find a third degree polynomial equation with rational coefficients that has roots -4 and 6+i
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1 Answer

A cubic equation always has three roots.

These three roots are:

1) three real roots

or,

2) 1 real root and two complex roots

If one of the two complex roots is a + ib, then the other complex root is a - ib.

We are given two of the roots as -4 and 6+i.

Since one of the roots is complex and equals 6+i, the the other complex root is 6-i.

Our three roots then are: -4, 6+i, 6-i.

Our three solutions to the cubic equation are: x = -4, x = 6+i, x = 6-i.

Which can be rewritten as: x + 4 = 0, x - (6+i) = 0, x - (6-i) = 0

Multiplying these together gives us the original cubic equation.

(x + 4)(x - (6+i))(x - (6-i)) = 0

Multiplying this out,

(x + 4)(x^2 - (6-i)x - (6+i)x + (6-i)(6+i)) = 0

(x + 4)(x^2 - 12x + 6^2 - i^2) = 0

(x + 4)(x^2 - 12x + 37) = 0

x^3 - 8x^2 - 11x +148 = 0

 

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