Question

Would someone please help me? The volume in cubic feet of a workshop’s storage chest can be expressed as the product of its three dimensions: . The depth is x + 1.
a. Find linear expressions with integer coefficients for the other dimensions.
b. If the depth of the chest is 6 feet, what are the other dimensions?Answer

Answer 1

do you have a picture of the volume? something appears to be missing

Answer 2

No there is no picture The problem is v(x)=x^3-5x^2-2x+24. The depth is x+2.

Answer 3

so really we want v(x) as a product of three linear factors

Answer 4

oh i though there must be something else !

Answer 5

I would assume correct. The word problem aga in v(x)=x^3-5x^2-2x+24. The depth is x +2. I need to find the linear expressions with integer coefficients for the other dimensions and if the depth of the chest is 75 feet what are the other dimensions as well?

I am really lost here

Answer 6

they are asking you to factor
\[x^3-5x^2-2x+24=(x+2)\times \text{something}\]

Answer 7

\[v(x)=x^3-5x^2-2x+24 = (x+2)(ax^2 + bx + c)\]

Answer 8

eigen got this, i gotta run but that is the idea.
divide to find the other factor

Answer 9

Please show the work-I am so confused

Answer 10

Are you still there eigenschmeigen?

Answer 11

there are a few different approaches, but i find this the easiest to explain well

we have a cubic v(x), and we want to express it as a product of three factors. a cubic can be written as the product of a linear factor and a quadratic, and that's a nice place to start:

we'll use a,b,c to represent the coefficients of this quadratic. we are going to find a,b and c
\[v(x)=x^3-5x^2-2x+24 = (x+2)(ax^2 + bx + c)\]

expanding out on the right we have:

\[(x+2)(ax^2 + bx + c) = ax^3 + bx^2 + cx + 2ax^2 + 2bx + 2c\]
collect together the terms and put it back into the previous equation
\[x^3-5x^2-2x+24 = ax^3 + (b + 2a)x^2 + (c+2b)x + 2c\]

now because the has to work for all x we can compare coefficients:

\[x^3 = ax^3 \text{ , } -5x^2 = (b+2a)x^2 \text{ , } -2x = (c+2b) \text{ , } 24 = 2c\]

that first equation there looks pretty straight forward:
\[x^3 = ax^3 \text{ , so: } a = 1 \]
now the last equation:
\[24 = 2c \text{ , so: } c = 12\]

now lets look at the second equation there:
\[-5x^2 = (b+2a)x^2\]
\[-5 = b+2a\]
since a = 1
\[-5 = b + 2\]
\[b = -7\]

Answer 12

that looks pretty long winded, once you have done many you need less and less of the method and it becomes quicker

so we have:
\[v(x)=x^3-5x^2-2x+24 = (x+2)(ax^2 + bx + c) = (x+2)(x^2 -7x +12)\]

Answer 13

you can check your answer by expanding it back out.


now we want to factorise the quadratic into two linear factors. are you ok doing that?
\[v(x)= (x+2)(x^2 -7x +12)\]

Answer 14

we get:
v(x)= (x+2)(x -3)(x-4)

Answer 15

Ok I am a little confused

Answer 16

which bit is confusing for you? i'll try and explain more

Answer 17

what would the linear expression with the integer coefficients be for the other dimensions? is that the b=-7?

Answer 18

before we break it down into linear expressions we need to get the quadratic

Answer 19

then we factorise the quadratic to get our other linear expressions

Answer 20

I am reviewing your notes - v(x)= (x+2)(x -3)(x-4) is the linear espressions with coefficients for other dimensions?

Answer 21

so depth is (x+2)

length and width are (x-3) and (x-4)