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Multiplication And
Division of Polynomials
The product of two monomials. We have already learned how to multiply
two or more monomials. Let us review this by multiplying - 5x2y by
7x3y2. The work may be arranged
either horizontally or vertically.
We multiply -5 by 7 to get -35;
we multiply x2 by x3 to get x5; we
multiply y by y2 to get y3; and
dually we indicate the product of
these quantities by writing -35x55y3.

Raising a monomial to a power,
This is a special case of multiplication of monomials; that is,
(3x)2 means (3X)(3X); (5X2)3 means (5x2)(5x2)(5x2). Thus to simplify
(2X2Y3)3, we cube the numerical
coefficient 2 to get 8; we cube x2 to
get x6 (to cube x2 we multiply x2 . x2 . x2); we cube y3 to get y9; and finally
we indicate the product of these quantities by writing 8x6y9.

REVIEW EXERCISES: Multiplication of monomials.
Multiply the upper monomial (the multiplicand) by the
lower monomial (the multiplier).


ANSWERS :
1. 18x ;
2.35y ;
3. 9a3 ;
4. 4b4 ;
5. 27x2y ;
6. -16a4 ;
7. -30x2y
;
8. -175a3b2 ;
9. 15abxy ;
10. -69a2b2 ;
11. -12x3y2z ;
12. -32a2b4 ;
13. 42x3y5 ;
14. -48x4y3 ;
15. 81a4b6 ;
16. -63a8b9 ;
17. 15x ;
18. -6a3 ;
19. 30x2 ;
20. 8y4 ;
21. -6x3y ;
22. - 72a3b ;
23. -24m6n ;
24. 10x5y4 ;
25. 9x2 ;
26. 25a4 ;
27. 25y2 ;
28. -8x3 ;
29. -24x2y2 ;
30. 42x4 ;
31. -64a6b3 ;
32. 24a5b3 ;
33. 32x10 ;
34. 625a8b12
Multiplying a polynomial by a monomial. We have already learned to
multiply a polynomial by a monomial. To multiply the polynomial 3x - 4y + 2 by the monomial 2x, we multiply each term of the polynomial
by 2x and indicate the algebraic sum of these products. This work may
be arranged either horizontally or vertically.

We may use our knowledge of arithmetic to interpret the multiplication
of a polynomial by a monomial. Thus we say that 6(a + 6 + 4) represents the area of a rectangle whose width is 6 and whose length is (a + b + 4).
Let us now divide the length of the rectangle into three parts a, b, and 4. The black vertical lines at the points of division divide the rectangle into three
smaller rectangles. Notice how the diagram below represents the multiplication
of the polynomial a + b + 4 by the monomial 6.


As you remember, to multiply a polynomial by a monomial, multiply each tTerm of the polynomial by the monomial.
REVIEW EXERCISES: Multiplying a polynomial by a monomial
Multiply as indicated:

Multiply the polynomial (the multiplicand) by the monomial (the multiplier).

Answers :
1. 3a +3b + 3c
2. -3x3 -2x2 + x
3. -7a3 + 6a2 - 4
4. x5y3 - x3 y4 + x2y6
5. 15x3 -24x
6. -12x2 + 30x3 -6x4
7. 4a4 -20a3 -28a2
8. -12x7 + 36x6
9. -40a3 -60a2b
10. 2a4b -a3b2 + 3a2b3
11. -18x7 + 21x10
12. 36y6 + 18y9
13. -4x2 + 12x - 48
14. 6x3y -18x2y2 + 3xy3
15. 2a3b - 4a2b2 + 2ab3
16. -9a5b2 + 12a3b4 - 3a2b7
17. -a2bcd + ab2cd + abc2d -abcd2
18. 15x4y4 -15x3y5 + 5x5y6
19. -a8 + a5b3
20. -a4b + 3a3b2 -3a2b3 + ab4
Multiplying a polynomial by another polynomial. Let us first consider how
to obtain the product of two binomials, each of which is numerical; for
example, let us multiply 7 + 3 by 9 - 2. In this problem we can simplify
each of the binomials and then find the product, thus:
(7 + 3)(9 —2)
(10) (7) or 70
Let us arrange the work in slightly different form, writing the first binomial
as the multiplicand and the second binomial as the multiplier:
7+3
9 -2
We multiply each term of the multiplicand, 7 + 3, by 9; then we multiply
each term of the multiplicand by -2; finally, we take the algebraic sum
of these partial products. Thus:

The first method shown above works well in arithmetic where the sum
or difference of two numbers can always be completed; however, in algebra,
when we are dealing with different literal numbers, their sums or their
differences cannot be completed but can be indicated only. Therefore,
in algebra, when we are using literal numbers, the second method shown
above is used.
ILLUSTRATIVE EXAMPLES: Multiplication of polynomials
1. Multiply as indicated: (3x — 4)(2x + 5)
Solution.

