Factoring Polynomials - Math
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? 5 + 2 4 7 3 + 14 2 10 + 20 can be factored into a product xxx x x
of a number, 1, a monic linear polynomial, 2, and two monic x
quadratic polynomials that don't have roots, 2 + 2 and 2 + 5. That
x
x
is 5 + 2 4 7 3 + 14 2 10 + 20 = ( 2)( 2 + 2)( 2 + 5).
xxx x x
xx x
(We can check the discriminants of 2 + 2 and 2 + 5 to see that
x
x
these two quadratics don't have roots.)
? 2 4 2 3 + 14 2 6 + 24 = 2( 2 + 3)( 2 + 4). Again, 2 + 3
xx xx
x xx
x
and 2 + 4 do not have roots. xx
Notice that in each of the above examples, the real number that appears in the product of polynomials ? 4 in the first example, 1 in the second, and 2 in the third ? is the same as the leading coe cient for the original polynomial. This always happens, so the Fundamental Theorem of Algebra can be more precisely stated as follows:
If ( ) = n +
n 1+???+ ,
p x anx an 1x
a0
then ( ) is the product of the real number ,
px
an
and a collection of monic quadratic polynomials that
do not have roots, and of monic linear polynomials.
Completely factored
A polynomial is
if it is written as a product of a real
completely factored
number (which will be the same number as the leading coe cient of the
polynomial), and a collection of monic quadratic polynomials that do not
have roots, and of monic linear polynomials.
Looking at the examples above, 4( 1)( 2) and ( 2)( 2 + 2)( 2 + 5)
xx
xx x
and 2( 2 + 3)( 2 + 4) are completely factored. x xx
One reason it's nice to completely factor a polynomial is because if you do, then it's easy to read o what the roots of the polynomial are.
Example. Suppose ( ) = 2 5 + 10 4 + 2 3 38 2 + 4 48. Written
px
x xx xx
in this form, its di cult to see what the roots of ( ) are. But after being px
completely factored, ( ) = 2( + 2)( 3)( 4)( 2 + 1). The roots of
px
xxxx
158
this polynomial can be read from the monic linear factors. They are 2, 3,
and 4.
(Notice that ( ) = 2( + 2)( 3)( 4)( 2 + 1) is completely factored
px
xxxx
because 2 + 1 has no roots.) x
*************
Factoring linears
To completely factor a linear polynomial, just factor out its leading coe cient:
+ = +b ax b a x
a
For example, to completely factor 2 + 6, write it as the product 2( + 3).
x
x
Factoring quadratics
What a completely factored quadratic polynomial looks like will depend on how many roots it has.
0 Roots. If the quadratic polynomial 2 + + has 0 roots, then it can ax bx c
be completely factored by factoring out the leading coe cient:
2+ + = 2+ b + c ax bx c a x x
aa
(The graphs of 2 + + and 2 + b + c dier by a vertical stretch or shrink ax bx c x x
that depends on . A vertical stretcah ora shrink of a graph won't change the a
number of -intercepts, so 2 + b + c won't have any roots since 2 + +
x
xx
ax bx c
aa
doesn't have any roots. Thus, 2 + b + c is completely factored.) xx
aa
Example. The discriminant of 4 2 2 +2 equals ( 2)2 4(4)(2) = 4 32 = xx
28, a negative number. Therefore, 4 2 2 + 2 has no roots, and it is xx
completely factored as 4( 2 1 + 1). xx
22
2 Roots. If the quadratic polynomial 2 + + has 2 roots, we can ax bx c
name them and . Roots give linear factors, so we know that (
)
1 2
x 1
159
and (
) are factors of 2 + + . That means that there is some
x 2
ax bx c
polynomial ( ) such that
qx
2 + + = ( )(
)(
)
ax bx c q x x 1 x 2
The degree of 2 + + equals 2. Because the sum of the degrees of the ax bx c
factors equals the degree of the product, we know that the degree of ( ) plus qx
the degree of (
) plus the degree of (
) equals 2. In other words,
x 1
x 2
the degree of ( ) plus 1 plus 1 equals 2. qx
Zero is the only number that you can add to 1 + 1 to get 2, so ( ) must qx
have degree 0, which means that ( ) is just a constant number. qx
Because the leading term of 2 + + ? namely 2 ? is the product of
ax bx c
ax
the leading terms of ( ), (
), and (
) ? namely the number ( ),
q x x 1
x 2
qx
, and ? it must be that ( ) = . Therefore,
xx
qx a
2+ + = (
)(
)
ax bx c a x 1 x 2
Example. The discriminant of 2 2 + 4 2 equals 42 4(2)( 2) = 16 + 16 = xx
32, a positive number, so there are two roots.
We can use the quadratic formula to find the two roopts, but pbefore we do, itp's best to simplify the square root of the discriminant: 32 = (4)(4)(2) = 4 2.
Now we use the quadratic formula to find that the roots are
p
4+4 2(2)
2=
p
4+4 4
2=
p 1+ 2
and
p
44 2(2)
2=
p
4
4 4
2=
1
p 2
Therefore, 2 2 + 4 2 is completely factored as
x
x p
p
p
p
2 ( 1 + 2)
( 1 2) = 2( + 1 2)( + 1 + 2)
x
x
x
x
1 Root. If 2 + + has exactly 1 root (let's call it ) then (
ax bx c
1
x
a factor of 2 + + . Hence,
ax bx c
2 + + = ( )(
)
ax bx c g x x 1
for some polynomial ( ). gx
160
) is 1
Because the degree of a product is the sum of the degrees of the factors,
g(x) must be a degree 1 polynomial, and it can be completely factored into
something of the form ( x
) where , 2 R. Therefore,
2 + + = ( )(
)
ax bx c x x 1
Notice that is a root of ( )(
), so is a root of 2 + +
x x 1
ax bx c
since they are the same polynomial. But we know that 2 + + has only ax bx c
one root, namely , so must equal . That means that
1
1
2+ + = (
)(
)
ax bx c x 1 x 1
The leading term of 2+ + is 2. The leading term of ( )( )
ax bx c ax
x 1 x 1
is 2. Since 2 + + equals (
)(
), they must have the same
x
ax bx c
x 1 x 1
leading term. Therefore, 2 = 2. Hence, = .
ax x
a
Replace with a in the equation above, and we are left with
2+ + = (
)(
)
ax bx c a x 1 x 1
Example. The discriminant of 3 2 6 +3 equals ( 6)2 4(3)(3) = 36 36 = xx
0, so there is exactly one root. We find the root using the quadratic formula:
p
( 6) + 2(3)
0
=
6 6
=
1
Therefore, 3 2 6 + 3 is completely factored as 3( 1)( 1).
xx
xx
Summary. The following chart summarizes the discussion above.
roots of 2 + + ax bx c
no roots
completely factored form of 2 + + ax bx c
( 2 + b + c) ax x
aa
2 roots: and 1 2
(
)(
)
a x 1 x 2
1 root: 1
(
)(
)
a x 1 x 1
161
................
................
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