polynomial division primer · polynomial division primer john banks and liz bailey i we can divide...
TRANSCRIPT
Polynomial Division Primer
John Banks and Liz Bailey
I We can divide polynomials using the same techniques of longdivision as we use with natural numbers – but we getpolynomials as our quotient and remainder.
I We first explain how this works before showing the quicktabular way to do the calculations.
I This slide presentation is best viewed on screen in full screenmode.
I Just view one slide at a time and try to understand what ishappening before moving on.
1 / 63
What are we trying to do?
I In the following examples, we divide the polynomial p(x) bythe polynomial d(x), obtaining an answer in the form
p(x)
d(x)= q(x) +
r(x)
d(x).
where r(x) has (strictly) smaller degree than d(x).
I This is achieved by repeatedly subtracting appropriatemultiples of d(x).
I We choose these multiples so that the “leading” term isremoved from p(x) by each subtraction.
I We can do this most efficiently in a tabular format, but firstlets see why it works.
2 / 63
First Example: Step 1
p(x) = x4 + 3x3 − 3x + 5, d(x) = x + 2.
I To remove the leading term x4 of p(x), subtract x3d(x) . . .
p(x)− x3d(x) = (x4 + 3x3 − 3x + 5)− (x4 + 2x3)
= x3 − 3x + 5
⇒ p(x)− x3d(x)
d(x)=
x3 − 3x + 5
d(x)
⇒ p(x)
d(x)= x3 +
x3 − 3x + 5
d(x)
sop(x)
d(x)is x3 plus a fraction
p1(x)
d(x)with numerator
p1(x) = x3 − 3x + 5 of smaller degree than p(x).
I We can do exactly the same thing to this new fraction!
3 / 63
First Example: Step 1
p(x) = x4 + 3x3 − 3x + 5, d(x) = x + 2.
I To remove the leading term x4 of p(x), subtract x3d(x) . . .
p(x)− x3d(x) = (x4 + 3x3 − 3x + 5)− (x4 + 2x3)
= x3 − 3x + 5
⇒ p(x)− x3d(x)
d(x)=
x3 − 3x + 5
d(x)
⇒ p(x)
d(x)= x3 +
x3 − 3x + 5
d(x)
sop(x)
d(x)is x3 plus a fraction
p1(x)
d(x)with numerator
p1(x) = x3 − 3x + 5 of smaller degree than p(x).
I We can do exactly the same thing to this new fraction!
4 / 63
First Example: Step 1
p(x) = x4 + 3x3 − 3x + 5, d(x) = x + 2.
I To remove the leading term x4 of p(x), subtract x3d(x) . . .
p(x)− x3d(x) = (x4 + 3x3 − 3x + 5)− (x4 + 2x3)
= x3 − 3x + 5
⇒ p(x)− x3d(x)
d(x)=
x3 − 3x + 5
d(x)
⇒ p(x)
d(x)= x3 +
x3 − 3x + 5
d(x)
sop(x)
d(x)is x3 plus a fraction
p1(x)
d(x)with numerator
p1(x) = x3 − 3x + 5 of smaller degree than p(x).
I We can do exactly the same thing to this new fraction!
5 / 63
First Example: Step 1
p(x) = x4 + 3x3 − 3x + 5, d(x) = x + 2.
I To remove the leading term x4 of p(x), subtract x3d(x) . . .
p(x)− x3d(x) = (x4 + 3x3 − 3x + 5)− (x4 + 2x3)
= x3 − 3x + 5
⇒ p(x)− x3d(x)
d(x)=
x3 − 3x + 5
d(x)
⇒ p(x)
d(x)= x3 +
x3 − 3x + 5
d(x)
sop(x)
d(x)is x3 plus a fraction
p1(x)
d(x)with numerator
p1(x) = x3 − 3x + 5 of smaller degree than p(x).
I We can do exactly the same thing to this new fraction!
6 / 63
First Example: Step 1
p(x) = x4 + 3x3 − 3x + 5, d(x) = x + 2.
