1. ## Cubics polynomial

hello all

y = -x^3 + x^2 + 3x -3

for this question i took out the - and got -(x^3 + x^2+3x - 3)
then i went ahead and found the common factor as (x-1) i wasnt sure when long dividing which form to use the one with or without the negative outside, and if i were to use the form with the negative sign outside, how would i represent this in the long division thing?

2.) A rectangular sheet of metal measuring 10cm * 12 cm is to be used to construct an open rectangular tray. The tray will be constructed by cutting out four squares from each corner of the sheet as shown in the diagram.

a.) Write down a rule for the volume, V cm^3, of the open tray in terms of x.

for part a the book says 4x(6-x)(5-x) however i did it like this 4x (-1)(x-6)(-1)(x-5) and ended up with 4x(x-6)(x-5) is this correct? gah i hate these negative sign business.

b.) Find the values for x for which V = 50
i don't know how to do b, its alot simpler with quadratics :P

2. Originally Posted by andrew2322
hello all

y = -x^3 + x^2 + 3x -3

for this question i took out the - and got -(x^3 + x^2+3x - 3)
then i went ahead and found the common factor as (x-1) i wasnt sure when long dividing which form to use the one with or without the negative outside, and if i were to use the form with the negative sign outside, how would i represent this in the long division thing?

2.) A rectangular sheet of metal measuring 10cm * 12 cm is to be used to construct an open rectangular tray. The tray will be constructed by cutting out four squares from each corner of the sheet as shown in the diagram.

a.) Write down a rule for the volume, V cm^3, of the open tray in terms of x.

for part a the book says 4x(6-x)(5-x) however i did it like this 4x (-1)(x-6)(-1)(x-5) and ended up with 4x(x-6)(x-5) is this correct? gah i hate these negative sign business.

b.) Find the values for x for which V = 50
i don't know how to do b, its alot simpler with quadratics :P
$\displaystyle y = -x^3 + x^2 + 3x -3$

Factor this by grouping

$\displaystyle y= -x^2(x-1)+3(x-1)=(-x^2+3)(x-1)=-(x^2-3)(x-1)$

for the next one your factorization is equivilent.

$\displaystyle 50=4x(x-6)(x-5)$

multiply out the right hand side
set the equation equal to zero
Factor again and use the zero factor priciple

Good luck.