SOLVING POLYNOMIAL
EQUATIONS
Objective: To recall how to solve polynomial equations by factoring.
Recall 1: If a polynomial
factors, then it is
divisible by any one of its
factors.
For example, the polynomial x^4 + 13x^2 +
36 factors as
(x - 2)(x + 2)(x - 3)(x + 3).
This
means that any one of the factors divides
x^4 + 13x^2 + 36 without
remainder.
Indeed, if you
divide x^4 + 13x^2 +
36 by (x - 2) you will find the result to
be x^3 + 2x^2 - 9x - 18,
exactly.
Note 1
Finding the factors of a polynomial equation is not
always
easy. The polynomials encountered in algebra textbooks are
generally
easier to factor than arbitrary ones . The next recall
suggests some
tricks for finding the factors.
Recall 2: Consider the polynomial
2x^3 - 3x^2 - 5x + 6.
By letting x = 1, we find that the polynomial becomes 0; that is:
2(1)^3 - 3(1)^2 - 5(1) + 6 = 0.
This means that (x - 1) is a factor of the polynomial. We can test this by performing long division.
2x^2
-x - 6
x-1)
2x^3 - 3x^2
- 5x + 6
2x^3
- 2x^2
-x^2
- 5x
-x^2
+ x
-6x
+ 6
-6x
+ 6
0
From this, we see that
2x^3 - 3x^2 - 5x + 6 = (x - 1)(2x^2 - x - 6).
The quadratic at the right side of this last equation factors as
2x^2 - x - 6 = (x - 2)(2x + 3),
which means that the original polynomial factors as
2x^3 - 3x^2 - 5x + 6 = (x - 1)(x - 2)(2x + 3)
Recall 3: (Solving polynomial equations by factoring.)
By what was said in Recall 1, the polynomial equation
2x^3 - 3x^2 - 5x + 6 = 0
may be rewritten as
(x - 1)(x - 2)(2x + 3) = 0.
Whenever a polynomial is fully factored into a product of degree one polynomials, the solutions to the polynomial equation can be read from the factors. This comes from the fact that if a product equals zero, then each factor of the product must also be zero. For the case of the polynomial above, it means that
x - 1 = 0, x - 2 = 0 and 2x + 3 = 0.
We therefore read the solutions from these last equations as:
x = 1, x = 2 and x = -2/3 .
Examples
Example 1. Find the solutions to the equation
x^3 - 3x^2 - 10x + 24 = 0.
Solution: Try the factors of 24. They are
1, - 1, 2, -2, 3, -3, 4, -4, 6, -6, 8, -8, 12, -12, 24, -24
x = 1 or -1 is not a solution; but x = 2 is a solution, since
(2)^3 - 3(2)^2 - 10(2) + 24 = 0.
If x = 2
is a solution, then x - 2 is a factor of
the
polynomial
x^3 - 3x^2 - 10x + 24
and so we may use long division to find the other factors.
x^2
- x - 12
Our polynomial factors as x^3 - 3x^2 - 10x +
24
= (x - 2)(x^2 - x - 12).
x^3 - 3x^2 - 10x
+ 24 = (x - 2)(x - 4)(x + 3).
We may now read the
solutions from the factors above;
they are:
x = 2,
x = 4 and x = -3.
Example
2. Find all solutions
to the
equation
6x^3 + 14x + 17x^2 + 3 =
0.
Solution: First, rearrange the
terms so that
they are in descending order of the powers of x. That
means rewrite
the equation as
6x^3 + 17x^2 + 14x +
3 = 0.
Next, test the quotients of the factors of
6 and 3; for
example, test
1, -1, 3, -3, 1/3,
-1/3, 3/6, -3/6, 3/2, -3/2 etc.
In this
case, we will not have to go far with the testing,
because x = -1 is a
solution. We see that
6(-1)^3 + 17(-1)^2 +
14(-1) + 3. = 0.
