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

Question

Could someone explain this as I'm very confused. Thanks

Answer 1

To find to polynomial from zeros.

1. Find the factors. If it is a positive zero, the factor will be (x- zero) If it is a negative zero if is (x+zero)
2. Multiply the factors together.

Zeros of -4 and 6+i would be:

1. Find the factors: (x+4)(x-6+i)(x-6-i) When you have one complex zero, it has to have a pair. The other one is the complex conjugate, same value different sign, 6+i and 6-i. Both are x- because the 6 is positive
2.  Multiply the factors together: Will start out with FOIL for (x-6+i)(x-6-i) = (x-6)(x-6)-i(x-6)+i(x-6)-i² This needs to be simplified. The 2 middle terms cancel out. x² -12x+36+1 = x² -12x+37

Then multiply by the last factor using the distributive property.

(x+4)(x² -12x+37) x(x²)+x(-12x)+x(37)+4(x²)+4(-12x)+4(37)

Simplify: x³-12x²+37x+4x²-48x+148

Combine like terms:

x³-60x²-11x+148