Answer:
The other two dimensions are either (x+9) or (x-7). other two dimensions are either 21 inches or 7 inches.
Step-by-step explanation:
The volume is defined by the function.
[tex]V(x)=x^3+5x^2-57x-189[/tex]
The width is (x+3).
Using synthetic or long division method divide [tex]V(x)=x^3+5x^2-57x-189[/tex] by (x+3).
[tex]V(x)=(x+3)(x^2+2x-63)[/tex]
[tex]V(x)=(x+3)(x^2+9x-7x-63)[/tex]
[tex]V(x)=(x+3)(x+9)(x-7)[/tex]
The volume of cuboid is
[tex]V=l\times b\times h[/tex]
Therefore other two dimensions are either (x+9) or (x-7).
If the width is 15 inches, then
[tex]x+3=15[/tex]
[tex]x=12[/tex]
The other two dimensions are
[tex]x+9=12+9=21[/tex]
[tex]x-7=12-7=5[/tex]
Therefore other two dimensions are either (x+9) or (x-7). other two dimensions are either 21 inches or 7 inches.
Answer: Other two dimensions are 21 and 5.
Height = 21 in.
Depth = 5 in.
Step-by-step explanation:
Since we have given that
[tex]x^3 + 5x^2 -57x-189=0[/tex]
As we have given that
Width = x+3
That means [tex]x+3=0\\\\x=-3[/tex]
So, by dividing, x^3 + 5x^2 -57x-189 by x+3 we get,
[tex]\frac{x^3 + 5x^2 -57x-189}{x+3}\\\\\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}x^3+5x^2-57x-189\mathrm{\:and\:the\:divisor\:}x+3\mathrm{\::\:}\frac{x^3}{x}=x^2\\\\Quotient=x^2\\\\\mathrm{Multiply\:}x+3\mathrm{\:by\:}x^2:\:x^3+3x^2\\\\\mathrm{Subtract\:}x^3+3x^2\mathrm{\:from\:}x^3+5x^2-57x-189\mathrm{\:to\:get\:new\:remainder}\\\\\mathrm{Remainder}=2x^2-57x-189\\\\=x^2+\frac{2x^2-57x-189}{x+3}[/tex]
[tex]\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}2x^2-57x-189\mathrm{\:and\:the\:divisor\:}x+3\mathrm{\::\:}\frac{2x^2}{x}=2x\\\\\mathrm{Quotient}=2x\\\\\mathrm{Multiply\:}x+3\mathrm{\:by\:}2x:\:2x^2+6x\\\\\mathrm{Subtract\:}2x^2+6x\mathrm{\:from\:}2x^2-57x-189\mathrm{\:to\:get\:new\:remainder}\\\\\mathrm{Remainder}=-63x-189\\\\=x^2+2x+\frac{-63x-189}{x+3}\\\\=x^2+2x-63[/tex]
Noe, we split the quadratic equation:
[tex]x^2+2x-63=0\\\\x^2+9x-7x-63=0\\\\x(x+9)-7(x+9)=0\\\\(x+9)(x-7)=0\\\\x=-9,7[/tex]
So, as we have given that
Width = 15
[tex]so, x+3=15\\\\x=15-3\\\\x=12\\\\Length=x-7=12-7=5\\\\\Height = x+9=12+9=21[/tex]
So, other two dimensions are 21 and 5.
Height = 21 in.
Depth = 5 in.
