homework help: I don’t usually post questions asking for homework help because I know how annoyed people get with people who make those type of posts. - Help.com

Pink Freud

Spokane, WA, US

I don’t usually post questions asking for homework help because I know how annoyed people get with people who make those type of posts.

However, I really do need help with a piece of my math homework that I have been trying to grasp for some time now and just can’t seem to get a handle on. I am not asking anyone here to do my homework for me, just that they help explain to me what needs to be done and/or how it is done as I seem to be a little lost when it comes to certain math concepts.

So, I have been trying to figure out how one evaluates a polynomial. I saw a tutor and discussed the matter with him and he made it look and sound pretty easy. I tried his approach and ended up more confused. I have even gone to websites like purplemath.com and the like but they also make it look very easy - too easy. However, when I try to do it, somewhere along the line I go wrong and I don’t quite know where and why. I have tried doing everything my tutor has suggested but I still ended up getting those problems wrong.

So here’s what I know. I know that evaluating a polynomial is where one replaces the letter variables with a given number. I also know that the solution is usually reached by following the order of operations. Is there anything else I should know or need to know to solve this type of problem b/c I’ve certainly come up empty all the other times I’ve tried.

So for example, how would I go about evaluating an expression that looks like this?

(b-10)(b+7)+ 10 - 8(-8+6b)

b = 3

Thanks!

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Since writing this post Pink Freud may have helped people, but has not within the last 4 days. Pink Freud is a verified member, has been around for 5 years, 3 months and has 126 posts and 19,878 replies to their name.

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Pink Freud invited 7 users to read this post 1 year, 2 months ago.

That is not an equation as there is no equal sign (sorry to be pedantic). You can evaluate the ‘expression’ though simply by substituting b=3 into the expression thus:
(3-10)(3+7)+10-8(-8+6×3)

Simplifies to
(-7)(10)+10-8(10)
and again to
-70+10-80
Which is simply 0.

Help me with: Insignificance.

fractal.scatter wrote:
That is not an equation as there is no equal sign (sorry to be pedantic). You can evaluate the ‘expression’ though simply by substituting b=3 into the expression thus:
(3-10)(3+7)+10-8(-8+6×3)

Simplifies to
(-7)(10)+10-8(10)
and again to
-70+10-80
Which is simply 0.

lol. That’s ok.

Thanks for the explanation, that makes a lot of sense. I see now that I was missing a step. Thanks for the help.

No problem :)

Help me with: Insignificance.

Aw! You beat me to it! :( But you were wrong. ;)

In more complicated problems, or if you are not given a value of b, (Or if you like math and want more fun) I would have done it by factoring:

(b-10)(b+7) + 10 + -8(-8+6b)

Factor the first two parentheses (b-10)(b+7) together using
FOIL(Firsts + Insides + Outsides + Lasts)

(b-10)(b+7)= b^2 + -10b + 7b + -70 simplified as b^2 - 3b - 70

First b*b = b^2
Inside -10b
Outside 7b
Last 7*-10 = -70

Then factor the -8 into (-8+6b) to get

([-8*-8] + (-8*6b)]

-8*-8 = 64
-8*6b = -48b

so put those two back into the ‘equation’ and you get

b^2 - 3b - 70 + 10 + 64 - 48b

Rearrange so like terms are together in descending power order:

b^2 + (-48b - 3b) + (-70 + 10 + 64)

simplify:

b^2 + -51b + 4

It’s a lot more complicated than Fractal’s way, but sometimes it isn’t that easy, or you aren’t given the variable value. If it’s too hard to do it the other way, do it like this, THEN you put the 3 in:

(3^2) + (-51*3) + 4
9 -153 + 4 = -140

Though in this situation it was obviously much more simple to just stick the 3 inside first… As long as you add and subtract properly in the end. :)

~CaitherrA~ wrote:
As long as you add and subtract properly in the end. :)

Oh, shut your pie hole :P

Help me with: Insignificance.

fractal.scatter wrote:

~CaitherrA~ wrote:
As long as you add and subtract properly in the end. :)

Oh, shut your pie hole :P

:)

~CaitherrA~ wrote:
Aw! You beat me to it! :( But you were wrong. ;)

In more complicated problems, or if you are not given a value of b, (Or if you like math and want more fun) I would have done it by factoring:

(b-10)(b+7) + 10 + -8(-8+6b)

