Question

PLEASE HELP ME!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
I WILL AWARD MEDAL!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

A tool box has a volume of x^3 + 8x^2 + 11x – 20 cm^3 and the height is x + 5 cm. Find the polynomial that would represent the area of the bottom of the tool box? Explain your reasoning.
@jim_thompson5910 @satellite73 @thomaster @Mertsj

Answer 1

@Mertsj @jim_thompson5910 @satellite73 @thomaster

Answer 2

divide

Answer 3

How.

Answer 4

\[( x^3 + 8x^2 + 11x – 20 )\div (x+5)\]

Answer 5

1) long division (ick)
2) synthetic division (very snappy)
3) thinking (it aint illegal yet)

Answer 6

Preferably synthetic division. Can you explain it please.

Answer 7

i can explain it but it is an amazing pita to write it here

Answer 8

how about we think?

Answer 9

Ok

Answer 10

you are dividing a polynomial of degree 3 by a polynomial of degree 1, leaving a polynomial of degree 2, i.e. a quadratic, and by the set up of the question is it pretty obvious it will go in there evenly so we start

Answer 11

\[ x^3 + 8x^2 + 11x – 20 =(x+5)(ax^2+bx+c)\]
now it should be rather obvious that i was being silly writing \(a\) when we know \(a=1\) since that is the only way we are going to get \(x^3\)
\[ x^3 + 8x^2 + 11x – 20 =(x+5)(x^2+bx+c)\]

Answer 12

also we know \(c\) right away , since the constant is \(-20\) and we have a \(+5\) in the first term, making \(c=-4\)

Answer 13

\[x^3 + 8x^2 + 11x – 20 =(x+5)(x^2+bx-4)\] so both \(a\) and \(c\) are clear from the start, i was only writing that other stuff to show what i was doing
the only think part is finding \(b\)

Answer 14

OK I am sorry can you slow do my thinking process is a little slow please.

Answer 15

ok really you need only look at this line \[x^3 + 8x^2 + 11x – 20 =(x+5)(x^2+bx-4)\] and see that it is clear that the first term on the quadratic must be \(x^2\) and the constant must be \(-4\)
when that is clear let me know and we can find \(b\)

Answer 16

where do you get b from.

Answer 17

we didn't get it yet, i was slowing down to make sure it is clear to you that the quadratic must be \(x^2+bx-4\) so all we need is \(b\) to complete it
is that clear? i.e. is it clear that it must start with \(x^2\) and end with \(-4\) ?

Answer 18

But the original equation does not even have a B.
That was what I was asking.

Answer 19

oh \(b\) is just some number we need to find
that is all the work , which is not hard at all

Answer 20

i just used \(b\) because frequently we write a quadratic as \(ax^2+bx+c\) so i used \(b\) for the coefficient of the \(x\) term

Answer 21

I have a question is a volume consisting of lxwxh so we only need the lxw no height right.

Answer 22

forget the verbiage of the question, which is nonsense , just trying to make it sound like a word problem
your job is to divide, that is all

Answer 23

the question is really asking you to divide
\[( x^3 + 8x^2 + 11x – 20 )\div (x+5)\]

Answer 24

Ok so all we do is divide So their is no 3 monomials right? That is the last stupid thing I will ask.

Answer 25

Oh ok.

Answer 26

no , it asks for the area which will be your quadratic

Answer 27

OK givem e a sek to divide.

Answer 28

there might be three monomials if your quadratic factors, but it doesn't ask for them

Answer 29

lol i though that is what we were doing

Answer 30

So it is x^2 +3x -4 with a remainder of 0.

Answer 31

yes

Answer 32

where did you get the 3 from? it is right, i am just asking

Answer 33

ok so just if I am not asking to much can you please word the problem answer for me.

Answer 34

When I divided that is what came up.

Answer 35

i guess i meant by what method, but nvm

Answer 36

Find the polynomial that would represent the area of the bottom of the tool box? Explain your reasoning.

Answer 37

the volume is length times width times height
since we know the volume is \(x^3+8x^2-11x-20\) and the height is \(x+5\) we know that length times width, the area of the base, is \((x^3+8x^2+11x-20)\div (x+5)\)

Answer 38

Is that the complete answer? @satellite73

Answer 39

Thank you very much @satellite73 I have only 3 more questions I hope you can help me with. Thank you again.