Solve Factorising Problem: Get A and B Values

[anthonyk2013]

Having trouble solving this problem S²+6S+13

For example I solved this one before and stuck on the above

8S+38/2S²+5S-3=8S=38/(2S-1)(S+3)=A/2S-1+B/s+3

A=12 B=-2

How do I factorise S²+6S+13 to get value for A and B

 

[adjacent]

Solve?Do you mean factorize as in the title? That's not an equation

 

[anthonyk2013]

Solve?Do you mean factorize as in the title? That's not an equation

ya sorry factorise

 

[Mark44]

Having trouble solving this problem S²+6S+13

As already noted, there's nothing to solve here - this is not an equation.

For example I solved this one before and stuck on the above

8S+38/2S²+5S-3=8S=38/(2S-1)(S+3)=A/2S-1+B/s+3

A=12 B=-2

You're missing lots of parentheses here. Whenever the numerator or denominator contains more than one term, put parentheses around the entire numerator or denominator.

How do I factorise S²+6S+13 to get value for A and B

This quadratic can't be factored into factors with integer or even rational coefficients. About the best you can do is complete the square to write it in the form (S + <something>)² + <something else>.

Note: Homework-type questions must be posted in the Homework & Coursework section, not in the technical math sections.

 

[adjacent]

the best you can do is complete the square to write it in the form (S + <something>)² + <something else>.

It's $(S+3)^2 + 4$

 

[anthonyk2013]

It's $(S+3)^2 + 4$

How do you get that?

 

[adjacent]

How do you get that?

wolframalpha.com but you should not use it to do all your homeworks.Learn to use your brain

 

[Mark44]

How do you get that?

See post #4.

About the best you can do is complete the square to write it in the form (S + <something>)² + <something else>.

 

[Ray Vickson]

Having trouble solving this problem S²+6S+13

For example I solved this one before and stuck on the above

8S+38/2S²+5S-3=8S=38/(2S-1)(S+3)=A/2S-1+B/s+3

A=12 B=-2

How do I factorise S²+6S+13 to get value for A and B

Your lack of parentheses makes your expressions unreadable (or, rather, readable and saying something you might not mean). Literally, your expression is
$$8S+\frac{38}{2S^2}+5S-3$$
when read using standard rules for mathematical expressions.
Did you mean
$$\frac{8S + 38}{2S^2 +5S -3} ?$$
Did you mean
$$\frac{8S + 38}{2S^2 +5S}-3 ?$$
If you meant the first, use parentheses like this: (8S+38)/(2S^2 + 5S - 3)---simple, and completely unambiguous. If you meant the second, write it as (8S+38)(/(2S^2 + 5S) - 3, which is unambiguous when read by standard rules (although some might engage in "overkill" and write, instead: [(8S+38)(/(2S^2 + 5S)] - 3).

 

[anthonyk2013]

Your lack of parentheses makes your expressions unreadable (or, rather, readable and saying something you might not mean). Literally, your expression is
$$8S+\frac{38}{2S^2}+5S-3$$
when read using standard rules for mathematical expressions.
Did you mean
$$\frac{8S + 38}{2S^2 +5S -3} ?$$
Did you mean
$$\frac{8S + 38}{2S^2 +5S}-3 ?$$
If you meant the first, use parentheses like this: (8S+38)/(2S^2 + 5S - 3)---simple, and completely unambiguous. If you meant the second, write it as (8S+38)(/(2S^2 + 5S) - 3, which is unambiguous when read by standard rules (although some might engage in "overkill" and write, instead: [(8S+38)(/(2S^2 + 5S)] - 3).

$$\frac{8S + 38}{2S^2 +5S -3}$$

Sorry but I can't write these expressions. my time is taken up studying them.

 

[.Scott]

Having trouble solving this problem S²+6S+13

There are a couple of ways to do this. The most reliable is using the solution to the quadratic equation: S²+6S+13 = 0 (A=1, B=6, C=13)

The quadratic is S = (-B +/- Sqrt(B²-4AC))/2A

Once you have the solutions, the factors will be (S-solution1)(S-solution2).

The reason this works is that you know that if (S-solution1)(S-solution2)=0, then at least one of terms must be zero and therefor S must be one of the solutions to the quadratic.

BTW: There is no nice [strike]integer[/strike] real solution to this.

 

[Mark44]

Sorry but I can't write these expressions. my time is taken up studying them.

If you want help from us, you need to clearly communicate what the problem is you're working on. No one is demanding that you become expert in using LaTeX to write these fractions - just use enough parentheses so that we're all on the same page as to what the problem is.

This means writing (8S+38)/(2S^2 + 5S - 3) instead of 8S+38/2S²+5S-3, as you wrote.

