Polynomial: (3z+2) is a factor of 3z^(3) - z^(2) + z(a+1) + a; find...

[Onigma]

Howdy Folks,

The question says: (3z+2) is a factor of 3z^(3) - z^(2) + z(a+1) + a, find 'a' and all the zeroes of the cubic.

I understand you will use what you know to make an equation..... 3z^(3) - z^(2) + z(a+1) + a = (3z+2)(z^(2) + bz + X) and then use equality to solve for a and b etc.

However, my problem is I don't know what value to put in place of X. To create the constant 'a', I need 2(X) to = a, I just can't work out what X should be. Or am I missing something and need to try something else?

Thanks

 

[mmm4444bot]

… The question says:

(3z+2) is a factor of 3z^3 - z^2 + z(a+1) + a …

Hmm. I'm not convinced that this is true.

(It's certainly not true, when a is zero.)

For the given cubic polynomial (as you have posted it), different Computer Algebra Systems' software that I tried could not find any factorization at all -- with or without a factor of (3z+2) -- including floating-point (approximated) factorizations or those including imaginary numbers (complex).

 

[Dr.Peterson]

The question says: (3z+2) is a factor of 3z^(3) - z^(2) + z(a+1) + a, find 'a' and all the zeroes of the cubic.

I understand you will use what you know to make an equation..... 3z^(3) - z^(2) + z(a+1) + a = (3z+2)(z^(2) + bz + X) and then use equality to solve for a and b etc.

However, my problem is I don't know what value to put in place of X. To create the constant 'a', I need 2(X) to = a, I just can't work out what X should be. Or am I missing something and need to try something else?

What the question means is, given that (3z+2) is a factor of 3z^(3) - z^(2) + z(a+1) + a, for some specific value of the constant a, find that value of a, and then find all the zeroes of the cubic. If your suggested method could work at all, it would be a very long way around.

If I gave you some polynomial, such as 3x^3 - x^2 + 4x + 3, and asked you whether (3x + 2) is a factor of it, what would you do? (I'd use synthetic division ...).

Do the same here, leaving a as an unknown constant, and you will end up with an equation in a. Solve it, and you'll know what a is.

Then if I asked you to find the other zeros of the polynomial, what would you do? (I'd use the results of the synthetic division.)

 

[Deleted member 4993]

Hmm. I'm not convinced that this is true.

(It's certainly not true, when a is zero.)

For the given cubic polynomial (as you have posted it), different Computer Algebra Systems' software that I tried could not find any factorization at all -- with or without a factor of (3z+2) -- including floating-point (approximated) factorizations or those including imaginary numbers (complex).

We need to calculate the value of 'a' when that is true. I get a nice integer number for 'a' when the given condition is satisfied.

 

[Deleted member 4993]

Howdy Folks,

The question says: (3z+2) is a factor of 3z^(3) - z^(2) + z(a+1) + a, find 'a' and all the zeroes of the cubic.

I understand you will use what you know to make an equation..... 3z^(3) - z^(2) + z(a+1) + a = (3z+2)(z^(2) + bz + X) and then use equality to solve for a and b etc.

However, my problem is I don't know what value to put in place of X. To create the constant 'a', I need 2(X) to = a, I just can't work out what X should be. Or am I missing something and need to try something else?

Thanks

You are going in the right direction. Now distribute the right hand side:

(3z+2)(z^(2) + bz + X) = 3z^3 + z^2(2+3b) + z(3X+2) + 2X

from above:

2 + 3b = -1 ..................................(1)

3X + 2 = a + 1..........................(2)

2X = a ..........................(3)

You can solve for 'a', 'b' and 'X' from the 3 equations.

 

[JeffM]

Howdy Folks,

The question says: (3z+2) is a factor of 3z^(3) - z^(2) + z(a+1) + a, find 'a' and all the zeroes of the cubic.

I understand you will use what you know to make an equation..... 3z^(3) - z^(2) + z(a+1) + a = (3z+2)(z^(2) + bz + X) and then use equality to solve for a and b etc.

However, my problem is I don't know what value to put in place of X. To create the constant 'a', I need 2(X) to = a, I just can't work out what X should be. Or am I missing something and need to try something else?

Thanks

You started out just fine. As Subhotosh explained, you proceed

\(\displaystyle 3z^3 - z^2 + z(a + 1) + a = (3z + 2)(z^2 + bz + x) \implies\)

\(\displaystyle 3z^3 - z^2 + z(a + 1) + a = 3z^3 + z^2(3b + 2) + z(3x + 2b) + 2x \implies\)

\(\displaystyle 3b + 2 = -\ 1 \text { and } 3x + 2b = a + 1 \text { and } 2x = a.\)

DO YOU SEE WHY? That is the important point of this exercise. If you don't get this, the exercise is a waste. If you do get it, all that is left is to solve a simple system of 3 linear equations in 3 unknowns.

 

[mmm4444bot]

We need to calculate the value of 'a' when that is true.

Yes, thank you, Subhotosh. I understood the exercise, but my paper-and-pencil efforts went off the rail, and then I committed the typical rookie-student mistake, by deferring to machines (and wolfram, too) instead of using my brain to try again.

In other words, I took the lazy way out, letting a machine do my thinking for me, and now I have egg all over my face!!

[video=youtube;xTt8IzilofM]https://www.youtube.com/watch?v=xTt8IzilofM[/video]