Asked by monica
vaule of x 3x2 minus 30x minus 72 = 0
Answers
Answered by
Anonymous
x^3 + x^2 - 30 x - 72 = 0 maybe, perhaps, possibly?
x^3 + x^2 - 30 x = 72
try x = -2
-8 + 4 + 60 = 56 not 72 so
try x = -1
-1 + 1 + 30 = 30
try x = 0
0 +0 + 0 = 0 , nope
try x = 2
8 + 4 - 60 = -48
headed wrong try x = -3
-27 + 9 + 90 = 72 !!! CARAMBA !!! x = -3 works
so we have one of the three roots
divide
(x^3 + x^2 - 30 x - 72) by (x+3)
wow now we have the quadratic to solve for the other two roots
x^2 -2x -24 = 0
(x-6)(x+4) = 0
x = 6 or x = 4 or x = -3
done
x^3 + x^2 - 30 x = 72
try x = -2
-8 + 4 + 60 = 56 not 72 so
try x = -1
-1 + 1 + 30 = 30
try x = 0
0 +0 + 0 = 0 , nope
try x = 2
8 + 4 - 60 = -48
headed wrong try x = -3
-27 + 9 + 90 = 72 !!! CARAMBA !!! x = -3 works
so we have one of the three roots
divide
(x^3 + x^2 - 30 x - 72) by (x+3)
wow now we have the quadratic to solve for the other two roots
x^2 -2x -24 = 0
(x-6)(x+4) = 0
x = 6 or x = 4 or x = -3
done
Answered by
Anonymous
I am sure you know a better way, but that was fun :)
Answered by
oobleck
x^3 + x^2 - 30 x - 72 = 0
x^2(x+3) - 2x^2 - 30x - 72 = 0
x^2(x+3) - 2x(x+3) - 24x - 72 = 0
x^2(x+3) - 2x(x+3) - 24(x+3) = 0
(x^2-2x-24)(x+3) = 0
(x-6)(x+4)(x+3) = 0
x = 6 or x = -4 or x = -3
x^2(x+3) - 2x^2 - 30x - 72 = 0
x^2(x+3) - 2x(x+3) - 24x - 72 = 0
x^2(x+3) - 2x(x+3) - 24(x+3) = 0
(x^2-2x-24)(x+3) = 0
(x-6)(x+4)(x+3) = 0
x = 6 or x = -4 or x = -3
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