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The polynomial p(x) = x^3 - x + 6. p(x)=0 has one root equal to -2. Show that the...

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jamiebullpresco | Student, Grade 10 | eNotes Newbie

Posted February 24, 2012 at 12:46 AM via web

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The polynomial p(x) = x^3 - x + 6. p(x)=0 has one root equal to -2. Show that the equation has no other real roots.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted February 24, 2012 at 1:06 AM (Answer #1)

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The polynomial p(x) = x^3 - x + 6 = 0 has one root equal to -2

x^3 - x + 6 = 0

=> x^3 + 2x^2 - 2x^2 - 4x + 3x + 6 = 0

=> x^2(x + 2) - 2x(x + 2) + 3(x + 2) = 0

=> (x^2 - 2x + 3)(x + 2) = 0

It can be seen that in the quadratic equation x^2 - 2x + 3, 2^2 - 4*1*3 = 4 - 12 = -8. As b^2 - 4ac < 0 the quadratic equation only has complex roots.

This shows that p(x) = 0 has only one real root x = -2

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