Polynomial Equation
\[\sum _{i=0}^na_{n-i}x^{n-i}=0\]
if \(a_0.a_n\ne0\)and \(a_0=ef\)
\[\sum _{i=0}^na_{n-i}e^{n-i}=0\]
\[\sum _{i=0}^{n-1}a_{n-i}e^{n-i}+ef=0\]
\[\sum _{i=0}^{n-1}a_{n-i}e^{n-i-1}+f=0\]
\[\{\sum _{i=0}^na_{n-i}x^{n-i}\}:(x-e)=a_nx^{n-1}+(a_{n-1}+ea_n)x^{n-2}+[a_{n-2}+e(a_{n-1}+ea_n)x^{n-3}+...\]
\[\{\sum _{i=0}^na_{n-i}x^{n-i}\}:(x-e)=\sum_{i=0}^n b_ix^{n-i-1}\]
\[b_i=\sum _{j=0}^ie^{i-j}a_{n-j}\]
\[\{\sum _{i=0}^na_{n-i}x^{n-i}\}:(x-e)=\sum_{i=0}^n \{\sum _{j=0}^ie^{i-j}a_{n-j}\}x^{n-i-1}\]
\[\sum _{i=0}^na_{n-i}x^{n-i}=(x-e)\sum_{i=0}^{n-1} \{\sum _{j=0}^ie^{i-j}a_{n-j}\}x^{n-i-1}\]
\[\sum _{i=0}^na_{n-i}x^{n-i}=(x-e_1)(x-e_2)\sum_{i=0}^n \{\sum _{j=0}^ie^{i-j}a_{n-j}\}x^{n-i-1}\]
First Degree Equation:
Quadratic Equation (Second Degree Equation):
\[ax^2+bx+c=0\]
\[x^2+\frac bax+\frac ca=0\]
\[x^2+2\frac b{2a}x+\frac ca+(\frac b{2a})^2-(\frac b{2a})^2=0\]
\[x^2+2\frac b{2a}x+(\frac b{2a})^2-[(\frac b{2a})^2-\frac ca]=0\]
\[(x+\frac b{2a})^2-[\frac {b^2-4ac}{(2a)^2}]=0\]
If \(b^2-4ac=\Delta\)so:
If \(\Delta >0\):
\[(x+\frac b{2a})^2-[\frac {\sqrt\Delta}{2a}]^2=0\]
\[A^2-B^2=(A+B)(A-B)\]
\[(x+\frac b{2a}+\frac {\sqrt\Delta}{2a})(x+\frac b{2a}-\frac {\sqrt\Delta}{2a})=0\]
\[(x+\frac {b+\sqrt\Delta}{2a})(x+\frac {b-\sqrt\Delta}{2a})=0\]
Either:
\[x+\frac {b+\sqrt\Delta}{2a}=0\]
\[x=\frac {-b-\sqrt\Delta}{2a}\]
Or:
\[x+\frac {b-\sqrt\Delta}{2a}=0\]
\[x=\frac {-b+\sqrt\Delta}{2a}\]
If \(\Delta =0\):
\[x=\frac {-b}{2a}\]
If \(\Delta <0\):
The equation does not have solutions in \(R\), but has to solutions in \(C\).
\(\sqrt \Delta =i\sqrt|\Delta|\); \(i = \sqrt{-1}\)
\[x=\frac {-b-i\sqrt|\Delta|}{2a}\]
\[x=\frac {-b+i\sqrt|\Delta|}{2a}\]
Cubic Equation (Third Degree Equation):
\[ax^3+bx^2+cx+d=0\]
if d is fctorizable \(d=ef\), and e is a solution:
\[ae^3+be^2+ce+ef=0\]
\[ae^3+be^2+e(c+f)=0\]
\(ae^2+be+(c+f)=0\) (I)
this is a quadratic equation whic determained by:
\[\Delta = b^2-4a(c+f)\]
\(e=\frac{-b\pm\sqrt {b^2-4a(c+f)}}{2a}\)’
e will have real values if and only if:
\[\Delta = b^2-4a(c+f)\ge0\]
\[(ax^3+bx^2+cx+d):(x-e)=ax^2+(b+ea)x+[c+e(b+ea)]+\frac {ef+e[c+e(b+ea)]}{x-e}\]
\[ef+e[c+e(b+ea)]=0\]
\(f+c+e(b+ea)=0\) which is the same term in (I)
\[(x-e)\{ax^2+(b+ea)x+[c+e(b+ea)]\}=0\]
the othe solutions:
\[ax^2+(b+ea)x+[c+e(b+ea)]=0\]
\[\Delta =(b+ea)^2-4a[c+e(b+ea)]\]
For the equation to have three real different solutions:
\[\Delta =(b+ea)^2-4a[c+e(b+ea)]\ge0\]