Factor the first two parentheses (b-10)(b+7) together using
FOIL(Firsts + Insides + Outsides + Lasts)

(b-10)(b+7)= b^2 + -10b + 7b + -70 simplified as b^2 - 3b - 70

First b*b = b^2
Inside -10b
Outside 7b
Last 7*-10 = -70

Then factor the -8 into (-8+6b) to get

([-8*-8] + (-8*6b)]

-8*-8 = 64
-8*6b = -48b

so put those two back into the ‘equation’ and you get

b^2 - 3b - 70 + 10 + 64 - 48b

Rearrange so like terms are together in descending power order:

b^2 + (-48b - 3b) + (-70 + 10 + 64)

simplify:

b^2 + -51b + 4

It’s a lot more complicated than Fractal’s way, but sometimes it isn’t that easy, or you aren’t given the variable value. If it’s too hard to do it the other way, do it like this, THEN you put the 3 in:

(3^2) + (-51*3) + 4
9 -153 + 4 = -140

Though in this situation it was obviously much more simple to just stick the 3 inside first… As long as you add and subtract properly in the end. :)

Awesome! Thanks! Those are actually the ones I seem to be struggling with the most.

Daft Puck wrote:

~CaitherrA~ wrote:
Aw! You beat me to it! :( But you were wrong. ;)

In more complicated problems, or if you are not given a value of b, (Or if you like math and want more fun) I would have done it by factoring:

(b-10)(b+7) + 10 + -8(-8+6b)

Factor the first two parentheses (b-10)(b+7) together using
FOIL(Firsts + Insides + Outsides + Lasts)

(b-10)(b+7)= b^2 + -10b + 7b + -70 simplified as b^2 - 3b - 70

First b*b = b^2
Inside -10b
Outside 7b
Last 7*-10 = -70

Then factor the -8 into (-8+6b) to get

([-8*-8] + (-8*6b)]

-8*-8 = 64
-8*6b = -48b

so put those two back into the ‘equation’ and you get

b^2 - 3b - 70 + 10 + 64 - 48b

Rearrange so like terms are together in descending power order:

b^2 + (-48b - 3b) + (-70 + 10 + 64)

simplify:

b^2 + -51b + 4

It’s a lot more complicated than Fractal’s way, but sometimes it isn’t that easy, or you aren’t given the variable value. If it’s too hard to do it the other way, do it like this, THEN you put the 3 in:

(3^2) + (-51*3) + 4
9 -153 + 4 = -140

Though in this situation it was obviously much more simple to just stick the 3 inside first… As long as you add and subtract properly in the end. :)

Awesome! Thanks! Those are actually the ones I seem to be struggling with the most.

Let me or Fractal know if you need more help. We are math geeks. :)

~CaitherrA~ wrote:

Daft Puck wrote:
~CaitherrA~ wrote:
Aw! You beat me to it! :( But you were wrong. ;)

In more complicated problems, or if you are not given a value of b, (Or if you like math and want more fun) I would have done it by factoring:

(b-10)(b+7) + 10 + -8(-8+6b)

Factor the first two parentheses (b-10)(b+7) together using
FOIL(Firsts + Insides + Outsides + Lasts)

(b-10)(b+7)= b^2 + -10b + 7b + -70 simplified as b^2 - 3b - 70

First b*b = b^2
Inside -10b
Outside 7b
Last 7*-10 = -70

Then factor the -8 into (-8+6b) to get

([-8*-8] + (-8*6b)]

-8*-8 = 64
-8*6b = -48b

so put those two back into the ‘equation’ and you get

b^2 - 3b - 70 + 10 + 64 - 48b

Rearrange so like terms are together in descending power order:

b^2 + (-48b - 3b) + (-70 + 10 + 64)

simplify:

b^2 + -51b + 4

It’s a lot more complicated than Fractal’s way, but sometimes it isn’t that easy, or you aren’t given the variable value. If it’s too hard to do it the other way, do it like this, THEN you put the 3 in:

(3^2) + (-51*3) + 4
9 -153 + 4 = -140

Though in this situation it was obviously much more simple to just stick the 3 inside first… As long as you add and subtract properly in the end. :)

Awesome! Thanks! Those are actually the ones I seem to be struggling with the most.

Let me or Fractal know if you need more help. We are math geeks. :)

lol. OK, thanks! :)

The one thing I notice that often needs to be done that I usually forget to do is re-order the polynomial left to right alphabetically starting with the highest term. Is there a way to know when this needs to be done?

