Question

Boys please help! ;)
4. Determine two different values of “b” in x2 + bx + 30 so that the expression can be factored into the product of two binomials. Explain how you determined those values and show each factorization. Explain how your process would change if the expression was 2x2 + bx + 30.

Answer 1

interesting how you appeal just to the boys. There are girls and boys that help out here on OS :)

Answer 2

I have to have an attention grabber. I don't WANT to exploit my gender for help, but this is urgent.

Answer 3

Yeah yeah, you have a timed (I'll save you grace and not mention what it really is). How much time do you have to get it done?

Answer 4

I am behind in my work, and I need to submit this assignment and many others if I want to pass to the next grade level.

Answer 5

What do you need help with? Do you know how to get started?

Answer 6

It's algebra. Here is the question : Determine two different values of “b” in x2 + bx + 30 so that the expression can be factored into the product of two binomials. Explain how you determined those values and show each factorization. Explain how your process would change if the expression was 2x2 + bx + 30.

I would like more than answers, I would like How to do it

Answer 7

So do you have any idea how to approach the question in mind?

Answer 8

No, I struggle with this subject.

Answer 9

Okay, do you understand what the question is asking for?

Answer 10

It wants to know 2 values of b that would allow it to be factored into the product of 2 binomials?

Answer 11

Yeah, but do you know what a binomial is?

Answer 12

Something with two factors - like 2b^2 + 12

Answer 13

I know what binomials, trinomials, and polynomials are

Answer 14

okay.

Answer 15

so we'll start from what form the answer will look like, and then approach the problem.
the product of two binomials will be in some form
(ax+c)(dx+e) where a, c, d, and e will all be constants
(I have skipped b to avoid confusion as it is already in the problem)
Does that make sense?

Answer 16

yes

Answer 17

the thing about this is though, if we expand this expression (with FOIL) we can actually turn it into something that looks like the original expression "x^2+bx+30"
Let's try it out.
Can you do it?

Answer 18

Would I start out by multiplying x^2 by bx and then 30?

Answer 19

no, I mean expanding the expression
(ax+c)(dx+e)

Answer 20

how did you expand the expression to that?

Answer 21

well, if you've learned FOIL, it's First, Outside, Inside, Last.
so
(ax)*(dx)=adx^2 the First two terms
(ax)*(e)=aex the Outside two terms
Can you continue for the inside terms and the last terms? After you get all your terms you add them up

Answer 22

(c)(dx)
(c)(e)

Answer 23

yeah, let's add them all up now...
adx^2+aex+cdx+ce
notice how ae and cd are constants, so we want to combine aex and cdx because they will just end up being numbers in front of the x.
so can you combine the x terms in adx^2+(ae)x+cd(x)+ce?

Answer 24

i believe so

Answer 25

what did you get?

Answer 26

So I combine aex and cdx?

Answer 27

yeah.

Answer 28

remember, treat all variables but x as constants, because we will soon be picking actual numbers for them

Answer 29

I'm not sure how to combine them. I am stumped. I've been up for 24 hours eugh >_<

Answer 30

okay. when I meant treat them like constants, I meant pretend they are numbers you know.
if you had 5x and 9x, when you add them together what you do in your head is (5+9)x=14x
so I want you to do the same thing with the letters... except you won't be able to get a result that's a single number.

Answer 31

So should I assign them real numbers? because I'm not sure how to write them out if i don't. I know that the answer would end in x^2 because you are squaring x?

Answer 32

no...
the main thing with algebra that had me stumped for a long time was the real meaning of all those letters. it's a more abstract concept than most are introduced to at first. When teachers say a variable is just some unknown number, they aren't really being entirely accurate. A variable represents a number, which may or may not be known. In this case, the variables that we don't know are b and x, although we will eventually know what values b can hold.

When I added in all those variables a, c, d, e, I used them as placeholders for numbers that we WILL determine in the near future.

The general idea in algebra when manipulating expressions and solving equations is to turn one side into a form similar to the other. There are many ways to do this, but in the upper levels, you actually have to analyze the problem from the SOLUTION and work backwards from there to get your problem. Then, you can work forwards.

What I'm basically doing here is constructing the solution, and turning it into the question. The reason for this is because it might be a lot easier for you than moving forward (since it's not always obvious what direction is the correct one for moving forward, but playign with a solution to make it the original expression is sometimes easier).

In this case I've given you (ax+c)(dx+e), not because I already know the answer, but because the problem ASKED for two binomials. the two binomials that make the most sense in this case are these two. all the a, c, d, and e are constants, and it all depends what values they will really hold. for instance, we could pick a=1,c=2,d=3,and e=4, but we could just as easily pick a=124235, c=-.000005, d=5, and e=0. This is why I HAD to use variables to represent the expression. Do all my words up until here make sense?

Answer 33

Okay, so what would the answer be? I partially understand what you are saying

Answer 34

lol, I thought you didn't want answers :)

Naggy life advice: you can pursue what seems to be the fast way, but it's not always the fastest in the long run.

