


Multiplication of Polynomials
Multiplication of Polynomials Deﬁnition and Notation The product of two natural numbers, 3and 4, is deﬁned by 3 x 4 = 4+4+4 three terms of 4 The following are some of the laws from 1. The commutative law of multiplication: ab = ba 2. The associative law of multiplication: a (bc) = (ab) c 3. The distributive law of multiplication: a (b+c) = (b+c) a =ab+ac 4. Multiplication of signed numbers: (+ a) (+ b) = + ab; (+ a) (− b) = − ab (− a) (+ b) = − ab; (− a) (− b) = + ab When we have 2·2·2·2. that is, four factors of 2, the notation 2^{4} is used, which reads. “two to the power four," or “two to the fourth power." Similarly, a·a·a·a·a = a^{5} means ﬁve factors of a. The a is called the base, and the 5 is called the exponent. When there is no exponent, as in x, we always mean x to the power 1. DEFINITION
Note the difference between (− 2^{4}) = (− 2) (− 2) (− 2) (− 2) = +16 and − 2^{4} = − (2^{4}) = (2·2·2·2) = − 16 Note also 2 a^{3} = 2 (a·a·a) While (2 a)^{3} = (2 a) (2 a) (2 a) =(2·2·2) (a·a·a) =2^{3} a^{3} = 8 a^{3} Remark a , a^{2} , a^{3},.... are not like terms. EXAMPLES 1. 7 a·a·a·a = 7 a^{4} 2. − (− 3) (− 3) (− 3) (− 3) = − (− 3)^{4} 3. (x1)^{3} = (x1) (x1) (x1) 4. − 2^{2}·3^{3} = − (2·2)·(3·3·3) = − 4·27 = − 108 5. 2^{2}+2^{3} = 2·2+2·2·2 = 4+8 = 12 6. 2^{3}2 = 2·2·22=82 = 6
Multiplication of Monomials We will discuss the multiplication of monomials, then the multiplication of a monomial and a polynomial, and ﬁnally the multiplication of two polynomials. From the deﬁnition of exponents we have a^{3}·a^{5} = (a·a·a) (a·a·a·a·a) =a·a·a·a·a·a·a·a =a^{8} =a^{3+5} THEOREM 1
Proof
EXAMPLES 1. 2^{3}·2^{5} = 2^{3+5} = 2^{8} 2. a^{2}·a^{4} = a^{2+4} = a^{6} 3. − 2^{4}·2^{3} = − 2^{4+3} = − 2^{7} 4. − 3 x^{3}·x^{2} = − 3 x^{3+2} = − 3 x^{5} 5. x^{5}·x = x^{5+1} = x^{6} 6. (a+1)^{2}·(a+1)^{3} = (a+1)^{2+3} = (a+1)^{5}
Let's see how our Polynomial solver simplifies this and similar problems. Click on "Solve Similar" button to see more examples.
Remark 2^{3}·2^{7} = 2^{3+7} = 2^{10} , and not 4^{10} Remark 2^{4}·3^{5} = 2^{4}·3^{5} , to find the product, multiply 2^{4} =16 by 3^{5} =243 ; that is 2^{4}·3^{5} = (16) (243) = 3888 Since the commutative and associative laws for multiplication hold for numbers, speciﬁc or general. we have
EXAMPLES 1. (2 ab^{2}) (3 a^{4} bc^{2})= (2·3) (a^{1}·a^{4}) (b^{2}·b^{1}) (c^{2}) = 6 a^{5} b^{3} c^{2} 2. (− 3 b^{2} c^{3}) (8 ab^{3} c) = (− 3·8) (b^{2}·b^{3}) (c^{3}·c) (a) =− 24 b^{5} c^{4} a 3. (− 3^{2} xy^{2}) (− 5 x^{2} y^{3}) = (− 9 xy^{2}) (− 5 x^{2} y^{3}) =(− 9) (− 5) (x·x^{2}) (y^{2}·y^{3})=45 x^{3} y^{5}
Let's see how our Polynomial solver simplifies this and similar problems stepbystep. Click on "Solve Similar" button to see more examples.
