Example - Solving a quartic polynomial
Solve: \(2{x^4} + 9{x^3} - 18{x^2} - 71x - 30 = 0\)
Solution
First, we need to find which number when substituted into the equation will give the answer zero.
\(f(1) = 2{(1)^4} + 9{(1)^3} - 18{(1)^2} - 71(1) - 30 = - 108\)
\(f( - 1) = 2{( - 1)^4} + 9{( - 1)^3} - 18{( - 1)^2} - 71( - 1) - 30 = 16\)
\(f(2) = 2{(2)^4} + 9{(2)^3} - 18{(2)^2} - 71(2) - 30 = - 140\)
\(f( - 2) = 2{( - 2)^4} + 9{( - 2)^3} - 18{( - 2)^2} - 71( - 2) - 30 = 0\)
Therefore \((x + 2)\) is a factor.
\((x + 2)(2{x^3} + 5{x^2} - 28x - 15) = 0\)
Now, we need to do the same thing until the expression is fully factorised.
\(f(3) = 2{(3)^3} + 5{(3)^2} - 28(3) - 15 = 0\)
Therefore \((x - 3)\) is a factor.
Factorise the quadratic until the expression is factorised fully.
\((x + 2)(x - 3)(2{x^2} + 11x + 5) = 0\)
\((x + 2)(x - 3)(x + 5)(2x + 1) = 0\)
\(x + 2 = 0\,so\,x = - 2\)
\(x - 3 = 0\,so\,x = 3\)
\(x + 5 = 0\,so\,x = - 5\)
\(2x + 1 = 0\,so\,x = - \frac{1}{2}\)