I see the word polynomial and I go O.o Then I remember that I know what those are. :D

If you are rearranging things, it can be done at ANY time. As long as you follow order of operations. For example, in the beginning of the problem, you cannot rearrange it.
(b-10)(b+7)+ 10 - 8(-8+6b)
is not the same as
(b-10) + 10 * (b+7) - 8(-8+6b)
Another general rule I usually forget is moving the subtractions properly. So where ever there is a minus, instead change it to adding a negative BEFORE rearranging things. So:
a - b = a + (-b)
5 - 10 = 5 + (-10)

So get rid of parentheses(usually means also getting rid of Multiplication and Division as well), convert subtraction to adding a negative, combine like terms, and then you are pretty much free to rearrange them as you want.
… Does that kind of make sense?

~CaitherrA~ wrote:
I see the word polynomial and I go O.o Then I remember that I know what those are. :D

If you are rearranging things, it can be done at ANY time. As long as you follow order of operations. For example, in the beginning of the problem, you cannot rearrange it.
(b-10)(b+7)+ 10 - 8(-8+6b)
is not the same as
(b-10) + 10 * (b+7) - 8(-8+6b)
Another general rule I usually forget is moving the subtractions properly. So where ever there is a minus, instead change it to adding a negative BEFORE rearranging things. So:
a - b = a + (-b)
5 - 10 = 5 + (-10)

So get rid of parentheses(usually means also getting rid of Multiplication and Division as well), convert subtraction to adding a negative, combine like terms, and then you are pretty much free to rearrange them as you want.
… Does that kind of make sense?

kinda. Let me see if I can find a problem where what I was referencing is needed. Hold on a sec.

That didn’t really make sense… Generally, you rearrange them at the very end. Once you have all the terms simplified as far as they can go, THEN you arrange them appropriately.

Okey dokey. :)

OK, when I tried evaluating this expression, the software I use said I had to re-order the polynomial in the manner I described above.

6 (4-4x)-4 (z+10) +4

x = 2
z= 0

Software? O.o You’ll have to elaborate on that part. But anyway, to evaluate this expression, I would start with factoring.

Also, I usually separate it into multiple different problems at the addition/subtraction signs.
6(4-4x)
-4(z+10)
4 (Easy! :D)

In those three problems, you can re order them as you please, since you are keeping the multiplication together. So

6 (4-4x)-4 (z+10) +4
is the same as
4+ -4(z+10) + 6(4-4x)
is the same as
-4(z+10) + 4 + 6(4-4x)

However, you CANNOT move the multiplication around.
So it CANNOT be:

6(z+10) + 4 + -4(4-4x)

Am I making sense so far? And is this the direction you are needing?

Well, the school has us use the my “mymathlab” and with that particular expression they suggest that we change the expression to 6(-4x + 4)-4 (z+10)+ 4

Does that make sense?

Anyway. Back to solving it:

6(4-4x) + -4(z+10) + 4

Factor the 6 and the 4 into their problems.

6(4-4x) = [(6*4) - 6*4x)] = (24-24x) = 24 + -24x
and the parentheses break as there are no more multiplication problems inside of it.

-4(z+10) = [(-4*z)+(-4*10)] = (-4z + -40) = -4z + -40

Now that we have factored/simplified those, we can put them back into the problem:
(24 + -24x) + (-4z + -40) + 4
The parentheses aren’t necessary as there is nothing being multiplied, I’m just demonstrating the problems separately.
So take the parentheses away and then combine like terms. Since we have different variables, that means pretty much only the whole numbers. First we will rearrange them so that they are next to their like terms. So x’s first, then z, then the whole numbers.
-24x + -4z + -40 + 24 + 4
And combine:
-24x + -4z + -12
can also be written as
-24x - 4z - 12
And substitute the variables in for their counter parts:
x = 2
z= 0

(-24*2) + (-4*0) + -12
-48 + 0 + -12
-48 - 12 = -60

Again, I’m not sure what software you are talking about? I hate calculators and computers, so I generally just stick with my own solving capabilities.

Change which expression to 6(-4x + 4)-4 (z+10)+ 4? What was the original?

~CaitherrA~ wrote:
Anyway. Back to solving it:

6(4-4x) + -4(z+10) + 4

Factor the 6 and the 4 into their problems.

6(4-4x) = [(6*4) - 6*4x)] = (24-24x) = 24 + -24x
and the parentheses break as there are no more multiplication problems inside of it.

