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If x^3-2x^2-5x+6 has x = 1 as one root find the other roots.
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`x^3-2x^2-5x+6 = 0`
`x^3-x^2-x^2-5x+6 = 0`
`x^2(x-1)-(x^2+5x-6) = 0`
`x^2(x-1)-(x^2+6x-x-6) = 0`
`x^2(x-1)-(x-1)(x+6) = 0`
`(x-1)(x^2-x-6) = 0`
`(x-1)(x^2-3x+2x+6) = 0`
`(x-1)(x(x-3)+2(x-3)) = 0`
`(x-1)(x-3)(x+2) = 0`
So the roots are x = 1, x = 3 and x = -2.
The other two roots is x = 3 and x = -2.
Posted by jeew-m on June 23, 2013 at 2:20 AM (Answer #1)
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