How can I find the values of b and c in a trinomial using factoring?

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In summary: I hope they can make sense out of it.In summary, the task was to find the values of b and c in the trinomial -0.5x2 + x -1= -0.5 (x2 - bx + c), which could be used to factor the expression and find the solution. After analyzing the given expression and manipulating it algebraically, it was determined that b=2 and c=2, which would result in the factored form (x-1)(0.5x-1) and the solution 0.5. However, it was also noted that this solution may not be easily understandable for students and further clarification or explanation may be needed.
  • #1
DaveC426913
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I'm helping edit a math textbook and have been asked to provide the answer for this:

Factor to find the value of b and c in the trinomial:
-0.5x2 + x -1= -0.5 (x2 - bx + c)

Here is what I've done:

-0.5x^2 + x -1= -0.5 (x^2 - bx + c)
-0.5x^2 + x -1= x^2 + .5bx - .5c
-0.5x^2 + x -1= x^2 + .5bx - .5c
x -1= 1.5x^2 + .5bx - .5c
-1 = 1.5x^2 - .5bx - .5c
1.5x^2 - .5bx - .5c +1 = 0

or, in latex:
[tex]-0.5x^2 + x -1= -0.5 (x^2 - bx + c)[/tex]
[tex]-0.5x^2 + x -1= x^2 + .5bx - .5c[/tex]
[tex]-0.5x^2 + x -1= x^2 + .5bx - .5c[/tex]
[tex]x -1= 1.5x^2 + .5bx - .5c[/tex]
[tex]-1 = 1.5x^2 - .5bx - .5c[/tex]
[tex]1.5x^2 - .5bx - .5c +1 = 0[/tex]

Ultimately, I'm going to end up with factors of the form (ax^2+b+?)(x^2+c+?) ? Oh shoot . I don't know what I'm supposed to end up with. The question wants values for b and c. It's been 25 years!

So, am I then supposed to use trial and error?

if b=1, c=3
1.5x^2 - .5bx - .5c +1 = 0
then 1.5x^2 – .5x - .5 = 0

So (1.5x + .5)(x - 1) (This isn't quite right, since it actually makes 1.5x2 – x -1=0, but I think I'm close.)

Even if I get the right answer, I'll need to hsow my work, I can't tellt he students to guess can I?
 
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  • #2
You've overcomplicated stuff. Go back to

-0.5x^2 + x -1= x^2 + .5bx - .5c

First, it's wrong. It should read

-0.5x^2 + x -1 = -.5x^2 + .5bx - .5c

Then, if it's true for all x, the coefficients next to each different x term must be the same. So -.5c = -1 for example. Or .5b = 1
 
  • #3
Oh shoot, I made a mistake. I'll rewrite.

-0.5x^2 + x -1= -.5x^2 + .5bx - .5c
x -1=.5bx - .5c
-1 = .5bx -.5c –x

So .5bx-x+.5c+1=0

But I didn't follow anything you said in the last line.

(bx-1)(.5c-1)=0. Nope, still lost.
 
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  • #4
Dude, you need to calm down on the algebraic errors :P

-1 = .5bx -.5c –x

So .5bx-x+.5c+1=0

should be
.5bx-x -.5c + 1 = 0

So it's true for all x. Suppose x=0. Then -.5c+1=0, and c=2. Now we have

.5bx-x-1+1=0 Or .5bx-x=0

EDIT: It should be noted you never actually factored. It should be

-0.5x2 + x -1 = -.5(-.5x^2/-.5 + x/-.5 -1/-.5) = -.5(x^2 -2x + 2)

which gives b and c trivially

So let x=1. .5b-1=0, or b=2.
 
  • #5
Sorry 'bout that. Need to get this back tomorrow - which was fine - until I got a phone call to go to a job interview TONIGHT. Sort of distracted me AND ate up my time.


