Q.Solve the cubic equations:

For a polynomial $f(x)=a_0x^n+a_1x^{n-1}+\cdots+a_n$ ($a_0\neq0$) and any number $a$, dividing $f(x)$ by the linear polynomial $x-a$ always leaves a constant remainder, and that constant is exactly $f(a)$. This is the Remainder Theorem:

for a unique quotient $q(x)$ of degree $n-1$.

Why it is true. Write $f(x)-f(a) = a_0(x^n-a^n)+a_1(x^{n-1}-a^{n-1})+\cdots+a_{n-1}(x-a)$. Every bracket $x^k-a^k$ factors as $(x-a)(x^{k-1}+x^{k-2}a+\cdots+a^{k-1})$, so it is divisible by $x-a$; hence $f(x)-f(a)$ is divisible by $x-a$, and calling the quotient $q(x)$ gives $f(x)=(x-a)q(x)+f(a)$. Setting $x=a$ on both sides confirms the constant term is indeed $f(a)$. …