Dividing Polynomials

Long DivisionSynthetic DivisionFactor TheoremRemainder Theorem

6.2 Dividing Polynomials

Algorithm – a systematic procedure for performing a computation.

Linear polynomial – largest degree is 1

Synthetic Division is an algorithm for long division.

ExamplesExamplesDividing Polynomial by a Monomial = = 2ab3+4a4b2-3a3

ab

bababa

2

684 43542

Simple division

• (Polynomial)/(Monomial)• Try this one using the method

explained in the previous slide.

• This method is only used for dividing when the divisor is a monomial. How about dividing a polynomial by another polynomial?

?7)72114( 23 xyxyzxyxyz

Answer 132 3 yzz

Dividing PolynomialsDividing Polynomials

Dividing a Polynomial is called long division, and it is the same skills used when you are dividing the monomials

You can use a process similar to long division to divide a polynomial by a polynomial with more than one term. The process is known as the division algorithm

Division Algorithm

• Does Mcdonalds Sell Burgers• Dad Mom Sister Brother• Divide Multiply Subtract Bring-down

Division Algorithm

2296412 2

xxxxxx 24)( 2 94 x24)( x

7

Eliminate the highest exponent by writing the quotient 2x since

xx

x2

2

4 2

Subtract and bring down the next term

Eliminate the highest exponent here by writing the quotient -2 since

22

4

x

x

Subtract and get the remainder 7

Multiply the 2x by the divisor 2x-1

Multiply the -2 by the divisor 2x-1

PRACTICE!

• Find the remainder of using division algorithm.Answer

)52()325( 23 xxxx

105320552 232

xxxxxx

xxx 25105)( 23

32310 2 xx502010)( 2 xx533 x

You should remember writing down in the dividend in order not to overlook the quadratic term. It still exists even though its coefficient is 0 and can’t be seen.

20x

Remainder

Shortcut-Synthetic Division조립제법

• Synthetic division is a shortcut for polynomial division when the divisor is a binomial. In this method, we only use the coefficients of the polynomial that we divide.

Process of synthetic division

• Let the binomial divisor take the form of

Ex) • Write down the coefficients of the

dividend from the highest degree term to the lowest. 1 4 -3 8

• Bring the first coefficient down.

)( ax )3()834( 23 xxxx

• Multiply the first coefficient by and write down the multiple just below the second coefficient. 1*3=3

• Add the multiple and the second coefficient and write the sum below the multiple. 3+4=7

• Next, multiply the sum by and write the multiple below the next(third) coefficient. 7*3=21

• Add the multiple and the next(third) coefficient and write the sum below the multiple.

-3+21=18• Continue the process until the last

column. 18*3=54 , 8+54=62

a

a

Not clear? Then look at this.

3 | 1 4 -3 8 _______________

834 23 xxx

13

72118

5462

3: xDivisor (Dividend)

(Example previously used)

• What do the numbers in the below mean then?

3 l 1 4 -3 8 3 21 54 1 7 18 62

Coefficients of the quotient

Remainder

The power of the first coefficient of the quotient is one less than the degree of the dividend since the divisor is a linear expression.

62:Re

187: 2

mainder

xxQuotient

3

62187: 2

x

xxQuotient

Divide 4x3 + 0x2 + x + 7 by (x – 2)

________________

x – 2 ) 4x3 + 0x2 + x + 7

Use the additive inverse of -2 as the divisor

2| 4 0 1 7 bring down the 4 (1st #)

+ 8 16 34

4 8 17 41 Multiply the number by the 2 (divisor) and add to the next column.

Your answer starts with one degree lower than the problem 4x2 + 8x + 17 + 41

With the last term as remainder x-2

What if the binomial divisor isn’t in the form of (x-a)?

• In this situation, we find the value that makes the binomial divisor 0. We use that value and later multiply the coefficients of the quotient (without the remainder) by a certain number to make the answer true.

• Let’s take an example!)12()6432( 23 xxxx

012

12 Use ½ here since it makes

the divisor 0 when substituted.

• ½ l 2 -3 4 -6 1 -1 3/2 2 -2 3 -9/2

2

9)2

3)(12(

2

9)322)(12(

2

1

64322

9)322)(

2

1(

22

232

xxxxxx

xxxxxx

As a result, the remainder remains the same but the coefficients of the quotient are multiplied by ½ since synthetic division was done by multiplying ½ to the divisor.

2

9:Re

2

3: 2

mainder

xxQuotient

How does it work?

• Then, why is it all right to use this method?

• For further development let’s look at the proof of synthetic division.

• When polynomial P(x) is divided by (x-k) and gets the quotient Q(x) and remainder R, it can be expressed this way. RxQkxxP )()()(

• If we let P(x) a cubic expression, we can generalize it as all polynomials, so we can let P(x) be a cubic expression.

• Since Q(x) is a quadratic expression, it can be expressed this way.

• Then,

dcxbxaxxP 23)(

rqxpxxQ 2)(

krdRkqcrkpbqap

Rkrdkqrckpqbpa

Rkrxkqrxkpqpx

Rrqxpxkxdcxbxax

,,,

,,,

)()(

))((23

223

Synthetic

division

k l a b c d kp kq kr a b+kp c+kq d+kr p q r R

Practice - Divide f(x) by g(x)

How do we do synthetic division when the divisor is a quadratic expression?

Here is one method although there are others.

Divide x4 + 8x3+ 15x2 + 4x + 1 by x2 + 3x + 2, using synthetic division:

Coefficients of dividend

Opposite of b term and opposite of c term

Add the next column

-10

add down and repeat steps

Remainder TheoremFor a polynomial P(x), the function value

P( r) is the remainder when P(x) is divided by x-r.

For P(x) = x3+8x2+8x-32

Find P(1)

P(1) = 1 + 8 + 8 -32 = -15

Therefore is we divide P(x) by x-1, we would get a remainder of -15.

Factor Theorem

• A polynomial f(x) has a factor x-k iff f(k)=0

if and only if (shortened iff) is a biconditional logical connective between statements;

either both statements are true, or both are false

Factor Theorem

For a polynomial P(x), if P(r) = 0, then the polynomial x-r is a factor of P(x).

For example P(x) = x3+8x2+8x-32

P(-4) = 0 and if we divide by x-(-4), x+4, we will get 0 as a remainder (no remainder).

Therefore, x+4 is a factor of x3+8x2+8x-32

Zeros

• x-intercepts are when the function equals 0, the y-value is 0.

• a(x-p)(x-q)=0 means the x-intercepts are at (p,0),(q,0).

• p and q are called zeros of the function.

Find the other zeros of f(x) given that f(-2)=0

Find the missing dimension

x+2

x+4

2x+5