I To remove the leading term x4 of p(x), subtract x3d(x) . . .
p(x)− x3d(x) = (x4 + 3x3 − 3x + 5)− (x4 + 2x3)
= x3 − 3x + 5
⇒ p(x)− x3d(x)
d(x)=
x3 − 3x + 5
d(x)
⇒ p(x)
d(x)= x3 +
x3 − 3x + 5
d(x)
sop(x)
d(x)is x3 plus a fraction
p1(x)
d(x)with numerator
p1(x) = x3 − 3x + 5 of smaller degree than p(x).
I We can do exactly the same thing to this new fraction!
7 / 63
First Example: Step 1
p(x) = x4 + 3x3 − 3x + 5, d(x) = x + 2.
I To remove the leading term x4 of p(x), subtract x3d(x) . . .
p(x)− x3d(x) = (x4 + 3x3 − 3x + 5)− (x4 + 2x3)
= x3 − 3x + 5
⇒ p(x)− x3d(x)
d(x)=
x3 − 3x + 5
d(x)
⇒ p(x)
d(x)= x3 +
x3 − 3x + 5
d(x)
sop(x)
d(x)is x3 plus a fraction
p1(x)
d(x)with numerator
p1(x) = x3 − 3x + 5 of smaller degree than p(x).
I We can do exactly the same thing to this new fraction!
8 / 63
First Example: Step 2
I To remove the leading term x3 of p1(x), subtract x2d(x) . . .
p1(x)− x2d(x) = (x3 − 3x + 5)− (x3 + 2x2)
= −2x2 − 3x + 5
⇒ p1(x)− x2d(x)
d(x)=−2x2 − 3x + 5
d(x)
⇒ p1(x)
d(x)= x2 +
−2x2 − 3x + 5
d(x)
⇒ p(x)
d(x)= x3 + x2 +
−2x2 − 3x + 5
d(x)
sop(x)
d(x)is x3 + x2 plus a fraction
p2(x)
d(x)with numerator
p2(x) = −2x2 − 3x + 5 of even smaller degree.
I So what do we do next . . .
9 / 63
First Example: Step 2
I To remove the leading term x3 of p1(x), subtract x2d(x) . . .
p1(x)− x2d(x) = (x3 − 3x + 5)− (x3 + 2x2)
= −2x2 − 3x + 5
⇒ p1(x)− x2d(x)
d(x)=−2x2 − 3x + 5
d(x)
⇒ p1(x)
d(x)= x2 +
−2x2 − 3x + 5
d(x)
⇒ p(x)
d(x)= x3 + x2 +
−2x2 − 3x + 5
d(x)
sop(x)
d(x)is x3 + x2 plus a fraction
p2(x)
d(x)with numerator
p2(x) = −2x2 − 3x + 5 of even smaller degree.
I So what do we do next . . .
10 / 63
First Example: Step 2
I To remove the leading term x3 of p1(x), subtract x2d(x) . . .
p1(x)− x2d(x) = (x3 − 3x + 5)− (x3 + 2x2)
= −2x2 − 3x + 5
⇒ p1(x)− x2d(x)
d(x)=−2x2 − 3x + 5
d(x)
⇒ p1(x)
d(x)= x2 +
−2x2 − 3x + 5
d(x)
⇒ p(x)
d(x)= x3 + x2 +
−2x2 − 3x + 5
d(x)
sop(x)
d(x)is x3 + x2 plus a fraction
p2(x)
d(x)with numerator
p2(x) = −2x2 − 3x + 5 of even smaller degree.
I So what do we do next . . .
11 / 63
First Example: Step 2
I To remove the leading term x3 of p1(x), subtract x2d(x) . . .
p1(x)− x2d(x) = (x3 − 3x + 5)− (x3 + 2x2)
= −2x2 − 3x + 5
⇒ p1(x)− x2d(x)
d(x)=−2x2 − 3x + 5
d(x)
⇒ p1(x)
d(x)= x2 +
−2x2 − 3x + 5
d(x)
⇒ p(x)
d(x)= x3 + x2 +
−2x2 − 3x + 5
d(x)
sop(x)
d(x)is x3 + x2 plus a fraction
p2(x)
d(x)with numerator
p2(x) = −2x2 − 3x + 5 of even smaller degree.