 

[.Scott]

BTW: There is no nice integer solution to this.

I should have said there is no nice real. Your solution will look like this:
(S+u+vi)(S+u-vi)

I'll check this thread in another 10 or 15 minutes. If you haven't solved for u and v by then, I'll just post the answer.

 

[Mark44]

I should have said there is no nice real. Your solution will look like this:
(S+u+vi)(S+u-vi)

I'm not sure, but I suspect that the posted problem is part of a larger problem involving an integral or a Laplace transform. If so, factorization as you show above won't be of much help, but writing the denominator as (x + number1)² + number2 will be useful

I'll check this thread in another 10 or 15 minutes. If you haven't solved for u and v by then, I'll just post the answer.

Please don't. Per forum rules (under Homework Help Guidelines at https://www.physicsforums.com/showthread.php?t=414380):

Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made.

 

[.Scott]

I'm not sure, but I suspect that the posted problem is part of a larger problem involving an integral or a Laplace transform. If so, factorization as you show above won't be of much help, but writing the denominator as (x + number1)² + number2 will be useful.

Only Anthony2013 knows for sure.

Please don't. Per forum rules (under Homework Help Guidelines at https://www.physicsforums.com/showthread.php?t=414380):

OK - but I pretty much out of hints.
Ohh, both u and v are integers.

 

[anthonyk2013]

Laplace guys, sorry about confusion on my part. Typo on my work (3) should be S(3)

 

[.Scott]

Laplace guys, sorry about confusion on my part.

Actually, I should have caught it when you were asking for the A and B.

 

[Mark44]

Just as I thought. In this case you don't want to factor your denominator. Instead, complete the square as already mentioned in posts #4 and #14. Since you either don't know or don't remember this technique, here's a link: http://www.purplemath.com/modules/sqrquad.htm. See the 2nd example on this page.

 

[anthonyk2013]

Just as I thought. In this case you don't want to factor your denominator. Instead, complete the square as already mentioned in posts #4 and #14. Since you either don't know or don't remember this technique, here's a link: http://www.purplemath.com/modules/sqrquad.htm. See the 2nd example on this page.

Don't remember 15 years out of the game, dropped back in lately. (old age as well)
Thanks for help

 

[Mark44]

Once you get your expression into the form I recommended, you should be able to look up the inverse Laplace in a table to get the solution of your differential equation.

 

[anthonyk2013]

S²+6S+13

S²+6S=-13

S²+6S+9=-13+9

(S+3)²=+-(-4)

S+3=2

S=-3+-2

S=-3+2 and -3-2

S= -1, 5

I know +- looks wrong but can't put minus under the positive sign

 

[Mark44]

S²+6S+13

S²+6S=-13

No, this is wrong.
You started with an expression. You can't turn it into an equation. All you can legitimately do is write the expression in some other form.

S²+6S+9=-13+9

(S+3)²=+-(-4)

S+3=2

S=-3+-2

S=-3+2 and -3-2

S= -1, 5

I know +- looks wrong but can't put minus under the positive sign

Let's start again.

S²+6S+13
= S² + 6S + 9 + 4
= (S + 3)² + 4
That's where you should end up.

 

[Chestermiller]

If you use the form of the denominator posted by anthonyk2013 in post #5, you should easily be able to find the inverse transform of your expression in your LT tables.

 

[Mark44]

If you use the form of the denominator posted by anthonyk2013 in post #5, you should easily be able to find the inverse transform of your expression in your LT tables.

anthonyk2013 is the OP. Post #5 was made by adjacent.

 

[Chestermiller]

anthonyk2013 is the OP. Post #5 was made by adjacent.

Oop. Sorry. I meant Adjacent.

Chet

 

[anthonyk2013]

Just as I thought. In this case you don't want to factor your denominator. Instead, complete the square as already mentioned in posts #4 and #14. Since you either don't know or don't remember this technique, here's a link: http://www.purplemath.com/modules/sqrquad.htm. See the 2nd example on this page.

Must have followed the wrong example from here, which one exactly should I follow?

 

[Mark44]

The second example, like I already said. Note that in that example they are solving an equation, which is different from what you're doing.

 

[HallsofIvy]

You should have seen immediately that, since 13 is a prime number, $x^2+ 6x+ 13$ won't factor into anything with integer coefficients. Instead, as others suggested, "complete the square":
$$x^2- 6x+ 13= x^2- 6x+ 9- 9+ 13= x^2- 6x+ 9+ 4= (x- 3)^2- (2i)^2$$
which, as the "difference of two squares", can be factored as the "sum and difference":
$$(x- 3+ 2i)(x- 3- 2i)$$