We illustrate the method of solving this situation in the next two examples.

**Partly Factored Polynomial**

**Quartic polynomial equations of the form ****( ax + b)(cx
+ d )( px + q)(rx + s)+ 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

(*x*^{2} - 5*x *+ 6)(*x*^{2} - 5*x *-14) +19 = 0 .

If we take *x*^{2} - 5*x *as *y *, then the equation
becomes ( *y *+ 6)( *y *-14) +19 = 0;

that is,

*y*^{2} - 8 *y *â€“ 65 = 0 .

Solving this we get solutions *y *= 13 and *y *= -5 .
Substituting this we get two quadratic equations

*x*^{2} - 5*x *-13 = 0 and *x*^{2} - 5*x *+ 5 = 0

which can be solved by usual techniques. The solutions obtained for these two equations together give solutions as

Solve the equation (2*x *- 3)(6*x *-1)(3*x *- 2)(*x
*-12) â€“ 7 = 0 .

The given equation is same as

(2*x *- 3)(3*x *- 2)(6*x *-1)(*x *-12) â€“ 7
= 0 .

After a computation, the above equation becomes

(6*x*^{2} -13*x *+ 6)(6*x*^{2} -13*x *+12) â€“ 7 = 0 .

By taking *y *= 6*x*^{2} -13*x*, the
above equation becomes,

( *y *+ 6)( *y *+12) - 7 = 0

which is same as

*y*^{2} +18 *y *+ 65 = 0 .

Solving this equation, we get *y *= -13 and *y *= -5 .

Substituting the values of *y *in *y *= 6*x*2 -13*x*,
we get

6*x*^{2} -13*x *+ 5 = 0

6*x*^{2} -13*x *+13 = 0

Solving these two equations, we get

as the roots of the given equation.

Tags : Solved Example Problems , 12th Mathematics : UNIT 3 : Theory of Equations

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12th Mathematics : UNIT 3 : Theory of Equations : Partly Factored Polynomial | Solved Example Problems

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