I So what do we do next . . .
12 / 63
First Example: Step 2
I To remove the leading term x3 of p1(x), subtract x2d(x) . . .
p1(x)− x2d(x) = (x3 − 3x + 5)− (x3 + 2x2)
= −2x2 − 3x + 5
⇒ p1(x)− x2d(x)
d(x)=−2x2 − 3x + 5
d(x)
⇒ p1(x)
d(x)= x2 +
−2x2 − 3x + 5
d(x)
⇒ p(x)
d(x)= x3 + x2 +
−2x2 − 3x + 5
d(x)
sop(x)
d(x)is x3 + x2 plus a fraction
p2(x)
d(x)with numerator
p2(x) = −2x2 − 3x + 5 of even smaller degree.
I So what do we do next . . .
13 / 63
First Example: Step 2
I To remove the leading term x3 of p1(x), subtract x2d(x) . . .
p1(x)− x2d(x) = (x3 − 3x + 5)− (x3 + 2x2)
= −2x2 − 3x + 5
⇒ p1(x)− x2d(x)
d(x)=−2x2 − 3x + 5
d(x)
⇒ p1(x)
d(x)= x2 +
−2x2 − 3x + 5
d(x)
⇒ p(x)
d(x)= x3 + x2 +
−2x2 − 3x + 5
d(x)
sop(x)
d(x)is x3 + x2 plus a fraction
p2(x)
d(x)with numerator
p2(x) = −2x2 − 3x + 5 of even smaller degree.
I So what do we do next . . .
14 / 63
First Example: Step 3
I . . . remove the leading term −2x2 of p2(x), by subtracting(−2x)d(x) of course!
p1(x) + 2xd(x) = (−2x2 − 3x + 5) + (2x2 + 4x)
= x + 5
⇒ p2(x) + 2xd(x)
d(x)=
x + 5
d(x)
⇒ p2(x)
d(x)= −2x +
x + 5
d(x)
⇒ p(x)
d(x)= x3 + x2 − 2x +
x + 5
d(x)
. . . sop(x)
d(x)is x3 + x2 − 2x plus a fraction in which the
numerator p3(x) = x + 5 has even smaller degree.
I This is so much fun. Lets do it again!
15 / 63
First Example: Step 3
I . . . remove the leading term −2x2 of p2(x), by subtracting(−2x)d(x) of course!
p1(x) + 2xd(x) = (−2x2 − 3x + 5) + (2x2 + 4x)
= x + 5
⇒ p2(x) + 2xd(x)
d(x)=
x + 5
d(x)
⇒ p2(x)
d(x)= −2x +
x + 5
d(x)
⇒ p(x)
d(x)= x3 + x2 − 2x +
x + 5
d(x)
. . . sop(x)
d(x)is x3 + x2 − 2x plus a fraction in which the
numerator p3(x) = x + 5 has even smaller degree.
I This is so much fun. Lets do it again!
16 / 63
First Example: Step 3
I . . . remove the leading term −2x2 of p2(x), by subtracting(−2x)d(x) of course!
p1(x) + 2xd(x) = (−2x2 − 3x + 5) + (2x2 + 4x)
= x + 5
⇒ p2(x) + 2xd(x)
d(x)=
x + 5
d(x)
⇒ p2(x)
d(x)= −2x +
x + 5
d(x)
⇒ p(x)
d(x)= x3 + x2 − 2x +
x + 5
d(x)
. . . sop(x)
d(x)is x3 + x2 − 2x plus a fraction in which the
numerator p3(x) = x + 5 has even smaller degree.
I This is so much fun. Lets do it again!
17 / 63
First Example: Step 3
I . . . remove the leading term −2x2 of p2(x), by subtracting(−2x)d(x) of course!
p1(x) + 2xd(x) = (−2x2 − 3x + 5) + (2x2 + 4x)
= x + 5
⇒ p2(x) + 2xd(x)
d(x)=
x + 5
d(x)
⇒ p2(x)
d(x)= −2x +
x + 5
d(x)
⇒ p(x)
d(x)= x3 + x2 − 2x +
x + 5
d(x)
. . . sop(x)
d(x)is x3 + x2 − 2x plus a fraction in which the
numerator p3(x) = x + 5 has even smaller degree.
