Solve Factorising Problem: Get A and B Values

  • Thread starter anthonyk2013
  • Start date
In summary, Anthony2013 is struggling with factoring the quadratic equation S2+6S+13 and needs help finding the values of A and B in the equation 8S+38/2S2+5S-3=8S=38/(2S-1)(S+3)=A/2S-1+B/s+3, which are A=12 and B=-2. Some forum members have pointed out that the lack of parentheses in the equation is causing confusion and have suggested using them to make the problem clearer.
  • #1
anthonyk2013
125
0
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
 
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  • #2
Solve?Do you mean factorize as in the title? That's not an equation
 
  • #3
adjacent said:
Solve?Do you mean factorize as in the title? That's not an equation

ya sorry factorise
 
  • #4
anthonyk2013 said:
Having trouble solving this problem S2+6S+13
As already noted, there's nothing to solve here - this is not an equation.
anthonyk2013 said:
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.
anthonyk2013 said:
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.
 
  • #5
Mark44 said:
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##
 
  • #6
adjacent said:
It's ##(S+3)^2 + 4##

How do you get that?
 
  • #7
anthonyk2013 said:
How do you get that?
wolframalpha.com but you should not use it to do all your homeworks.Learn to use your brain
 
  • #8
anthonyk2013 said:
How do you get that?
See post #4.
Mark44 said:
About the best you can do is complete the square to write it in the form (S + <something>)2 + <something else>.
 
  • #9
anthonyk2013 said:
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
[tex]8S+\frac{38}{2S^2}+5S-3 [/tex]
when read using standard rules for mathematical expressions.
Did you mean
[tex] \frac{8S + 38}{2S^2 +5S -3} ?[/tex]
Did you mean
[tex] \frac{8S + 38}{2S^2 +5S}-3 ? [/tex]
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).
 
  • #10
Ray Vickson said:
Your lack of parentheses makes your expressions unreadable (or, rather, readable and saying something you might not mean). Literally, your expression is
[tex]8S+\frac{38}{2S^2}+5S-3 [/tex]
when read using standard rules for mathematical expressions.
Did you mean
[tex] \frac{8S + 38}{2S^2 +5S -3} ?[/tex]
Did you mean
[tex] \frac{8S + 38}{2S^2 +5S}-3 ? [/tex]
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).


[tex] \frac{8S + 38}{2S^2 +5S -3} [/tex]

Sorry but I can't write these expressions. my time is taken up studying them.
 
  • #11
anthonyk2013 said:
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.
 
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  • #12
anthonyk2013 said:
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.
 
  • #13
.Scott said:
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.
 
  • #14
.Scott said:
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
.Scott said:
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.
 
  • #15
Mark44 said:
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.
Mark44 said:
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.
 
  • #16
Laplace guys, sorry about confusion on my part. Typo on my work (3) should be S(3)
 

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  • #17
anthonyk2013 said:
Laplace guys, sorry about confusion on my part.
Actually, I should have caught it when you were asking for the A and B.
 
  • #18
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.
 
  • #19
Mark44 said:
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
 
  • #20
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.
 
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  • #21
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
 
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  • #22
anthonyk2013 said:
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.
anthonyk2013 said:
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.
 
  • #23
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.
 
  • #24
Chestermiller said:
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.
 
  • #25
Mark44 said:
anthonyk2013 is the OP. Post #5 was made by adjacent.

Oop. Sorry. I meant Adjacent.

Chet
 
  • #26
Mark44 said:
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?
 
  • #27
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.
 
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  • #28
You should have seen immediately that, since 13 is a prime number, [itex]x^2+ 6x+ 13[/itex] won't factor into anything with integer coefficients. Instead, as others suggested, "complete the square":
[tex]x^2- 6x+ 13= x^2- 6x+ 9- 9+ 13= x^2- 6x+ 9+ 4= (x- 3)^2- (2i)^2[/tex]
which, as the "difference of two squares", can be factored as the "sum and difference":
[tex](x- 3+ 2i)(x- 3- 2i)[/tex]
 

What is factorising and why is it important?

Factorising is the process of breaking down a mathematical expression into its simplest form by finding its factors. It is important because it allows us to simplify complex expressions and solve problems more easily.

How do I solve a factorising problem?

To solve a factorising problem, you need to identify the common factors in the given expression and then use the distributive property to expand the expression. Finally, you can rearrange the terms to get the simplified form of the expression.

What are the A and B values in factorising?

In factorising, the A and B values refer to the coefficients of the two terms that make up the expression. For example, in the expression 2x + 4y, A=2 and B=4.

What are some common methods for solving factorising problems?

Some common methods for solving factorising problems include the difference of squares method, the grouping method, and the quadratic formula. Each method may be more suitable for different types of expressions.

Are there any tips for practicing and improving my factorising skills?

Yes, some tips for practicing and improving your factorising skills include familiarizing yourself with different methods, breaking down complex expressions into simpler ones, and practicing with a variety of problems. It is also helpful to check your answers and understand any mistakes you may have made.

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