Can all cubic and quartic equations be solved by factoring? Explain.

[JSmith]

Can all cubic and quartic equations be solved by factoring? Explain. Create an example to use in your explanation.

 

[srmichael]

Can all cubic and quartic equations be solved by factoring? Explain. Create an example to use in your explanation.

Sounds like a question for YOU to answer? We need to see some work on your part to determine where you may be stuck so we know where we can help you. So I ask, Can all cubic and quartic equations be solved by factoring?

 

[JSmith]

I really have no idea to be honest...

 

[JeffM]

Can all cubic and quartic equations be solved by factoring? Explain. Create an example to use in your explanation.

Is this the question in its entirety? If so, is there context to the question?

I ask this because the question, as you have posed it, is a very bad question. The technically correct answer is "Yes." Providing a rigorous demonstration was beyond the powers of three centuries of very great mathematicians, including Gauss.

http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra

On the basis of the theorem discussed in the article cited above, you should be able to construct a non-rigorous explanation of the technically correct answer.

Hint: How many real and complex roots do cubic and quartic equations have respectively? What does that imply?

 

[JSmith]

Is this the question in its entirety? If so, is there context to the question?

I ask this because the question, as you have posed it, is a very bad question. The technically correct answer is "Yes." Providing a rigorous demonstration was beyond the powers of three centuries of very great mathematicians, including Gauss.

http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra

On the basis of the theorem discussed in the article cited above, you should be able to construct a non-rigorous explanation of the technically correct answer.

Hint: How many real and complex roots do cubic and quartic equations have respectively? What does that imply?

No there was no other context, so I will just give it my best shot

 

[HallsofIvy]

That depends upon what you mean by "solved by factoring". Normally, when we solve just a quadratic by "factoring", we mean "factoring with integer coefficients". In that sense even the quadratics \(\displaystyle x^2- 2= 0\) and \(\displaystyle x^2+ 1= 0\) cannot be "solved by factoring". Of course, those can be factored as \(\displaystyle (x- \sqrt{2})(x+ \sqrt{2})\) and \(\displaystyle (x- i)(x+ i)\). If you simply mean "is it possible to factor all cubic and quartic polynomials", then the answer is clearly yes - and even higher degree polynomial equations. If, perhaps through some other methods, you can find all solutions to an nth degree polynomial, \(\displaystyle x_0\), \(\displaystyle x_1\), \(\displaystyle x_2\), ..., \(\displaystyle x_n\), you can then factor it: \(\displaystyle a(x- x_0)(x- x_1)\cdot\cdot\cdot(x- x_n)\).

 

[mmm4444bot]

I will just give it my best shot

Does this statement go without saying?



That is, I am wondering about the sorts of exercises with which you do not give your best shot.


It's okay to ask for hints or clarifications about anything, anytime!