I This is so much fun. Lets do it again!
18 / 63
First Example: Step 3
I . . . remove the leading term −2x2 of p2(x), by subtracting(−2x)d(x) of course!
p1(x) + 2xd(x) = (−2x2 − 3x + 5) + (2x2 + 4x)
= x + 5
⇒ p2(x) + 2xd(x)
d(x)=
x + 5
d(x)
⇒ p2(x)
d(x)= −2x +
x + 5
d(x)
⇒ p(x)
d(x)= x3 + x2 − 2x +
x + 5
d(x)
. . . sop(x)
d(x)is x3 + x2 − 2x plus a fraction in which the
numerator p3(x) = x + 5 has even smaller degree.
I This is so much fun. Lets do it again!
19 / 63
First Example: Step 4
I To remove the leading term x of p3(x), subtract 1× d(x):
p1(x)− d(x) = (x + 5)− (x + 2) = 3
⇒ p3(x) + 2xd(x)
d(x)=
3
d(x)
⇒ p3(x)
d(x)= −2x +
3
d(x)
⇒ p(x)
d(x)= x3 + x2 − 2x + 1 +
3
d(x)
. . . expressingp(x)
d(x)as a polynomial x3 + x2 − 2x + 1 plus a
fraction in which the numerator p4(x) = 3 has even smallerdegree.
I Can we do this again?
20 / 63
First Example: Step 4
I To remove the leading term x of p3(x), subtract 1× d(x):
p1(x)− d(x) = (x + 5)− (x + 2) = 3
⇒ p3(x) + 2xd(x)
d(x)=
3
d(x)
⇒ p3(x)
d(x)= −2x +
3
d(x)
⇒ p(x)
d(x)= x3 + x2 − 2x + 1 +
3
d(x)
. . . expressingp(x)
d(x)as a polynomial x3 + x2 − 2x + 1 plus a
fraction in which the numerator p4(x) = 3 has even smallerdegree.
I Can we do this again?
21 / 63
First Example: Step 4
I To remove the leading term x of p3(x), subtract 1× d(x):
p1(x)− d(x) = (x + 5)− (x + 2) = 3
⇒ p3(x) + 2xd(x)
d(x)=
3
d(x)
⇒ p3(x)
d(x)= −2x +
3
d(x)
⇒ p(x)
d(x)= x3 + x2 − 2x + 1 +
3
d(x)
. . . expressingp(x)
d(x)as a polynomial x3 + x2 − 2x + 1 plus a
fraction in which the numerator p4(x) = 3 has even smallerdegree.
I Can we do this again?
22 / 63
First Example: Step 4
I To remove the leading term x of p3(x), subtract 1× d(x):
p1(x)− d(x) = (x + 5)− (x + 2) = 3
⇒ p3(x) + 2xd(x)
d(x)=
3
d(x)
⇒ p3(x)
d(x)= −2x +
3
d(x)
⇒ p(x)
d(x)= x3 + x2 − 2x + 1 +
3
d(x)
. . . expressingp(x)
d(x)as a polynomial x3 + x2 − 2x + 1 plus a
fraction in which the numerator p4(x) = 3 has even smallerdegree.
I Can we do this again?
23 / 63
First Example: Step 4
I To remove the leading term x of p3(x), subtract 1× d(x):
p1(x)− d(x) = (x + 5)− (x + 2) = 3
⇒ p3(x) + 2xd(x)
d(x)=
3
d(x)
⇒ p3(x)
d(x)= −2x +
3
d(x)
⇒ p(x)
d(x)= x3 + x2 − 2x + 1 +
3
d(x)
. . . expressingp(x)
d(x)as a polynomial x3 + x2 − 2x + 1 plus a
fraction in which the numerator p4(x) = 3 has even smallerdegree.
I Can we do this again?
24 / 63
First Example: The end!
I No! The leading term 3 of p4(x) = 3 is now too small (indegree) to express as a multiple of d(x).