From the deﬁnition of exponents we have (a^{2})^{3} = (a^{2}) (a^{2}) (a^{2}) =(a^{2}·a^{2}) (a^{2}) =a^{2+2}·a^{2} =a^{2+2+2} =a^{3·2} = a^{2·3} =a^{6}
THEOREM 2
Proof
EXAMPLES 1. (3^{2})^{4} = 3^{2·4} = 3^{8} 2. (a^{3})^{5} = a^{3·5} = a^{15} 3. (− 3^{2})^{3} = − 3^{2·3} = − 3^{6} 4. (− a^{3})^{2} = a^{3·2} = a^{6}
Note 2^{3}·2^{4} = 2^{3+4}=2^{7}, While (2^{3})^{4} = 2^{3·4}=2^{12} From the deﬁnition of exponents we have 6^{4} = (2·3)^{4} = (2·3) (2·3) (2·3) (2·3) = (2·2·2·2) (3·3·3·3) =2^{4}·3^{4} THEOREM 3
Proof
Note a and b are factors. If a=3, b=x, and m=5,(3 x)^{5}=3^{5} x^{5} Do not forget to raise the number 3 to the power 5 Applying Theorem 3 repeatedly, we obtain (abcd)^{m}=[(ab) (cd)]^{m} =(ab)^{m} (cd)^{m} =a^{m}b^{m}c^{m}d^{m} Remark 21^{2} = (3·7)^{2} = 3^{2}·7^{2} = 9 x49 = 441 Remark The quantity (5+3)^{2} = (8)^{2} = 64, but 5^{2}+3^{2} = 25+9 = 34 If we consider (a+b) as one quantity, then (a+b)^{5} = (a+b) (a+b) (a+b) (a+b) (a+b) The method of calculating the product will be explained later COROLLARY (a^{m}) (b^{n})^{k} = [(a^{m}) (b^{n})]^{k} =(a^{m})^{k} (b^{n})^{k} =a^{mk}b^{nk} EXAMPLE Perform the following multiplication: (2 x^{2} yz^{3})(− 4 x^{3} y^{2}) Solution (2 x^{2} yz^{3})(− 4 x^{3} y^{2}) = (2) (− 4) (x^{2}·x^{3}) (y·y^{2}) (z^{3}) = − 8 x^{5} y^{3} z^{3}
Let's see how our Polynomial solver simplifies this and similar problems, showing explanations for each step. Click on "Solve Similar" button to see more solved examples. EXAMPLE Perform the following multiplication: (5 a^{2} b)^{3} Solution (5 a^{2} b)^{3} = (5)^{3} (a^{2})^{3} (b)^{3} = 5^{3} a^{6} b^{3} = 125 a^{6} b^{3}
EXAMPLE Perform the following multiplication: − 2^{2} a^{3} (ab^{3})^{2} Solution − 2^{2} a^{3} (ab^{3})^{2} = − 4 a^{3} (a^{2} b^{6}) = − 4 (a^{3}·a^{2}) (b^{6}) = − 4 a^{5} b^{6}
EXAMPLES Perform the following multiplication: (3 x^{2} y)^{2} (2 xy^{3})^{3} Solution (3 x^{2} y)^{2} (2 xy^{3})^{3} = (3^{2} x^{4} y^{2}) (2^{3} x^{3} y^{9}) = (3^{2}·2^{3}) (x^{4}·x^{3}) (y^{2}·y^{9}) =(9·8) x^{7} y^{11} = 72 x^{7} y^{11} Remark Perform the outside exponents first
EXAMPLE Perform the following multiplication: (− 2 ab^{2})^{2} (− 3 a^{2} b)^{3} (− bc^{2})^{4} Solution (− 2 ab^{2})^{2} (− 3 a^{2} b)^{3} (− bc^{2})^{4} = (− 2)^{2} a^{2} b^{2}·(− 3)^{3} a^{6} b^{3}·(− 1)^{4} b^{4} c^{8} =(− 2)^{2} (− 3)^{3} (− 1)^{4} (a^{2}·a^{6}) (b^{4}·b^{3}·b^{4}) (c^{8}) =(4) (− 27) (+ 1) a^{8} b^{11} c^{8} =− 108 a^{8} b^{11} c^{8}
EXAMPLE Perform the indicated operations and simplify: (2 ab)^{4} (− a^{3} b)^{2}(− 3 a^{2})^{3} (a^{2} b^{3})^{2} Solution (2 ab)^{4} (− a^{3} b)^{2}(− 3 a^{2})^{3} (a^{2} b^{3})^{2} = (16 a^{4} b^{4}) (a^{6} b^{2})(− 27 a^{6}) (a^{4} b^{6}) =16 a^{10} b^{6}+27 a^{10} b^{6} =43 a^{10} b^{6}
Let's see how our Polynomial calculator explains all simplification steps
Note: To evaluate expressions involving exponents. ﬁrst replace each letter by its indicated speciﬁc value. Use grouping symbols where necessary so as not to confuse operation signs with number signs.
