Mathematics

1. Multiplication and
Division of
Polynomial
Analyn M. Delos Santos

2. Target Competency
⦁ Define polynomials and the
remarks and theorems
involving polynomials
⦁ Multiply and divide
polynomials

3. Basic Concepts
An expression of the form
𝑎0𝑥𝑛
+ 𝑎1𝑥𝑛−1
+ . . . + 𝑎𝑛 in
which 𝑎0, 𝑎1 , . . . , 𝑎𝑛 are
given numbers (real or
imaginary) and with x as the
variable is called an integral
rational function of x or a
polynomial in x.

4. Basic Concepts
The constants 𝑎0, 𝑎1 , . . . , 𝑎𝑛
are called coefficients, and
the single monomials 𝑎0𝑥𝑛
,
𝑎1𝑥𝑛−1
, . . . , 𝑎𝑛 are called
the terms of the polynomial.
If 𝑎0 ≠ 0, the polynomial is of
degree n and 𝑎0𝑥𝑛
is the
leading term.

5. Examples
f (x) = 3𝑥3
– x + 2
g (x) = 4𝑥4
– 𝑥3
+ 2x – 1
h (x) = 2𝑥
2
– (3 + 2)x + 4
f (-1) = 0
g (i) = 3 + 3i
h (1) = 1

6. Multiplication of Polynomials
Example 1.
Multiply 𝑥2 - x + 1 and 𝑥2 + x + 1
(𝑥2
- x + 1) x (𝑥2
+ x + 1)
𝑥4 - 𝑥3+𝑥2
𝑥3 - 𝑥2+ x
𝑥2 - x + 1 0
𝑥4 + 𝑥2 + 1

7. 1 -1 1 x 1 1 1
1 -1 1
1 -1 1
1 -1 1 0
1 0 1 0 1
𝑥4 + 0 𝑥3 + 𝑥2 + 0x + 1 = 𝑥4 + 𝑥2 + 1

8. Example 2.
Multiply 𝑥5 + 𝑥3 – 2𝑥2 + 3 by 2𝑥4 – 3𝑥3 + 4𝑥2 – 1
1 0 1 -2 0 3 x 2 -3 4 0 -1
2 0 2 -4 0 6
-3 0 -3 6 0 -9
4 0 4 -8 0 12
-1 0 -1 2 0 -3
2 -3 6 -7 9 -2 -10 14 0 -3
2𝑥9
– 3 𝑥8
+ 6𝑥7
– 7𝑥6
+ 9𝑥5
– 2 𝑥4
– 10𝑥3
+ 14𝑥2
– 3

9. Example 3.
Multiply 2𝑥4 – 3𝑥3 + x – 1by 𝑥3 + 3𝑥2 – 1
2 -3 0 1 -1 x 1 3 0 -1
2 -3 0 1 -1
6 -9 0 3 -3
-2 3 0 -1 1 1
2 3 -9 -1 5 -3 -1 1
2𝑥7 + 3𝑥6 – 9𝑥5 – 𝑥4 + 5𝑥3 – 3𝑥2 – x + 1

10. Remark:
If two non - identically vanishing polynomials
f(x) and g(x) have the leading terms 𝑎0𝑥𝑛
and 𝑏0𝑥𝑚, the leading term of the product
will be 𝑎0𝑏0𝑥𝑛+𝑚 and the coefficient differs
from 0; hence, f(x)g(x) is a non - identically
vanishing polynomial. Consequently, if f(x)
g(x) = 0 one of the factors must be an
identically vanishing polynomial.

11. Division of Polynomials
Let f(x) = 𝒂𝟎𝒙𝒏 + 𝒂𝟏𝒙𝒏−𝟏+ . . . + 𝒂𝒏 g(x) =𝒃𝟎 𝒙𝒎 + 𝒃𝟏𝒙𝒎−𝟏+ . . . + 𝒃𝒎
be two polynomials of degrees n and m, respectively, so
that 𝑎0 ≠ 0, 𝑏0≠ 0, and assume that n ≥ m. By choosing
properly a constant 𝑐0 we can obtain the polynomial f(x)
– 𝑐0𝑥𝑛−𝑚
g(x) = 𝑓1(x) which, if not vanishing identically,
will have degree 𝑛1 < n; for this it suffices to take 𝑐0 =
𝑎0
𝑏0 .
As long as 𝑛1 ≥ m, a constant 𝑐1 can be found so that
𝑓1 (x) – 𝑐1𝑥𝑛1−𝑚 g(x) = 𝑓2 (x) which, if not vanishing
identically, will have degree 𝑛2 < 𝑛1.

