Q.Form the monic cubic equation with rational coefficients having $3-\sqrt2$ and $4$ among its roots.

Step 1. Since the equation is to have rational coefficients and $3-\sqrt2$ is to be a root, its conjugate $3+\sqrt2$ must also be a root (Irrational Conjugate Roots theorem) -- it cannot be given alone.

Step 2. The conjugate pair contributes the rational quadratic factor

Step 3. The required cubic is $(x^2-6x+7)(x-4)=0$.

Step 4. Expand:

$(x^2-6x+7)(x-4)=x^3-6x^2+7x-4x^2+24x-28=x^3-10x^2+31x-28.$ …

Forming an equation from given roots, using the irrational-conjugate theorem to introduce the req …

• Writing the cubic with $3-\sqrt2$ and $4$ only, omitting the forced conjugate $3+\sqrt2$, which would leave irrational coeff …