Polynomial (1 Viewer)

[mirakon]

Fully factorise over the complex field

2z^3+3z^2+8z+5

Please provide a fully worked solution. Thanks in advance.

 

[Affinity]

factors are not nice, you will need to use the cubic formula:

Cubic Formula -- from Wolfram MathWorld

 

[khorne]

If it has any real solutions, you could begin with testing factors such as 5/2, 1/2, 5 etc

 

[mirakon]

Yeah I tried -5/2, -1/2 etc.

By fundamental theorem there has to be at least one root as complex roots come in conjugate pairs.

Our teacher wrote this question on the board and then she realised that it was supposed to be -3z^2 in the actual question instead of +3z^2. But the guys in our class were thinking logically that the above question should be able to be factorised too.

 

[mirakon]

factors are not nice, you will need to use the cubic formula:

Cubic Formula -- from Wolfram MathWorld

Yeah, but syllabus maths would be preferable. Thanks anyway though.

 

[Affinity]

If it has any real solutions, you could begin with testing factors such as 5/2, 1/2, 5 etc

you mean *rational*
all cubics with real coefficients have atleast 1 real solution, just that this one doesn't look particularly nice

 

[khorne]

No, I meant real...It could easily have an irrational factor

 

[mirakon]

No, I meant real...It could easily have an irrational factor

That's what I'm assuming it is, but do you know an easy way to do it?

 

[khorne]

If you test out the -3z^2, z = -0.5 is a root, and from that you get the factore (2z+1)...then it's doable

 

[Affinity]

Yeah I tried -5/2, -1/2 etc.

By fundamental theorem there has to be at least one root as complex roots come in conjugate pairs.

Our teacher wrote this question on the board and then she realised that it was supposed to be -3z^2 in the actual question instead of +3z^2. But the guys in our class were thinking logically that the above question should be able to be factorised too.

Every poly nomial factorizes..In the HSC they load it so that you get a nice rational factor (a cubic doesn't usually factorize so nicely!).. and that's why trying those numbers work.. but for your original cubic, the factors are not nice so you'll need to use the cubic formula.

f you change the cubic to 2z^3-3z^2+8z+5 then after a bit of testing you find that -1/2 is a zero,

so one of hte facotrs is (2z+1). then you divide the cubic by this to get

2z^3-3z^2+8z+5 = (2z+1)*(z^2 - 2z + 5) = (2z+1)(z - (1+2i))(z-(1-2i))

 

[mirakon]

yeah we got that one easily but i was just wodnering if the other, original one, is doable with syllabus mathematics

 

[Affinity]

yeah we got that one easily but i was just wodnering if the other, original one, is doable with syllabus mathematics

Yes and No.. the actual steps doesn't use much beyond year 10 advanced maths + complex number knowledge, roughly:

1. divide through by 2.
2. let z = x - 1/2, substitute and you will get the equation: x^3 + 13/4 x + 3/4 = 0
3. do another subsitution, letting x = w - 13/(12w) you will get
w^6 + 3/4 w^3 - 2197/1728 = 0, this is a quadratic in w^3 which you can solve,
giving w^3 = 1/72 * [ - 27 +- 2*sqrt(1830) ] or w = cubert(1/72 * [ - 27 +- 2*sqrt(1830) ]), note that there are 6 possible values for w (every number has 3 cube roots)

4. from each w you can get an x and hence z. there will some repeats leaving you with 3 roots.

5. write down the factorization

 

[mirakon]

Yes and No.. the actual steps doesn't use much beyond year 10 advanced maths + complex number knowledge, roughly:

1. divide through by 2.
2. let z = x - 1/2, substitute and you will get the equation: x^3 + 13/4 x + 3/4 = 0
3. do another subsitution, letting x = w - 13/(12w) you will get
w^6 + 3/4 w^3 - 2197/1728 = 0, this is a quadratic in w^3 which you can solve,
giving w^3 = 1/72 * [ - 27 +- 2*sqrt(1830) ] or w = cubert(1/72 * [ - 27 +- 2*sqrt(1830) ]), note that there are 6 possible values for w (every number has 3 cube roots)

4. from each w you can get an x and hence z. there will some repeats leaving you with 3 roots.

5. write down the factorization

Excellent! Thank you so much! Same to you khorne!