12. Division of Polynomials
If 𝑛2m, the same process can be repeated. Now
the degrees of the “partial remainders” 𝑓1 (x),
𝑓2 (x) . . . form a decreasing sequence so that
there will be some first partial remainder
𝑓𝑘+1(x) that either vanishes identically or is of
degree 𝑛𝑘+1 < m.

13. Division of Polynomials
By eliminating 𝑓1(x), 𝑓2(x), . . . , 𝑓𝑘(x) from the
identities f(x) – 𝑐0𝑥𝑛−𝑚 g(x) = 𝑓1 (x), 𝑓1 (x) –
𝑐1𝑥𝑛1−𝑚g(x) = 𝑓2(x), . . . , 𝑓𝑘(x) – 𝑐𝑘𝑥𝑛𝑘−𝑚g(x) =
𝑓𝑘+1 (x) and setting for brevity 𝑐0𝑥𝑛−𝑚
+
𝑐1𝑥𝑛1−𝑚
+ . . . + 𝑐𝑘𝑥𝑛𝑘−𝑚
= q(x), 𝑓𝑘+1 (x) = r(x) we
obtain the identity f(x) = g(x) q(x) + r(x) in which
r(x) has a degree < m or vanishes identically.
The polynomials q(x) and r(x) are called the
quotient and the remainder in the division of f(x)
by g(x).

14. Dividend Divisor
1 1 0 0 3 0 0 0 -1 1 -3 0 4 1
1 -3 0 4 1 1 4 12 32 82 quotient
` 4 0 -4 2 0
4 -12 0 16 4
12 -4 -14 -4 0
12-36 0 48 12
32 -14 -52 -12 0
32 -96 0 128 32
82 -52 -140 -32 -1
82 -246 0 328 82
194 -140 -360 -83 remainder
Example 1.
Divide 𝑥8 + 𝑥7 +3𝑥4 – 1 by 𝑥4 – 3𝑥3 + 4x + 1

15. Thus, the quotient and the
remainder are
𝑥4
+ 4𝑥3
+ 12𝑥2
+ 32x + 82, quotient
194 𝑥3
– 140 𝑥2
– 360 x – 83,
remainder and identically
𝑥8
+ 𝑥7
+3𝑥4
– 1 = (𝑥4
– 3𝑥3
+ 4x +
1)( 𝑥4 + 4𝑥3+ 12𝑥2 + 32x + 82) +
194𝑥3 – 140𝑥2 – 360 x – 83.

16. Dividend Divisor
1 3 0 0 -2 3 -1 1 1 0 0 -1 1
1 0 0 -1 1 1 3 0 1 quotient
‘ 3 0 1 -3 3
3 0 0 -3 3
0 1 0 0 -1
0 0 0 0 0
1 0 0 -1 1
1 0 0 -1 1
0 remainder
Example 2.
Divide 𝑥7 + 3𝑥6 - 2𝑥3 +3𝑥2 – x + 1 by 𝑥4 – x + 1

17. Thus, the quotient and the
remainder are
𝑥3
+3𝑥2
+ 1, quotient
0, remainder and identically
𝑥7
+ 3𝑥6
- 2𝑥3
+3𝑥2
– x + 1 = (𝑥4
– x
+ 1)(𝑥3
+3𝑥2
+ 1).

18. Remark1: If the remainder in the division of f(x) by g(x)
is 0, that is, if f(x) = g(x) q(x) where q(x) is a polynomial,
it is said that f(x) is divisible by g(x) or that g(x) is a
divisor of f(x).
Remark 2: If two polynomials f and 𝑓1 are divisible by g,
then for arbitrary polynomials l and 𝑙1 the polynomial lf +
𝑙1𝑓1 will be divisible by g. In fact, by hypothesis f =gq
𝑓1=g𝑞1 where q and 𝑞1 are polynomials; hence, lf + 𝑙1𝑓1
= g (lq + 𝑙1𝑞1) is divisible by g.