Evaluate − a^{2} b^{3}, give that a = − 3 and b = 2 Solution − a^{2} b^{3} = − (− 3)^{2} (2)^{3} = − (9) (8) = − 72
EXAMPLE Evaluate the expression b^{2}a^{2} (c^{3}b^{3}), give that a = − 2 , b = 3, and c = − 1 Solution b^{2}a^{2} (c^{3}b^{3}) = (3)^{2}(− 2)^{2} [((− 1)^{3}(3)^{3})] =9(+ 4) [(− 1)(27)] =94 (− 127) =94 (− 28) =9+112 = 121 Multiplication of a Polynomial by a Monomial The extended distributive law of multiplication
is used to multiply a monomial by a polynomial EXAMPLE Multiply 3 x^{2}+x2 by x Solution x (3 x^{2}+x2) = x (3 x^{2})+x (x)+x (− 2) =3 x^{2}+x^{2}2 x EXAMPLE Multiply x^{2}x+4 by − 2 x^{2} Solution (− 2 x^{2}) (x^{2}x+4) = (− 2 x^{2}) (x^{2}) + (− 2 x^{2}) (− x) + (− 2 x^{2}) (4) =− 2 x^{4}+2 x^{3}8 x^{2}
EXAMPLE Multiply a^{2} b2 b^{2} c+5 c^{2} a by 3 a^{2} b Solution 3 a^{2} b (a^{2} b2 b^{2} c+5 c^{2} a) =3 a^{2} b (a^{2} b)+3 a^{2} b (− 2 b^{2} c)+3 a^{2} b (5 c^{2} a) =3 a^{4} b^{2}6 a^{2} b^{3} c+15 a^{3} bc^{2} EXAMPLE Multiply 3 x24  2 x16 by 12 Solution 121[3 x242 x16] = 121[3 x24]  121 [2 x16] =3 (3 x2)2 (2 x1) =9 x64 x+2 =5 x4
Multiplication of Polynomials Multiplication of two polynomials is the same as multiplication of a monomial and a polynomial where the ﬁrst polynomial is considered as one quantity. To multiply (x+2) by (x3), consider (x+2) as one quantity and apply the distributive law:
Then reapply the distributive law =x^{2}+2 x3 x6 =x^{2}x6 Notice that each term of the second polynomial has been multiplied by each term of the ﬁrst polynomial. The same result can be obtained by arranging the polynomials in two rows and multiplying the upper polynomial by each term of the lower polynomial. Arrange like terms of the product in the same column so that addition is easier.+
EXAMPLE Multiply (3 x4)^{2} Solution (3 x4)^{2} = (3 x4) (3 x4)
Notes 1. (a+b)^{2} = a^{2}+2 ab+b^{2} 2. (ab)^{2} = a^{2}2 ab+b^{2} 3. (a+b) (ab) = a^{2}b^{2} EXAMPLE Multiply (x^{2}2 x+1) by (2 x3) Solution
Therefore (x^{2}2 x+1) (2 x3) = 2 x^{3}7 x^{2}+8 x3 EXAMPLE Perform the indicated operations and simplify (2 x3) (x+4)(x+2) (x6) Solution (2 x3) (x+4)(x+2) (x6) =(2 x^{2}+ 5 x12)(x^{2}4 x12) =2 x^{2}+5 x12x^{2}+4 x+12 =x^{2}+9 x
Grouping Symbols Grouping symbols, such as parentheses ( ), braces { }, and brackets [ ], are used to designate, in a simple manner, more than one operation. When we write the binomial 3 a+5 b as (3 a+5 b), we are considering the sum of 3 a and 5 b as one quantity. The expression a(b+c) means the sum of b and c is to be subtracted from a. The statement, three times x minus four times the sum of y and z, can be written in algebraic notation as 3 x4 (y+z)Removal of the grouping symbols means performing the operations that these symbols indicate. Remove the symbols one at a time, starting with the innermost, following the proper order of operations to be performed. EXAMPLE Remove the grouping symbols and combine like terms 2 x(5 x2 y)+(x6 y) Solution 2 x(5 x2 y)+(x6 y) = 2 x5 x+2 y+x6 y =(2 x5 x+x)+(2 y6 y) =− 2 x4 y
Let's see how our polynomial solver simplifies this and similar problems. Clicking on "Solve Similar" button will show more stepbystep solved examples. EXAMPLE Remove the grouping symbols and combine like terms 7 a+2 [2 b3 (3 a5 b)] Solution 7 a+2 [2 b3 (3 a5 b)] = 7 a+2 [2 b9 a+15 b] =7 a+4 b18 a+30 b =34 b11 a
Let's see how our polynomial calculator explains all the steps for this and similar problems. Click on "Solve Similar" button to see more examples solved.
EXAMPLE Remove the grouping symbols and combine like terms 6 a{2 b+[3(a+b)+(5 a2)]} Solution 6 a{2 b+[3(a+b)+(5 a2)]} =6 a{2 b+[3ab+5 a2]} =6 a{2 b+3ab+5 a2} =6 a2 b3+a+b5 a+2 =(6 a+a5 a)+(− 2 b+b)+(− 3+2) =2 ab1 Let's see how our polynomial calculator solves this and similar problems. Click on "Solve Similar" button to see more examples. It is sometimes necessary to group some of the terms of an expression. This can be accomplished by use of a set of parentheses. When the grouping symbol is preceded by a plus sign. we keep the signs of the terms the same when it is preceded by a minus sign, we use the additive inverses (negatives) of the terms.
EXAMPLE Group the last three terms of the polynomial 3a  5b + c  2 with a grouping symbol in two ways, one preceded by a plus sign, the second preceded by a minus sign. Solution
