Question

The area of a rectangular desktop is `6x^2- 3x -3`. The width of the desktop is `2x+1`. What is the length of the desktop?

Answer 1

The length of the desktop is `3(x-1)`

Explanation:

Area of the rectangle is `A=l*w` , where `l ,w` are length and width of rectangle respectively.

So `l=A/w or l = (6x^2-3x-3)/(2x+1) or (3(2x^2-x-1))/(2x+1) or (3(2x^2-2x+x-1))/(2x+1) or (3(2x(x-1)+1(x-1)))/(2x+1) or (3cancel((2x+1))(x-1))/cancel((2x+1)) or 3(x-1)`

The length of the desktop is `3(x-1)` [Ans]

Answer 2

Length is `(3x-3)`

Explanation:

Note that LHS is left hand side and RHS is right hand side

The way the question is worded means we have to have the initial condition of:

`(2x+1)(?+?)=6x^2-3x-3 ........................Equation(1)`

`color(blue)("Consider the "x^2" term:")`

We have `2x xx?=6x^2`

To end up with `x^2` we must have:

`2x xx?x=6x^2`

To end up with the 6 from `6x^2` we must have:

`2x xx3x=6x^2`

So we now have:

`(2x+1)(3x+?)=6x^2-3x-3.....................Equation(1_a)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)("Consider the constant of "color(red)(-3)" in "6x^3-3xcolor(red)(-3))`

We already have 1 in the `(2x+1)` and `1xx(-3)=-3`

This implies that we have:`" "(2x+1)(3x-3)`

So we need to test:

`color(blue)((2x+1))color(green)((3x-3)) = 6x^2-3x-3`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)("Consider just the brackets")`

Multiply the 2nd brackets by everything in the 1st brackets

`color(green)(color(blue)(2x)(3x-3)color(blue)(" "+1)(3x-3))`

`6x^2-6x" "+color(white)(..)3x-3`

`6x^2-3x-3 = "LHS of the equation"`

So LHS = RHS of the equation, thus the answer is:

`"Width "xx" Length"`

`(2x+1)xx(3x-3)`

Answer 3

3x-3

Explanation:

Area of a rectangle=W*L
`6x^2-3x-3 =(2x+1)*L`
`=(6x^2-3x-3)/(2x+1)`
`=3(2x^2-x-1)/(2x+1)`
`=3((2x+1)(x-1))/((2x+1))`
cancel out 2x+1
Then length= 3x-3
check
`3(x-1)(2x+1)`
`(3x-3)(2x+1)`
`6x^2-3x-3=3(x-1)(2x+1)`
`6x^2-3x-3=6x^2-3x-3`

Answer 4

`color(red)("Alternative method - polynomial division")`

`"Length"= 3x-3`

Explanation:

We have: `" width"xx"length"=6x^2-3x-3`

`=>"length "=(6x^2-3x-3)/("width")" "=" "(6x^2-3x-3)/(2x+1)`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("The division")`

`" "color(white)(.)6x^2-3x-3`
`color(red)(3x)(2x+1)-> ul(6x^2+3x) larr" subtract"`
`" "0color(white)(.)-6xcolor(white)(.)-3`
`color(red)(-3)(2x+1)->ul(" "-6xcolor(white)(.)-3) larr" subtract"`
`" "0color(white)(.)+color(white)(.)0`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`=>"length" = color(red)(3x-3) = (6x^2-3x-3)/(2x+1)`