19. The Remainder Theorem
The remainder obtained in
dividing f(x) by x – c is the
value of the polynomial f(x) for
x = c, that is f(c).

20. Proof. Since the divisor is of the first degree, the
remainder will be a constant r. Calling the quotient
q(x), we have the identity f(x) = (x – c) q(x) + r. On
substituting the number c in place of x into this identity
we must get equal numbers. Now, since r is a constant,
it is not affected by this substitution and the value of the
right-hand member for x = c will be (c – c) q(c) + r = r
whereas the value of the left-hand member is f(c);
hence, r = f(c), which means also that identically in x f(x)
= (x – c) q(x) + f(c). It follows from this theorem that f(x)
is divisible by x – c if and only if f(c) = 0.

21. x – c = x + 3
c = -3
f(-3) = -27 + 9 + 15 + 3 = 0
f(x) is divisible by x + 3
To show that 𝑥𝑛 – 𝑐𝑛 is divisible by x – c. This is true
since 𝑐𝑛 – 𝑐𝑛 = 0; the quotient found by ordinary division
is 𝑥𝑛−1
+ 𝑐𝑥𝑛−2
+ 𝑐2
𝑥𝑛−3
+ . . . + 𝑐𝑛−1
.
Example 1.
Divide f(x) = 𝑥3 + 𝑥2 - 5x + 3 by x + 3

22. x – c = x + 2
c = -2
f(-2) = 16 – 24 + 12 – 6 + 2 = 0
f(x) is divisible by x + 2
Example 2.
Divide f(x) = 𝑥4
+ 3𝑥3
+ 3𝑥2
+ 3x + 2 by x + 2

23. Synthetic Division
From the identity, f(x) = (x – c) q(x) + r let us
substitute q (x) = 𝑏0 𝑥𝑛−1
+ 𝑏1 𝑥𝑛−2
+ . . . + 𝑏𝑛−1
where 𝑏0 , 𝑏1 , . . . , 𝑏𝑛−1 are coefficients to be
determined. Performing the multiplication, we
have (x – c) q(x) = 𝑏0 𝑥𝑛
+ (𝑏1 − 𝑐𝑏0) 𝑥𝑛−1
+
(𝑏2 − 𝑐𝑏1) 𝑥𝑛−2
+ . . . + (𝑏𝑛−1 − 𝑐𝑏𝑛−2) 𝑥 − 𝑐𝑏𝑛−1,
and
(x – c) q(x) + r = 𝑏0 𝑥𝑛 + (𝑏1 − 𝑐𝑏0) 𝑥𝑛−1 +
(𝑏𝑛−1 − 𝑐𝑏𝑛−2) 𝑥 + 𝑟 − 𝑐𝑏𝑛−1.

24. Since this polynomial must
be identical to 𝑎0 𝑥𝑛 + 𝑎1 𝑥𝑛
+ . . . + 𝑎𝑛 to determine
𝑏0, 𝑏1, . . . , 𝑏𝑛−1 and r we
equate coefficients of like
powers of x, getting the set
of equations
𝒃𝟎 = 𝒂𝟎
𝒃𝟏 – 𝒄𝒃𝟎 = 𝒂𝟏
𝒃𝟐 – 𝒄𝒃𝟏 = 𝒂𝟐
. . . ,
𝒃𝒏−𝟏 – 𝒄𝒃𝒏−𝟐 = 𝒂𝒏−𝟏
𝒓 − 𝒄𝒃𝒏−𝟏 = 𝒂𝒏
from which it follows that
𝑏0, 𝑏1, . . . , 𝑏𝑛−1 , r are found
one after another as follows:
𝑏0 = 𝑎0
𝑏1 = 𝑎1 + 𝑐𝑏0
𝑏2 = 𝑎2 + 𝑐𝑏1
. . . ,
𝑏𝑛−1 = 𝑎𝑛−1 + 𝑐𝑏𝑛−2
𝑟 = 𝑎𝑛 + 𝑐𝑏𝑛−1