I So we cannot repeat the procedure . . . but we now have ananswer in which all possible polynomial terms have beenextracted:
p(x)
d(x)= x3 + x2 − 2x + 1 +
3
x + 2
I Our job is done!
I That was fun, but involved a lot of writing.
I We can carry out the division much more efficiently in atabular format.
25 / 63
First Example: The end!
I No! The leading term 3 of p4(x) = 3 is now too small (indegree) to express as a multiple of d(x).
I So we cannot repeat the procedure . . .
but we now have ananswer in which all possible polynomial terms have beenextracted:
p(x)
d(x)= x3 + x2 − 2x + 1 +
3
x + 2
I Our job is done!
I That was fun, but involved a lot of writing.
I We can carry out the division much more efficiently in atabular format.
26 / 63
First Example: The end!
I No! The leading term 3 of p4(x) = 3 is now too small (indegree) to express as a multiple of d(x).
I So we cannot repeat the procedure . . . but we now have ananswer in which all possible polynomial terms have beenextracted:
p(x)
d(x)= x3 + x2 − 2x + 1 +
3
x + 2
I Our job is done!
I That was fun, but involved a lot of writing.
I We can carry out the division much more efficiently in atabular format.
27 / 63
First Example: The end!
I No! The leading term 3 of p4(x) = 3 is now too small (indegree) to express as a multiple of d(x).
I So we cannot repeat the procedure . . . but we now have ananswer in which all possible polynomial terms have beenextracted:
p(x)
d(x)= x3 + x2 − 2x + 1 +
3
x + 2
I Our job is done!
I That was fun, but involved a lot of writing.
I We can carry out the division much more efficiently in atabular format.
28 / 63
First Example: The end!
I No! The leading term 3 of p4(x) = 3 is now too small (indegree) to express as a multiple of d(x).
I So we cannot repeat the procedure . . . but we now have ananswer in which all possible polynomial terms have beenextracted:
p(x)
d(x)= x3 + x2 − 2x + 1 +
3
x + 2
I Our job is done!
I That was fun, but involved a lot of writing.
I We can carry out the division much more efficiently in atabular format.
29 / 63
First Example: The end!
I No! The leading term 3 of p4(x) = 3 is now too small (indegree) to express as a multiple of d(x).
I So we cannot repeat the procedure . . . but we now have ananswer in which all possible polynomial terms have beenextracted:
p(x)
d(x)= x3 + x2 − 2x + 1 +
3
x + 2
I Our job is done!
I That was fun, but involved a lot of writing.
I We can carry out the division much more efficiently in atabular format.
30 / 63
Same Example in Tabular Format
x + 2) x4 + 3x3 + 0x2 − 3x + 5
31 / 63
Same Example in Tabular Format
x + 2) x4 + 3x3 + 0x2 − 3x + 5Easier to maintain alignmentif we add any “missing powers”
32 / 63
Same Example in Tabular Format
x + 2) x4 + 3x3 + 0x2 − 3x + 5Need a product of x + 2that removes x4 . . .
33 / 63
Same Example in Tabular Format
x3
x + 2) x4 + 3x3 + 0x2 − 3x + 5. . . so multiply by x3.
34 / 63
Same Example in Tabular Format
x3
x + 2) x4 + 3x3 + 0x2 − 3x + 5
x4 + 2x3
x3(x + 2) = x4 + 2x3.
35 / 63
Same Example in Tabular Format
x3
x + 2) x4 + 3x3 + 0x2 − 3x + 5
x4 + 2x3
x3 + 0x2 − 3x + 5
Subtracting gives
x3 + 0x2 − 3x + 5.
36 / 63
Same Example in Tabular Format
x3
x + 2) x4 + 3x3 + 0x2 − 3x + 5
x4 + 2x3
x3 + 0x2 − 3x + 5 Need a product of x + 2that removes x3 . . .
37 / 63
Same Example in Tabular Format
x3 + x2
x + 2) x4 + 3x3 + 0x2 − 3x + 5
x4 + 2x3
x3 + 0x2 − 3x + 5 . . . so multiply by x2.
