Partly Factored Polynomial
Quartic polynomial equations of the form (ax + b)(cx
+ d )( px + q)(rx + s)+ k = 0 , k ≠0 which can be rewritten
in the form (αx2 + β x +λ) (αx2 + β x + µ)+
k = 0
We illustrate the method of solving this situation in the next
two examples.
Solve the equation
(x - 2)(x - 7)(x - 3)(x + 2) +19 =
0.
We can solve this fourth degree equation by rewriting it
suitably and adopting a technique of substitution. Rewriting the equation as
(x - 2)(x - 3)(x - 7)(x + 2) +19 = 0
.
the given equation becomes
(x2 - 5x + 6)(x2 - 5x -14) +19 = 0 .
If we take x2 - 5x as y , then the equation
becomes ( y + 6)( y -14) +19 = 0;
that is,
y2 - 8 y – 65 = 0 .
Solving this we get solutions y = 13 and y = -5 .
Substituting this we get two quadratic equations
x2 - 5x -13 = 0 and x2 - 5x + 5 = 0
which can be solved by usual techniques. The solutions obtained for these two equations together give solutions as
Solve the equation (2x - 3)(6x -1)(3x - 2)(x
-12) – 7 = 0 .
The given equation is same as
(2x - 3)(3x - 2)(6x -1)(x -12) – 7
= 0 .
After a computation, the above equation becomes
(6x2 -13x + 6)(6x2 -13x +12) – 7 = 0 .
By taking y = 6x2 -13x, the
above equation becomes,
( y + 6)( y +12) - 7 = 0
which is same as
y2 +18 y + 65 = 0 .
Solving this equation, we get y = -13 and y = -5 .
Substituting the values of y in y = 6x2 -13x,
we get
6x2 -13x + 5 = 0
6x2 -13x +13 = 0
Solving these two equations, we get
as the roots of the given equation.
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