25. 𝑎0 𝑎1 𝑎2 . . . 𝑎𝑛−1 𝑎𝑛
𝑏0𝑐 𝑏1𝑐 . . . 𝑏𝑛−2𝑐 𝑏𝑛−1𝑐
𝐚𝟎 = 𝐛𝟎 𝐛𝟏 𝐛𝟐 . . . 𝐛𝐧−𝟏 𝐫 remainder
The calculation is of a recursive nature and in
practice can be arranged more conveniently thus:
coefficients of the quotient

26. The independent expressions for
𝑏0, 𝑏1, . . . , 𝑏𝑛−1 , r obtained by
successive substitutions are
𝑏0 = 𝑎0
𝑏1 = 𝑎0𝑐 + 𝑎1 ,
𝑏2 = 𝑎0𝑐2 + 𝑎1𝑐 + 𝑎2 ,
. . . ,
𝑏𝑛−1 = 𝑎0𝑐𝑛−1
+ 𝑎0𝑐𝑛−2
+ . . . +
𝑎𝑛−1
and
𝑟 = 𝑎0𝑐𝑛
+ 𝑎1𝑐𝑛−1
+ . . . + 𝑎𝑛 = f(c)
which gives the second proof of
the remainder theorem.
Considering the sequence of
polynomials
𝑓0 = 𝑎0 ,
𝑓1 = 𝑥𝑓0 + 𝑎1 ,
𝑓2 = 𝑥𝑓1 + 𝑎2 ,
. . . ,
𝑓𝑛 (x) = 𝑥𝑓𝑛−1(x) + 𝑎𝑛
It is clear that
𝑓𝑖(𝑥) = 𝑎0𝑥𝑖
+ 𝑎1𝑥𝑖 −1
+ . . . + 𝑎𝑖
Hence
𝑏𝑖 = 𝑓𝑖(𝑐) , i = 0, 1, 2, . . . , i – 1
and, moreover,
𝑓𝑖(𝑥) = (x – c) ⦋ 𝑓0(𝑐)𝑥𝑖−1 +
𝑓1(𝑐)𝑥𝑖 −2 + . . . + 𝑓𝑖−1 (𝑐)⦌ + 𝑓𝑖(𝑐)

27. -2⃓ 3 -7 5 0 -1 - 6 -8
-6 26 -62 124 -246 504
3 -13 31 -62 123 -252 496
Hence, the quotient is
3𝑥5 – 13𝑥4 + 31𝑥3 – 62𝑥2 + 123x – 252
And the remainder is
r = 496
Example 1
Divide 3𝑥6 – 7𝑥5 + 5𝑥4 – 𝑥2 – 6x – 8 by 𝑥 + 2

28. 1 ⃓ 5 0 -7 6 -2 4
5 5 -2 4 2
5 5 -2 4 2 6
Hence, the quotient is
5𝑥4 + 5𝑥3 – 2𝑥2 + 4x + 2
And the remainder is
r = 6
Example 2
Divide 5𝑥5
– 7𝑥3
+ 6𝑥2
–2x + 4 by 𝑥 – 1

29. -2⃓ 2 -6 7 -5 1
-4 20 -54 118
2 -10 27 -59 119
Hence, the quotient is
2𝑥3 - 10𝑥2+ 27x - 59
And the remainder is
r = 119
Example 3
Divide 2𝑥4 – 6𝑥3 + 7𝑥2 –5x + 1 by 𝑥 + 2

30. The derivative of
f(x) = 𝑎0𝑥𝑛 + 𝑎1𝑥𝑛−1+ . . . +
𝑎𝑛−1 x + 𝑎𝑛 is
f '(x) = 𝑛𝑎0𝑥𝑛−1
+(n – 1)
𝑎1𝑥𝑛−2+ . . . + 𝑎𝑛−1
Since f’(x) is a polynomial,
we can consider its
derivative, which is called the
second derivative, f ’’(x), of
f(x). Similarly, the derivative
of the second derivative is
the third derivative, f ’’’(x), of
f(x), etc. Now take the
expansion
f(x) = 𝐴0+ 𝐴1 𝑥 − 𝑐 +
𝐴2 (𝑥 − 𝑐 )2 + . . . +
𝐴𝑛 (𝑥 − 𝑐 )𝑛