38 / 63
Same Example in Tabular Format
x3 + x2
x + 2) x4 + 3x3 + 0x2 − 3x + 5
x4 + 2x3
x3 + 0x2 − 3x + 5
x3 + 2x2 x2(x + 2) = x3 + 2x2.
39 / 63
Same Example in Tabular Format
x3 + x2
x + 2) x4 + 3x3 + 0x2 − 3x + 5
x4 + 2x3
x3 + 0x2 − 3x + 5
x3 + 2x2
− 2x2 − 3x + 5 Subtracting gives −2x2 − 3x + 5.
40 / 63
Same Example in Tabular Format
x3 + x2
x + 2) x4 + 3x3 + 0x2 − 3x + 5
x4 + 2x3
x3 + 0x2 − 3x + 5
x3 + 2x2
− 2x2 − 3x + 5 Need a product of x + 2that removes −2x2 . . .
41 / 63
Same Example in Tabular Format
x3 + x2 − 2x
x + 2) x4 + 3x3 + 0x2 − 3x + 5
x4 + 2x3
x3 + 0x2 − 3x + 5
x3 + 2x2
− 2x2 − 3x + 5 . . . so multiply by −2x .
42 / 63
Same Example in Tabular Format
x3 + x2 − 2x
x + 2) x4 + 3x3 + 0x2 − 3x + 5
x4 + 2x3
x3 + 0x2 − 3x + 5
x3 + 2x2
− 2x2 − 3x + 5
−2x2 − 4x −2x(x + 2) = −2x2 − 4x .
43 / 63
Same Example in Tabular Format
x3 + x2 − 2x
x + 2) x4 + 3x3 + 0x2 − 3x + 5
x4 + 2x3
x3 + 0x2 − 3x + 5
x3 + 2x2
− 2x2 − 3x + 5
−2x2 − 4x
x + 5 Subtracting gives x + 5.
44 / 63
Same Example in Tabular Format
x3 + x2 − 2x
x + 2) x4 + 3x3 + 0x2 − 3x + 5
x4 + 2x3
x3 + 0x2 − 3x + 5
x3 + 2x2
− 2x2 − 3x + 5
−2x2 − 4x
x + 5 Need a product of x + 2that removes x . . .
45 / 63
Same Example in Tabular Format
x3 + x2 − 2x + 1
x + 2) x4 + 3x3 + 0x2 − 3x + 5
x4 + 2x3
x3 + 0x2 − 3x + 5
x3 + 2x2
− 2x2 − 3x + 5
−2x2 − 4x
x + 5 . . . so multiply by 1.
46 / 63
Same Example in Tabular Format
x3 + x2 − 2x + 1
x + 2) x4 + 3x3 + 0x2 − 3x + 5
x4 + 2x3
x3 + 0x2 − 3x + 5
x3 + 2x2
− 2x2 − 3x + 5
−2x2 − 4x
x + 5
x + 2 1× (x + 2) = x + 2.
47 / 63
Same Example in Tabular Format
x3 + x2 − 2x + 1
x + 2) x4 + 3x3 + 0x2 − 3x + 5
x4 + 2x3
x3 + 0x2 − 3x + 5
x3 + 2x2
− 2x2 − 3x + 5
−2x2 − 4x
x + 5
x + 2
3 Subtracting gives 3and we stop.
48 / 63
Same Example in Tabular Format
x3 + x2 − 2x + 1
x + 2) x4 + 3x3 + 0x2 − 3x + 5
x4 + 2x3
x3 + 0x2 − 3x + 5
x3 + 2x2
− 2x2 − 3x + 5
−2x2 − 4x
x + 5
x + 2
3
Answer is:
polynomial plus remainder
p(x)
d(x)= x3 + x2 − 2x + 1 + 3
x+2
49 / 63
Same Example in Tabular Format
x3 + x2 − 2x + 1
x + 2) x4 + 3x3 + 0x2 − 3x + 5
x4 + 2x3
x3 + 0x2 − 3x + 5
x3 + 2x2
− 2x2 − 3x + 5
−2x2 − 4x
x + 5
x + 2
3
Answer is:
polynomial plus remainder
p(x)
d(x)= x3 + x2 − 2x + 1 + 3
x+2
50 / 63
Same Example in Tabular Format
x3 + x2 − 2x + 1
x + 2) x4 + 3x3 + 0x2 − 3x + 5
x4 + 2x3
x3 + 0x2 − 3x + 5
x3 + 2x2
− 2x2 − 3x + 5
−2x2 − 4x
x + 5
x + 2
3
Answer is:
polynomial plus remainder
p(x)
d(x)= x3 + x2 − 2x + 1 + 3
x+2
When we subtract, we canleave out unnecessary terms.