-4(z+10) = [(-4*z)+(-4*10)] = (-4z + -40) = -4z + -40

Now that we have factored/simplified those, we can put them back into the problem:
(24 + -24x) + (-4z + -40) + 4
The parentheses aren’t necessary as there is nothing being multiplied, I’m just demonstrating the problems separately.
So take the parentheses away and then combine like terms. Since we have different variables, that means pretty much only the whole numbers. First we will rearrange them so that they are next to their like terms. So x’s first, then z, then the whole numbers.
-24x + -4z + -40 + 24 + 4
And combine:
-24x + -4z + -12
can also be written as
-24x - 4z - 12
And substitute the variables in for their counter parts:
x = 2
z= 0

(-24*2) + (-4*0) + -12
-48 + 0 + -12
-48 - 12 = -60

Again, I’m not sure what software you are talking about? I hate calculators and computers, so I generally just stick with my own solving capabilities.

I take my classes on line, hence the software. Just ignore that fact it isn’t really relvent to anything.

Change this

6 (4-4x)-4 (z+10) +4

to this:

6(-4x + 4)-4 (z+10)+ 4

Am I taking you in the right direction? I’m not sure what exactly you need help with. :P

~CaitherrA~ wrote:
Am I taking you in the right direction? I’m not sure what exactly you need help with. :P

I think you are. Everything you have said has been helpful.

My only question is why do they have us change it? It doesn’t seem like it is really necessary. Is it just a weird way of solving?

Daft Puck wrote:
Change this

6 (4-4x)-4 (z+10) +4

to this:

6(-4x + 4)-4 (z+10)+ 4

That wouldn’t make sense unless you were to change it from
6 (4-4x)-4 (z+10) +4
to
-6 (-4+4x)-4 (z+10) +4

You are basically reverse factoring in that case. You are taking OUT a -1 instead of putting in the 6. So
(4-4x) = -1(-4+4x) and then you have the 6 and -1 on the outside which combine to just -6.

Daft Puck wrote:

~CaitherrA~ wrote:
Am I taking you in the right direction? I’m not sure what exactly you need help with. :P

I think you are. Everything you have said has been helpful.

My only question is why do they have us change it? It doesn’t seem like it is really necessary. Is it just a weird way of solving?

It MIGHT be because mymathlab only accepts addition in the parentheses, I guess? I don’t see any point in rearranging it like that, but it is possible to do.

Yea, I didn’t understand the reasoning there either. Thanks for all the explanations. It really has been helpful!

Oh wait! I read it backwards! Sorry!!! I get it. It wants the variable value first!
So you put the x in the parentheses first. I totally looked over that. I was wrong. Forget my last post. Haha.

So:
6 (4-4x) = 6(-4x + 4) Is true. Because you didn’t factor out a -1, you just moved the insides around.

So always have the variable first in the parentheses. Because you are inside the parentheses, and there is no multiplication inside of it, you can rearrange it to your will.
Look at it like this

So now you can switch it around to
6(-4x + 4)

When doing things like FOIL, you’ll need the biggest variables first. I understand what you meant earlier too. Anytime there is ONLY addition/subtraction, you can rearrange. You always want variables first, and then in biggest power to smallest power.

~CaitherrA~ wrote:
Oh wait! I read it backwards! Sorry!!! I get it. It wants the variable value first!
So you put the x in the parentheses first. I totally looked over that. I was wrong. Forget my last post. Haha.

So:
6 (4-4x) = 6(-4x + 4) Is true. Because you didn’t factor out a -1, you just moved the insides around.

So always have the variable first in the parentheses. Because you are inside the parentheses, and there is no multiplication inside of it, you can rearrange it to your will.
Look at it like this

So now you can switch it around to
6(-4x + 4)

When doing things like FOIL, you’ll need the biggest variables first. I understand what you meant earlier too. Anytime there is ONLY addition/subtraction, you can rearrange. You always want variables first, and then in biggest power to smallest power.

Ohhh…ok. Thanks! That clarifies things. :D

You’re welcome. :) It was just confusing because they were both 4 so it’s really easy to just mix them up. XD

~CaitherrA~ wrote:
You’re welcome. :) It was just confusing because they were both 4 so it’s really easy to just mix them up. XD

yea, I understand.

Pink Freud closed this post: Resolved. Thanks for all of the help! :)

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