Solving Polynomial Equations
Factoring is useless in itself. However, in solving polynomial equations it is an
invaluable tool and is crucial for success.
For Example: Which of the following equations is easier to solve?
i)
2 x 3  x 2  72 x  36  0
or
Obviously option ii) is easier to solve
x–6=0
or
x+6=0
Therefore, the roots are 6, -6 and -
ii)
( x  6)( x  6)( 2 x  1)  0
or 2x + 1 = 0
1
1
(or the solutions are x = 6, x = -6 or x = - )
2
2
To solve a polynomial equation, bringing it to factored form is a must.
Ex.
Solve the following equation by factoring 4 x 3 12 x 2  x  3  0
First, let f(x) = 4 x 3 12 x 2  x  3
Test
x
f(x)
** Possible b values are ±1,±3
Possible a values are ±1,±2, ±4
Conclusion
b
values are
a
1
3
1 3
 1,  3,  ,  ,  , 
2
2
4 4
Therefore, Possible
Therefore,
is a factor
Now, use polynomial division to find the other factors.
Now we can factor fully and solve the equation.
What do we do if we can’t fully factor a polynomial equation but still want to solve it?
****REMEMBER – The quadratic formula that will solve ANY quadratic
If the quadratic is of the form ax  bx  c  0 then
2
x
 b  b 2  4ac
2a
Let’s try to solve x 3  9 x 2 13x  5  0
Notice that we cannot factor x 2  8 x  5 any further, but there are still real solutions if we
set x 2  8 x  5 = 0 and solve.
Complex Roots/Solutions: Recall that complex numbers include the square-roots of
negative numbers.
Simply,
i  1
i2  1
Therefore,
 36
  1  36
i 6
 6i
Ex. Solve and include any complex solutions: x 3  2 x 2  45  0