51 / 63
Another example
I When we subtract, we sometimes get lucky!
p(x) = 8x3 − 4x2 − 2x + 1, d(x) = 2x − 1.
2x − 1) 8x3 − 4x2 − 2x + 1
52 / 63
Another example
I When we subtract, we sometimes get lucky!
p(x) = 8x3 − 4x2 − 2x + 1, d(x) = 2x − 1.
2x − 1) 8x3 − 4x2 − 2x + 1
53 / 63
Another example
I When we subtract, we sometimes get lucky!
p(x) = 8x3 − 4x2 − 2x + 1, d(x) = 2x − 1.
2x − 1) 8x3 − 4x2 − 2x + 1
54 / 63
Another example
I When we subtract, we sometimes get lucky!
p(x) = 8x3 − 4x2 − 2x + 1, d(x) = 2x − 1.
4x2
2x − 1) 8x3 − 4x2 − 2x + 1
55 / 63
Another example
I When we subtract, we sometimes get lucky!
p(x) = 8x3 − 4x2 − 2x + 1, d(x) = 2x − 1.
4x2
2x − 1) 8x3 − 4x2 − 2x + 1
8x3 − 4x2
56 / 63
Another example
I When we subtract, we sometimes get lucky!
p(x) = 8x3 − 4x2 − 2x + 1, d(x) = 2x − 1.
4x2
2x − 1) 8x3 − 4x2 − 2x + 1
8x3 − 4x2
0 − 2x + 1
57 / 63
Another example
I When we subtract, we sometimes get lucky!
p(x) = 8x3 − 4x2 − 2x + 1, d(x) = 2x − 1.
4x2
2x − 1) 8x3 − 4x2 − 2x + 1
8x3 − 4x2
0 − 2x + 1
58 / 63
Another example
I When we subtract, we sometimes get lucky!
p(x) = 8x3 − 4x2 − 2x + 1, d(x) = 2x − 1.
4x2 − 1
2x − 1) 8x3 − 4x2 − 2x + 1
8x3 − 4x2
0 − 2x + 1
59 / 63
Another example
I When we subtract, we sometimes get lucky!
p(x) = 8x3 − 4x2 − 2x + 1, d(x) = 2x − 1.
4x2 − 1
2x − 1) 8x3 − 4x2 − 2x + 1
8x3 − 4x2
0 − 2x + 1
60 / 63
Another example
I When we subtract, we sometimes get lucky!
p(x) = 8x3 − 4x2 − 2x + 1, d(x) = 2x − 1.
4x2 − 1
2x − 1) 8x3 − 4x2 − 2x + 1
8x3 − 4x2
0 − 2x + 1
−2x + 1
61 / 63
Another example
I When we subtract, we sometimes get lucky!
p(x) = 8x3 − 4x2 − 2x + 1, d(x) = 2x − 1.
4x2 − 1
2x − 1) 8x3 − 4x2 − 2x + 1
8x3 − 4x2
0 − 2x + 1
−2x + 1
0
62 / 63
Another example
I When we subtract, we sometimes get lucky!
p(x) = 8x3 − 4x2 − 2x + 1, d(x) = 2x − 1.
4x2 − 1
2x − 1) 8x3 − 4x2 − 2x + 1
8x3 − 4x2
0 − 2x + 1
−2x + 1
0
Answer:
p(x)
d(x)= 4x2 − 1 +
0
2x − 1
= 4x2 − 1
63 / 63