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

*************

Factors in Z

Recall that the factors of an integer n are all of the integers k such that

= for some third integer .

n mk

m

Examples.

? 12 = 3 ? 4, so 4 is a factor of 12.

? 30 = 2 ? 15, so 15 is a factor of 30.

? 1, 1, and are all factors of an integer . That's because

n

n

n

n = n ? 1 and n = ( n)( 1).

Important special case. If 1, 2, . . . n 2 Z, then each of these numbers are factors of the product ? ? ? . For example, 2, 10, and 7 are each

12 n

factors of 2 ? 10 ? 7 = 140.

Check factors of degree 0 coe cient when searching for roots

If , , and are all integers, then the polynomial

k 1

2

( )= (

)(

)= 2 ( + ) +

q x k x 1 x 2 kx k 1 2 x k12

has and as roots, and each of these roots are factors of the degree 0

1 2

coe cient of ( ). (The degree 0 coe cient is

.)

qx

k12

More generally, if k, 1, 2, . . . , n 2 Z, then the degree 0 coe cient of the polynomial

( )= (

)(

)???(

)

g x k x 1 x 2 x n

equals

? ? ? . That means that each of the roots of ( ) ? which are

k12 n

gx

the ? are factors of the degree 0 coe cient of ( ).

i

gx

Now it's not true that every polynomial has integer roots, but many of the

polynomials you will come across do, so the two paragraphs above oer a

powerful hint as to what the roots of a polynomial might be.

162

When searching for roots of a polynomial whose coe cients are all integers,

check the factors of the degree 0 coe cient.

Example. 3 and 7 are both roots of 2(x 3)(x + 7).

Notice that 2( 3)( + 7) = 2 2 + 8 42, and that 3 and 7 are both

xx

xx

factors of 42.

Example. Suppose ( ) = 3 4 + 3 3 3 2 + 3 6. This is a degree 4 px x x x x

polynomial, so it will have at most 4 roots. There isn't a really easy way to

find the roots of a degree 4 polynomial, so to find the roots of ( ), we have px

to start by guessing.

The degree 0 coe cient of ( ) is 6, so a good place to check for roots is px

in the factors of 6.

The factors of 6 are 1, 1, 2, 2, 3, 3, 6, and 6, so we have eight

quick candidates for what the roots of ( ) might be. A quick check shows

px

that of these eight candidates, exactly two are roots of p(x) ? namely 1 and

2. That is to say, (1) = 0 and ( 2) = 0.

p

p

*************

Factoring cubics

It follows from the Fundamental Theorem of Algebra that a cubic polynomial is either the product of a constant and three linear polynomials, or else it is the product of a constant, one linear polynomial, and one quadratic polynomial that has no roots.

In either case, any cubic polynomial is guaranteed to have a linear factor, and thus is guaranteed to have a root. You're going to have to guess what that root is by looking at the factors of the degree 0 coe cient. (There is a "cubic formula" that like the quadratic formula will tell you the roots of a cubic, but the formula is di cult to remember, and you'd need to know about complex numbers to be able to use it.)

Once you've found a root, factor out the linear factor that the root gives you. You will now be able to write the cubic as a product of a monic linear

163

factors of --3 are 1, --1, 3, and --3. Check these factors to see if any of them

are roots.

After checking, you'll see that 1 is a root. That means that x --1 is a factor

of 2x3 -- 3x2 + 4x -- 3. Therefore, we can divide 2x3 -- 3x2 + 4x -- 3 by x -- 1

ptoolygneot mainaoltahnedr paoqlyunadormaitaicl polynomial. Completely factor the quadratic and tpthhoeenlynnyyooomuuiwawlilialllnhhdaavaveeqcucoaomdm2prxpal3elteit--cetelpyl3yoxflafy2acn+ctoto4omrxreei--dadl.t3thhCe=eoc2mcuxubpb2ilciec-- .t.elxy+fa3ctor the quadratic and

Problem. Completely factor 2 3 3 2 + 4 3.

PThrousb,lem.

Completely

factor

xx 2x3 -- 3~2

x + 4x -- 3.

Solution. Start 2byx3gu--es3sixn2g+a4rxo-- ot.3T=h(xe --deg1r)e(e20x2co--e xc+ie3n)t is 3, and the

fSacotolurstioonf . 3Staarret 1b,y g1u,e3s,sianngda ro3.otC. hTehcke tdheegsreeefa0ctcoorseftfiociseenet iifsa--ny3,oaf ntdhetmhe

afraectroorostso.f --3 are 1, --1, 3, and --3. Check these factors to see if any of them

arAeftroerotcsh.ecking, you'll see that 1 is a root. That means that 1 is a factor

x

of A2 ft3er c3he2c+kin4g, yo3u.'lTl sheeerethfoarte,1wise acaronodt.ivTidhea2t m3ean3s 2th+at4x --13 ibsya fact1or

xxx

xxx

x

toof g2ext3 a-- no3txh2e+r p4oxly-- no3m. Tiahlerefore, we can divide 2x3 -- 3x2 + 4x -- 3 by x -- 1

to get another polyno2m3ial 3 2 + 4 3 The discriminant of2x2xx32---- x3xxx2+1+34xxe--qu3al=s=(22_xx122)2--x--x+4+(323)(3) = 1 -- 24 = --23,

Tahnues,gative number. Therefore, 2x2 -- x + 3 has no roots, so to completely

Tfahcutos,r 2x2 -- x +32 w3 e ju3st2 h+av4e to 3fac=tor( out 1th)e(2le2ading+co3e)fficient as follows: 2x2--x-1-3=2(x2x2x--3--~x3x+x2~+).x4x--3=(xx--1)(2xx2--xx+3)

2x3-3x1 ~~+z-3

/\

(x-l) (zx2_x+3')

/N

a The discriminant of 2 2 + 3 equals ( 1)2 4(2)(3) = 1 24 = 23, xx a nTehgeatdivisecrniumminbaenrt. oTf h2exr2ef-- orex, +2 32 equal+s 3(_h1a)2s n--o4r(o2o)(t3s), s=o t1o--co2m4 p=let--e2ly3,

xx

faacnteogr a2tiv2e nu+mb3ewr.e jTuhsterheafoveret,o2fxa2ct-- orxou+t 3thhealseandoinrgoocotse, scoietnot acsomfopllloewtesl:y

xx

2faxcT2thoerx2fix+n2a3-- l =axn+s2w3xe2rweis

ju1 st+ha3ve. x

to

factor

out

the

leading

coefficient

as

follows:

~+~) 2x2--x-1-3=2(x2-- 2~x+~22 )(.x_1)(x2_

2x3-3x1 1~37~+z-3

/\

(x-l) (zx2_x+3')

/N

a

The final answer is The final answer is

1 3

~+~) 2( 1) 2

+

2(xx_1)(xx1264_ 2x 2

137

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