. Then I would use the rational root theorem to find all possible rational roots. Then test them until you find two of them. This will bring you to quadratic equation which you can solve using various methods that you have learned.

After arriving at 108x^4+504x^3+879x^2+679x+190=0 what have you done? Show us your work so that we can guide you.

 

[JeffM]

Please help me to solve this quartic equation.
(6x+7)^2(3x+4)(x+1)=6


I managed to come this far:
108x^4+504x^3+879x^2+679x+190=0
and also:
3x^2(6x+7)^2+7x(6x+7)^2+4(6x+7)^2=6

You could use the quartic formula (UGH).

You could notice that $\displaystyle x = 0, \implies f(0) = 392 \implies f(0) - 6 >> 0.$

And $\displaystyle x = -\ 1 \implies f(x) = 0 \implies f(x) - 6 < 0.$

Therefore, if there is a rational root to f(x) - 6, one of them must lie between -1 and 0.

Now try either the rational root theorem or the Newton-Raphson method to find that root. Now you are down to a cubic.

 

[Steven G]

Thank you very much. Yes, I did use the rational root theorem and I found -2/3 and -5/3, but it was tedious because I had to check many possibilities. I wonder if there is any other method to solve it more algebraically.

Yes, it is tedious. This is why you must try first to find the gcd so you can reduce the number of possibilities. I do admit that in your case the gcd was 1 so you had to check all the possible numbers that you found (until you found 2 of them!)

Consider this problem 6x²+8x+4. Leaving it the way the possible rational roots are +/-1, +/-2, +/-4, +/-(1/2), +/-(1/3), +/-(2/3), +/-(4/3), +/-(1/6). So you have to check up to 8 of them.

On the other hand if you realized that 6x²+8x+4 = 2(3x²+4x+2) then the possible rational roots would be +/-1, +/-2, +/-(1/3), +/-(2/3). So you would only have to try up to 4 of them. This is one half the number we got above (it is a coincidence that I factored out a 2 and got half as many). Not that the smaller set of possible rational roots are ALL in the larger set of possible rational roots. Do you know why that is?? You really should!

You asked if there is an easier way to do your problem then by using the rational root theorem. With your problem the answer is most certainly no. If there is an easier way this easier method will NOT work with all quartic equation so basically you are stuck with the rational root theorem