Oh @#$!&*, I feel like I'm asking to be spoonfed. See, I need to give an authoritative answer (it's the answer that I'm providing for the textbook), and I need to show the work. But before begging for you to peel my grapes for me, I will try to take what you've demonstrated and write it out as a correct and succinct answer to the problem...


-.5x^2 + x - 1 = -0.5 (x^2 - bx + c)
-.5x^2 + x - 1 = -.5x^2 + .5bx - .5c
x^2 + 2x - 2 = x^2 + bx - c
2x - 2 = bx - c
0 = bx - c - 2x - 2
bx - 2x - c + 2 = 0

At x=0:
b(0)-2(0)-c+2=0
-c+2=0
c=2

At x=-1:
b(-1)-2(-1)-c+2=0
-b+2-c+2=0
-b-c+4=0
-b-(2)+4=0
-b+2=0
b=2

OK, well that gets us the answer, but
1] it doesn't do it by factoring, which is what was asked. How are students supposed to figure this out??
2] Have I written the full form of the answer correctly? This is going in the answers section of the book.
 
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  • #6
DaveC426913 said:
I'm helping edit a math textbook and have been asked to provide the answer for this:

Factor to find the value of b and c in the trinomial:
-0.5x2 + x -1= -0.5 (x2 - bx + c)
This question is meaningless (for starters, a trinomial is an expression, not an equation...but that's hardly all that's wrong with it) and certainly NOT what you want to put into a textbook. If it's not too late, please ask the author to remove it entirely.
 
  • #7
DaveC426913 said:
OK, well that gets us the answer, but
1] it doesn't do it by factoring, which is what was asked. How are students supposed to figure this out??
2] Have I written the full form of the answer correctly? This is going in the answers section of the book.

1.) That's where the last part of my post came in. Factor -.5 from the expression -.5x2 + x - 1. You get -.5*(x2 - 2x + 2).

But this needs to equal -.5*(x2 - bx + c). Now, it should instinctually leap to you, that this implies x2 - bx + c = x2 - 2x + 2. And from here, let's analyze polynomials in general.

If I told you ax+b = 3x+2, what are a and b? Trivially, if a=3, and b=2, we're done. Now, it can be shown this is a unique solution, but I don't think you're worried about that. Going back to our example up here, x2 - bx + c = x2 - 2x + 2 is true if b=2 and c=2 (plug it in, it should be obvious). This is because -b is next to the 'x' term, and so is -2 on the other side. And 2 is the constant term on one side, and so is c. So -b=-2, and c=2. This is probably what they're trying to drive at
 
  • #8
I thought answers in a textbook only need the final solution and not the work itself, so why concern yourself with that?

Get Maple, and let it do it for you.
 
  • #9
JasonRox said:
I thought answers in a textbook only need the final solution and not the work itself, so why concern yourself with that?
You thought wrong.:rolleyes:

Anyway, it's OK. Thnaks everyone. The work has been passed to the appropriate person for the job (I was not it).
 

What is factoring a trinomial?

Factoring a trinomial is the process of finding the two binomials that, when multiplied together, produce the given trinomial. This is useful in simplifying algebraic expressions and solving equations.

How do I factor a trinomial?

To factor a trinomial, you can use the FOIL method or the grouping method. In the FOIL method, you multiply the first terms, the outer terms, the inner terms, and the last terms, and then combine like terms. In the grouping method, you group the terms with common factors and factor them out.

Why is factoring a trinomial important?

Factoring a trinomial is important because it helps in solving quadratic equations, finding the roots of a function, and simplifying algebraic expressions. It also helps in understanding the relationship between factors and the original expression.

What are the common mistakes when factoring a trinomial?

Some common mistakes when factoring a trinomial include forgetting to factor out a common factor, making errors in the FOIL method, and not considering all possible factor pairs. It is also important to check if the factored trinomial can be further simplified.

Can all trinomials be factored?

Not all trinomials can be factored. Some trinomials may have complex or irrational roots, which cannot be expressed as factors. In addition, some trinomials may not have any real solutions, meaning they cannot be factored using real numbers.

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