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# Help. ​

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Help.

NotSoSmart  Dec 7, 2017
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#1
+79741
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(x - 1) (x -2) (x - 3) (x + 3)  =

[ x^2 - 3x + 2] [ x^2 - 9 ] =

x^4  -  3x^3 + 2x^2

+                 - 9x^2 +  27x - 18

________________________

x^4   -  3x^3  -  7x^2 + 27x - 18

4x^3  -  20x^2  +  24x   =  0       divide through by  4

x^3  -  5x^2  + 6x  = 0      factor

x (x^2 - 5x + 6)  = 0

x( (x - 3) (x -2)  = 0

The zeroes  are    0, 2 , 3

x^3  -  216  =  0        factor as a difference of cubes

(x - 6) (x^2  + 6x  +  36)

6 is a root

And using the quadratic formula  the other  roots will be complex

So we will have

[ -6 ±  √ [ 36 - 144] ] / 2

[ -6 ±  √ [  - 108] ] / 2

[ -6 ±  √ [  - 36 * 3 ] ] / 2

[ -6 ±  6i√ [  3] ] / 2

-3  + 3i√ 3       and  -3  - 3i√ 3

CPhill  Dec 7, 2017
edited by CPhill  Dec 7, 2017
#2
+1146
0

which choices are those? It don't have what you have for each one you solved..?

3.

4.

5.

NotSoSmart  Dec 8, 2017
#3
+79741
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CPhill  Dec 8, 2017
#4
+1146
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Ah ok thanks!!

NotSoSmart  Dec 8, 2017

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