Question

find a cubic equations with integer coefficients whose roots are 2+√-2 and 1​

Answer 1

Answer:

We have a complex roots for this equation and complex roots ALWAYS form a conjugate therefore we want an equation with roots;

3,1+2i,1−2i

From the factor theorem which states that;

If x=a where a is a real number is a root of f(x) for some function f then (x−a) is a factor of f(x) .

We already know it's a cubic equation because of three roots and since we have a real root x=3 we can write our equation in the form;

(x−3)(ax2+bx+c)=0

Now the quadratic equation in the second brackets have the roots; 1+2i and 1−2i

So let's try forming an equation with this roots;

Let α and β be the solutions of our quadratic equation where;

α=1+2i

β=1−2i

α+β=2

αβ=5

So and as we know;

α+β=−ba and αβ=ca our quadratic equation becomes;

x2−2x+5

Now we have our full equation;

(x−3)(x2−2x+5)=0

Expand;

x3−2x2+5x−3x2+6x−15=0

x3−5x2+11x−15=0

I know this method is long but it's a good practice because we also form a quadratic equation.

Hope I was helpful :)