Factorise problem

[Anthonyk2013]

I'm doing a laplace problem and I have got stuck in a simple area.

I have got as far as making L(y) the subject but need to factorise S²+6S+13 so I can move on.

S*S=s²
can solve the other values.

 

[pka]

I'm doing a laplace problem and I have got stuck in a simple area.

I have got as far as making L(y) the subject but need to factorise S²+6S+13 so I can move on.

\(\displaystyle s^2+6s+13\) cannot be factored in the real number field.

It can be factored using complex numbers. See here.

 

[Anthonyk2013]

\(\displaystyle s^2+6s+13\) cannot be factored in the real number field.

It can be factored using complex numbers. See here.

For Example we factorised 8s+38/2s²+5s-3=8s+38/(2s-1)(s+2)=A/2s+1+B/s+3

A=12 and B=-2

Is it possible to get a value for A and B with s²+6s+13?

 

[lookagain]

For Example we factorised 8s+38/2s²+5s-3=8s+38/(2s-1)(s+2)=A/2s+1+B/s+3

A=12 and B=-2

Anthonyk2013, here are a couple of facts and a suggestion:

1) Certain denominators are incorrect.

2) You are missing required grouping symbols for the denominators, and the first two numerators need them.

3) You should spread out your characters within a line for better readability. And one of the ways
\(\displaystyle \ \ \ \ \ \)of doing this is to put separate expressions on separate lines with a space between each line.


\(\displaystyle (8s + 38)/(2s^2 + 5s - 3) \ = \)

\(\displaystyle (8s + 38)/[(2s - 1)(s + 3)] \ = \)

\(\displaystyle A/(2s - 1) \ + \ B/(s + 3)\)


A = 12 \(\displaystyle \ \) and \(\displaystyle \ \) B = -2

 

[Anthonyk2013]

Anthonyk2013, here are a couple of facts and a suggestion:

1) Certain denominators are incorrect.

2) You are missing required grouping symbols for the denominators, and the first two numerators need them.

3) You should spread out your characters within a line for better readability. And one of the ways
\(\displaystyle \ \ \ \ \ \)of doing this is to put separate expressions on separate lines with a space between each line.


\(\displaystyle (8s + 38)/(2s^2 + 5s - 3) \ = \)

\(\displaystyle (8s + 38)/[(2s - 1)(s + 3)] \ = \)

\(\displaystyle A/(2s - 1) \ + \ B/(s + 3)\)


A = 12 \(\displaystyle \ \) and \(\displaystyle \ \) B = -2

Thanks I remember that