We use long division to find the
other factors.
6x^2
+ 11x + 3
This tells us that
the polynomial 6x^3 + 17x^2 + 14x +
3 factors as
(x + 1)(6x^2 + 11x + 3).
The quadratic
expression on the right in-turn factors
as
(2x +
3)(3x + 1),
which means that the polynomial
equation becomes
(x + 1)(2x + 3)(3x + 1) =
0.
We may read the solutions from the factors
above; they
are:
x
= -1, x = -2/3 and x = -1/3.
Note
1, -1, 3, -3, 1/3, -1/3, 3/6,
-3/6, 3/2, -3/2
etc.
Terminology: A value for a variable
that makes a polynomial zero is
called a root or
zero of the polynomial.
Note
Example 3. Find
all solutions to the equation
x^3 - 2x^2 + 1 =
0.
Solution:
Here again, x = 1 is a solution; so,
we may use long division to find the
other factors of the polynomial x^3
- 2x^2 + 1.
We
use long division to find the other factors.
x2
-
x
-
1
This tells us that the polynomial x^3 - 2x^2 + 1 factors
as (x -
1)(x^2 - x - 1).
The
solutions to this last equation will also be the remaining
solutions to
the original equation. From the quadratic formula,
we
have
x =(1 + sqrt(5))/2 or (1 -
sqrt(5))/2
as the solutions of the
quadratic. Therefore, the
solutions for the original equation
are
Note
EXERCISES
In exercises 1 through 11,
find the real solutions to the
polynomial equations.
1. x^2 - x - 2=0
2. x^4 + 2x^3 + x^2=0
3. x^3 - x^2 - x +
1=0
4. 3x^3 +
5x^2 + 2x=0
5.
16x^4 - 1=0
6.
x^5 - 2x^4 - x^3 + 2x^2 + x
- 2=0
7. x^4 - x^3 + 5x^2 -
5x=0
8. 8x^4 +
10x^2 - 3=0
9. x^4 -
2x^3 + 2x^2 - 2x + 1=0
10. x^4 + 13x^2 +
36=0
11. 4x^4 -
25x^2 + 36=0
x - 2) x^3 - 3x^2 - 10x + 24
x^3
- 2x^2
-x^2
- 10x
-x^2
+ 2x
-12x
+ 24
-12x
+ 24
0
Next, notice that the
quadratic on the right side of
the above equation factors as x^2 - x - 12
= (x - 4)(x + 3); so,
x + 1) 6x^3 + 17x^2 +
14x + 3
6x^3
+ 6x^2
11x^2
+ 14x
11x^2
+ 11x
3x
+ 3
3x
+ 3
0
All
the solutions are among the quotients of factors
that we listed at the
beginning,
Not every polynomial can
be factored as easily
as the ones that we have just encountered. The
Fundamental Theorem
of Algebra says that any polynomial can be factored
into a product of first
degree or second degree polynomials. The
second degree polynomials
may not have real numbers for roots. For
example x^2 + x + 1 does
not have a real number as a root. Rest
assured that we shall not
encounter such a case in this
Workout.
x - 1) x^3 - 2x^2 + 0x +
1
x^3
- x^2
-x^2
+ 0x
-x^2
+ 1x
-x
+ 1
-x
+ 1
0
Our task is now to factor the
quadratic polynomial
x^2 - x - 1. But this has no obvious
factors. We must use the
quadratic formula to get the solutions
to
x^2 - x - 1. = 0.
x =(1 + sqrt(5))/2 or (1 -
sqrt(5))/2 and
x = 1.
The real solutions to a
polynomial equation is
the set of real number values that satisfy the
equation. For example,
the set of real solutions to the equation x^3
- 8 = 0 is x = 2, because
the factors of x^3 - 8 are x - 2 and x^2 + 2x +
4. Note that the
quadratic factor does not have any real roots,
according to the quadratic
formula.