Question 999752

The volume of a rectangular box is 

{{{V(x)=(6x^3+29x^2-7x-10)cm^3}}}. 

The box has a length of {{{(2x+1)cm}}} and a width of{{{ (x+5)cm}}}.

 Determine the height of the box.

{{{V(x)=(6x^3+29x^2-7x-10)cm^3}}}...since {{{V(x)=length*width*height }}}, means {{{(2x+1)cm}}} and {{{ (x+5)cm}}} two of factors and the height is third factor; so,
 factor given equation  completely

{{{V(x)=(6x^3+29x^2-7x-10)cm^3}}}...write {{{29x^2}}} as {{{33x^2-4x^2}}} and {{{-7x}}} as {{{15x-22x}}}

{{{V(x)=6x^3+33x^2-4x^2+15x-22x-10}}}...group

{{{V(x)=(6x^3-4x^2)+(33x^2-22x)+(15x-10)}}}

{{{V(x)=2x^2(3x^3-2)+10x(3x-2)+5(3x-2)}}}

{{{V(x)=(3x-2) (2x^2+11x+5)}}}

{{{V(x)=(3x-2) (2x^2+x+10x+5)}}}

{{{V(x)=(3x-2) ((2x^2+10x)+(x+5))}}}

{{{V(x)=(3x-2) (2x(x+5)+(x+5))}}}

{{{V(x)=(3x-2) (2x+1) (x+5)}}}


so, the height is: {{{(x+5)cm}}}