2. Multiply 3x2 - x + 2 by 2x - 3.
Check the result
.
Check. To check the result obtained, we let the
literal quantity, here x, equal a positive number
other than the number 1. Therefore, we shall let
x = 2 and substitute 2 for x in each polynomial and
in the product. Thus:

3. In multiplying x + 5 by x + 2, let us represent the multiplication of one polynomial by another polynomial as finding the area of a rectangle whose length and width are given.
Solution :

Let us divide the length of the rectangle into two segments of x and 5,
and the width into two segments of x and 2. The black lines divide the
rectangle into four smaller rectangles. The area of the original rectangle,
of course, equals the sum of the four smaller rectangles. Notice how the
diagram below represents the multiplication of x + 5 by x + 2.

EXERCISES: Multiplication of polynomials
Multiply the upper polynomial (the multiplicand) by the
lower polynomial (the multiplier).


Multiply as indicated and check the result.

Answers
1. 6x2 -11x -10.
2. 35a2 + ab -12b2
3. 6y2 -32y -70
4. 21 -26a + 8a2
5. y2 -2y + 1
6. 18x2 + 27xy -5y2
7. 14a2 -60ab + 16b2
8. 9x2 -y2
9. 4a2 + 12ab + 9b2
10. 2x3 -11x2 + 17x -6
11. 20a3 -29a2 + 56a -60
12. 3x5 + 25x4 -25x3 -63x2
13. x4 -10x3 -26x2 -6x + 5
14. x4 -2x3 + 2x2 -2x + 1
15. x4 -3x2 + 2
16. a3 + 9a2 + 23a + 12
17. 4a3 -16a2b + 17ab2 -3b3
18. 3x4 + 2x3 -19x2 -12x + 8
19. 3x2 -39x + 90
20. 2x2 + 8x -42
21. 24a2 + 34a -10
22. 4a2 -13ab + 3b2
23. a3 +1
24. 2x3 -15x2 + 31x -15
25. 3x3 -x2 +x -35.
26. 5a3 + 29a2b -11ab2 + b3
Power of a binomial. Squaring a binomial, or raising
it to another power, is a special case of the multiplication
of polynomials. We know that (2a)2 means (2a)(2a),
and ( -3x2)3 means ( -3x2)( -3x2)( -3x2); similarly,
(a -3)2means (a -3)(a -3) and (2x -y)3 means
(2x -y)(2x -y)(2x -y). Thus to simplify an expression such as (3x -2y)2, we multiply 3x -2y by 3x -2y as in the preceding exercises. Thus:


To simplify (2x -y + 3)2 we multiply 2x -y + 3 by 2x -y + 3, as previously explained. Thus:

Answer
EXERCISES: Raising a polynomial to a power
Raise each of the following to the indicated power:
- (2x - y)2
- (3x + y)2
- (x - 2y)2
- (a - 4b)2
- (3a - 2b)2
- (6x - 4y)2
- (2m + 5n)2
- (5x -7y)2
- (x - 9y)2
- (8a + 3b)2
- (a + b)3
- (2x - y)3
- (m + 3 n)3
- (x - y + z)2
- (3a + 4b - 2c)2
Answers .

Special cases in the multiplication of polynomials. The special cases we
shall consider are examples in which one or both polynomials are not
written in descending (or ascending) powers of some letter and in which
one or more terms in the sequence of exponents is missing.
ILLUSTRATIVE EXAMPLE: Multiplication of polynomials
Multiply x3 + 7 - 3x by 5x - 2 + 4x2
Solution. In the first polynomial the term containing x2 in the sequence
of exponents is missing. First we arrange each polynomial in descending
powers of x. We therefore leave a space for the missing term. Thus the
first polynomial is written x3 - 3x + 7, and the second polynomial is
written 4x2 + 5x - 2. Making the second polynomial the multiplier, we
write:
Answer
EXERCISES: Multiplication of polynomials
First arrange each polynomial according to the descending powers of
the literal quantity and then multiply as illustrated above. Check the
results obtained.
1. (2 + 3x3 — 5x)(2x — 5)
2. (2x — 4 + x2)(3x2 + 3 - 2x)
3. (5a2 + a3 - 8)(6 + 4a2 - a)
4. (2y - y2 + 4)(8 - 3y)
5. (x2 - 1 + 2x)(3 - 2x)
6. (x2 - 5 + 2x)2
7. (3x - 2 + 7x2)2
8. (8x - 1 + x3)(6 - 5x)
9. (x3 - 2 + 3x)2
10. (5x2 - x3 +2x - 2)(3x +x2 - 4)
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