Bobosharif S. answered • 03/04/18

PhD in Math, MS's in Calulus

^{3}+6x

^{2}+11x+6 this is the Volume of the rectangle, which is product of its length, (x+3), whidth (x+2) and height, h that should be found.

^{3}+6x

^{2}+11x+6=(x+1)(x+2)(x+3)

Cam C.

asked • 03/04/18The volume of a rectangle box is (X^3 +6x^2 +11x +6) cm^3. The box is (×+3) cm long and (×+2) cm wide. How high is the box?

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Bobosharif S. answered • 03/04/18

PhD in Math, MS's in Calulus

x^{3}+6x^{2}+11x+6 this is the Volume of the rectangle, which is product of its length, (x+3), whidth (x+2) and height, h that should be found.

You can factorize Volume, that is x^{3}+6x^{2}+11x+6=(x+1)(x+2)(x+3)

h(x+3)(x+2)=(x+1)(x+2)(x+3) (divide both sides by (x+2)(x+3) )

h=x+1.

Michael P. answered • 03/04/18

Quick Answers to Math, Chemistry, and Physics Problems

Let L(x) = length of box (cm) = x + 3

W(x) = width of box (cm) = x + 2

H(x) = height of box (cm)

V = volume of box (cm^{3}) = L(x)*W(x)*H(x) = x^{3} + 6*x^{2} + 11*x + 6

Substitute for L(x) and W(x):

(x + 3)*(x + 2)*H = x^{3} + 6*x^{2} + 11*x + 6

(x^{2} + 5*x + 6)*H = x^{3} + 6*x^{2} + 11*x + 6

H(x) = height of box = [x^{3} + 6*x^{2} + 11*x + 6]/((x^{2} + 5*x + 6) for x ≥ 0

Of course I suppose you wish to simplify this ( ha ha) therefore; finding the rational roots (from rational root theorem) of cubic: H = (x + 1)*(x +2)*(x + 3)/[(x + 2)*(x + 3)] = x + 1 (cancel like terms)

Kenneth S. answered • 03/04/18

Expert Help in Algebra/Trig/(Pre)calculus to Guarantee Success in 2018

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