I'm going to give you two choices that I will honor.
1. Take the answer and a brief explanation as required for your question, but having to go on openstudy again next time you meet this stuff.
2. Hear me out, and perhaps gain a better understanding of algebra, and more importantly, problem solving in general, while attaining my support if you have some other problems.

It's your call.

Answer 35

I'll take number 2. I apologize, I just rush all the time.

Answer 36

Okay (you can change your mind at any time; I'm not paid to do this, so it doesn't matter to me either way).

once you have (ax+c)(dx+e) (which is essentially what the question is asking for but without determined constants a, c, d, and e), the next step is to attempt to make it look like the problem, which was the expression
x^2+bx+30

Now given prior experience with multiplying binomials, (you have heard of the FOIL rule right?), one quickly sees that if you expand (ax+c)(dx+e) it should look kind of like x^2+bx+30

Answer 37

so that is why my next suggestion was to multiply out (aka expand) the expression, because it would lead to something that looked like the form of the original problem we were given.

Answer 38

we determined each individual term to be adx^2, aex, cdx, and ce.
So as with FOIL, we just add the terms up as usual.
we get adx^2+aex+cdx+ce.

The only problem is, we are looking for something with three terms, because we had
x^2+bx+30.

so our goal is to rewrite the expression adx^2+aex+cdx+ce so it only has 3 terms.
Upon closer observation, in x^2+bx+30, there's one term with an x^2, one term with an x, and one term with a constant.

in adx^2+aex+cdx+ce, however, there is 1 term with an x^2, 2 terms with an x, and one term with a constant (that is, ce, because those numbers are soon to be picked).

so how do you think we can make the two expressions like more similar (so they both have the same amount of terms, or "parts")?

Answer 39

can we multiply aex and cdx to get aecdx^2?

Answer 40

no, close, but then the expression wouldn't be the same as (ax+c)(dx+e).
what we actually want to do is rewrite them in some sort of way so we can combine them.
I'll do that for you, and let's see if you can combine them.

adx^2+aex+cdx+ce=adx^2+ (ae)*x + (cd)*x +ce
remember, all we care about is the result having just an x after it for the aex and cdx

Answer 41

I can't. for the life of me think of a way to combine them without squaring the x

Answer 42

I always think these things too hard and the answer is right in front of me

Answer 43

okay. the expression actually appears more complicated than you would think possible, but in reality it's the one we are looking for.
adx^2+(ae+cd)x+ce is equivalent to the previous expression.
now you might think that ae+cd isn't acceptable, but give it a minute of thought.
we can multiply two pairs of known numbers together and add them right? it's a more complicated way of expressing a single number, but it's definitely valid.
Does that make sense to you?

Answer 44

what I just did was combining like terms. It doesn't look like they are like terms on first look, but upon closer examination that's just what they are, which makes it entirely possible for them to be combined.

Answer 45

alright, so it is ae+cd, but what about the x's?

Answer 46

well, we weren't ever supposed to know what they were in the first place. the problem doesn't want to know what the value of x is (and rightfully so, for you cannot find the value seeing as this is not an equation); it merely wants you to turn one expression into another.
we've gotten
adx^2+(ae+cd)x+ce. let's compare it to
x^2+bx+30.

so let's look at the numbers in front of the two x^2 terms. the first one would be ad (ad times x^2 is adx^2, right), and the second would be 1 (1*x^2=x^2, right?)
but we know that these two expressions are the SAME, because we've already combined like terms, and we KNOW that we DEFINED our solution to be what we are looking for. so let's just pick two easy values for a and d so that ad=1.

Answer 47

pick two numbers :)

Answer 48

.2 and .5?

Answer 49

those actually multiply out to be .1.
Really when I meant easy values, I meant the easiest thing possible.
I'd pick a=1, and d=1.
that way, we can replace all those other a's and d's with one so that we don't have to do as much work.

Answer 50

I understand. So D and A both = 1. What would that make the final answer?

Answer 51

well, let's see. our expression was
adx^2+(ae+cd)x+ce
we picked a and d to both be 1. Can you put those numbers in and get us something less messy to work with?

Answer 52

(1)(1)x^2+(1e+c1)x+ce?

Is that even right? and if it is, I need to go further correct

Answer 53

yes, and yeah, you can get rid of all the ones essentially

Answer 54

x^2+(e+c)x+ce

Answer 55

yup. let's compare once more.
x^2+(e+c)x+ce
x^2+bx+30
so, b=e+c, and 30=ce right?

Answer 56

Right!

Answer 57

okay, well now it's as easy as picking two numbers that multiply out to 30, and we know that if we add the two numbers we will get b, right?

Answer 58

15 and 2?

Answer 59

yeah. so that's one set.
(x+15)(x+2)=x^2+bx+30. what's b?

Answer 60

17?

Answer 61

yup. so that's one b value, and one set of binomials.
But we need anotehr one cause the problem asks for two.
Can you pick another set?