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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