31. Form successive derivatives of both members
f ’(x) = 𝐴1+ 2𝐴2 𝑥 − 𝑐 + 3𝐴3 (𝑥 − 𝑐 )2
+ . . . +
𝑛𝐴𝑛 (𝑥 − 𝑐 )𝑛−1
f ’’(x) = 2𝐴2+ 3 • 2𝐴3 𝑥 − 𝑐 + . . . + 𝑛(𝑛

32. in the general
𝐴𝑖 =
𝑓𝑖(c)
1•2•3•𝑖
Thus, the expansion of f(x) in powers of x – c
takes the form
f (x) = f (c) +
𝑓′(c)
1
(x –c) +
𝑓′′(c)
1•2
(x – c)2 + . . . +
𝑓𝑛(c)
1•2•3 . . .• 𝑛
(x – c)𝑛

33. f (x) = f (1) +
𝑓′(1)
1
(x –1) +
𝑓′′(1)
1•2
(x – 1)2
+
𝑓′′′(1)
1•2•3
(x – 1)3
f(x) = 𝑥3 + 𝑥2 + x + 1
f(1) = (1)3 + (1)2 + 1 + 1
f(1) = 4
𝑓′(x) = 3𝑥2 + 2x + 1
𝑓′
(1) = 3(1)2
+ 2(1) + 1
𝑓′
(1) = 6
𝑓′′(x) = 6x + 2
𝑓′′(1) = 6 (1) + 2
𝑓′′(1) = 8
𝑓′′′
(x) = 6
𝑓′′′(1) = 6
Example 1
Find the taylor’s formula of 𝑥3 + 𝑥2 + x + 1 at x = 1
f (x) = 4 +
6
1
(x –1) +
8
1•2
(x – 1)2
+
6
1•2•3
(x – 1)3
f (x) = 4 + 6(x – 1) + 4 (x – 1)2
+ (x – 1)3

34. f (x) = f (1) +
𝑓′(1)
1
(x –1) +
𝑓′′(1)
1•2
(x – 1)2 +
𝑓′′′(1)
1•2•3
(x – 1)3
f(x) = − 𝑥4 + 6𝑥3 + x –1
f(1) = −(1)4
+ 6(1)3
+ 1 – 1
f(1) = 5
𝑓′
(x) = −4𝑥3
+ 18𝑥2
+ 1
𝑓′(1) = −4(1)3 + 18(1)2 + 1
𝑓′(1) = 15
𝑓′′
(x) = −12𝑥2
+ 36x
𝑓′′(1) = −12(1)2 + 36 (1)
𝑓′′
(1) = 24
𝑓′′′
(x) = - 24x + 36
𝑓′′′(1) = -24 (1) + 36
𝑓′′′(1) = 12
Example 2
Find the taylor’s formula of − 𝑥4
+ 6𝑥3
+ x –1 at x = 1
f (x) = 5 +
15
1
(x –1) +
24
1•2
(x – 1)2
+
12
1•2•3
(x – 1)3
f (x) = 5 + 15(x – 1) + 12 (x – 1)2
+ 2(x – 1)3

35. Thank you
for
listening!

36. Activity
1. Multiply 𝑥4
+ 4𝑥3
– 5𝑥2
– 2 by 𝑥4
– 4𝑥3
– 5𝑥2
– 2
using the method of detached coefficients.
2. Divide 2𝑥7 – 3𝑥6 + 𝑥5 – 3𝑥4 + 5𝑥3 – 4𝑥2 + 2x –
1 by 2𝑥3
– 3𝑥2
+ x – 1
3. Without actual division show that 2𝑥4
– 7𝑥3
–
2𝑥2 + 13x + 6 is divisible by 𝑥2 – 5x + 6.
4. By synthetic division find the quotient and the
remainder in the division of 6𝑥3
– 10𝑥2
+ 5x + 3
by x – 12.