Asked by Dee
A slice 1 unit thickness is removed from one see of a cube. Use the rational zeroes theorem and synthetic division to find the original dimensions of the cube, if the remaining volume is 48 cm^3
v= x^3
v=x*x*x
x=(x-1)*x*x
48=(x-1)*x*x
v= x^3
v=x*x*x
x=(x-1)*x*x
48=(x-1)*x*x
Answers
Answered by
Reiny
carry on ...
x^2(x-1) = 48
x^3 - x^2 - 48 = 0
some quick mental gymnastic will show that
x = 4 works. (64-16 = 48)
so it factors into
(x-4)(x^2 +3x+12) = 0
some trivial calculations will show the remaining quadratic has no real roots, so
x = 4
the original was 4 by 4 by 4
x^2(x-1) = 48
x^3 - x^2 - 48 = 0
some quick mental gymnastic will show that
x = 4 works. (64-16 = 48)
so it factors into
(x-4)(x^2 +3x+12) = 0
some trivial calculations will show the remaining quadratic has no real roots, so
x = 4
the original was 4 by 4 by 4
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