


Division of Polynomials
Division of Polynomials The following are some of the properties pertaining to fractions. These properties are discussed in Chapter 2. 1. ab = a cb c 2. a+bc = ac+bc 3. ab·cd = a cb d 4. ab÷cd = ab·dc Note Since division by zero is not deﬁned, all denominators are assumed different from zero First we will discuss division of monomials, then division of a polynomial by a monomial, and ﬁnally division of two polynomials.
Division of Monomials From the properties of fractions and the rules governing exponents we have a^{8}a^{5} = a^{5}·a^{3}a^{5}·1 = a^{3}1 = a^{3} =a^{85} a^{4}a^{4}= 1 a^{7}a^{10} = a^{7}·1a^{7}·a^{3} = 1a^{3} =1a^{107}
THEOREM 4
Proof a^{m}a^{n} = a^{n}·a^{mn}a^{n}·1 = a^{mn} When m>n a^{m}a^{n} = a^{n}a^{n} = 1 When m=n a^{m}a^{n} = a^{m}·1a^{m}·a^{nm} = 1a^{nm} When {m?<s t r o n g>e X A M P L e<?s t r o n g>1. [[2^{6}2^{2}]] = 2^{62} = 2^{4} 2. a^{7}a^{5} = a^{75} = a^{2} 3. (a1)^{4}(a1)^{3} = (a1)^{43} = (a1) 4. 5^{4}5^{4} = 1 5. (x+1)^{3}(x+1)^{3} = 1 6. 3^{8}3^{12} = 13^{128} = 13^{4} 7. a^{3}a^{9} = 1a^{93} = 1a^{6} 8. (x+2)^{4}(x+2)^{6} = 1(x+2)^{64} = 1(x+2)^{2} From the properties of fractions and the deﬁnition of exponents we have (23)^{4} = 23·23·23·23 = 2·2·2·23·3·3·3 = 2^{4}3^{4}
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THEOREM 5
Proof
COROLLARY then by the use of Theorem 2 and Theorem 3 and theorem 5, we have (a^{m} b^{n}c^{p} d^{q})^{k} = (a^{m} b^{n})^{k}(c^{p} d^{q})^{k} = a^{m k} b^{n k}c^{p k} d^{q k}
EXAMPLE Simplify − 30 a^{3} b^{2}12 a^{2} b^{4} by applying the laws of exponents. Solution − 30 a^{3} b^{2}12 a^{2} b^{4} = − (2·3·5 a^{3} b^{2}2·2·3 a^{2} b^{4}) =− (2·32·2)·(52)·(a^{3}a^{2})·(b^{2}b^{4}) =− (52)·(a1)·(1b^{2}) =− (5 a2 b^{2})
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EXAMPLE By applying the laws of exponents, simplify [2 x^{4} y z6 x y^{2}]^{3} Solution We can simplify the fraction first before applying the outside exponent. [2 x^{4} y z6 x y^{2}]^{3} = [x^{3} z3 y]^{3} = x^{9} z^{3}3^{3} y^{3} = x^{9} z^{3}27 y^{3}
EXAMPLE 12^{4}18^{3} = (2^{2}·3)^{4}(2·3^{2})^{3} = 2^{8}·3^{4}2^{3}·3^{6} =(2^{8}2^{3})·(3^{4}3^{6}) = (2^{5}1)·(13^{2}) = 329
EXAMPLE Simplify (2 a^{2} b c^{3})^{3}(3 a b^{2})^{2} by applying the laws of exponents Solution Here we cannot simplify ﬁrst. since the numerator and the denominator have different powers. Apply the outside exponents ﬁrst. then simplify, (2 a^{2} b c^{3})^{3}(3 a b^{2})^{2} = 2^{3} a^{6} b^{3} c^{9}3^{2} a^{2} b^{4} = 8 a^{4} c^{9}9 b
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EXAMPLE Perform the indicated operations and simplify: 16 a^{4} b^{3}÷(− 2 a b)^{3} + 36 a^{5} b^{2}÷(− 3 a^{2} b)^{2} Solution 16 a^{4} b^{3}÷(− 2 a b)^{3} + 36 a^{5} b^{2}÷(− 3 a^{2} b)^{2} =16 a^{4} b^{3}(− 2 a b)^{3} + 36 a^{5} b^{2}(− 3 a^{2} b)^{2} =16 a^{4} b^{3}− 2^{3} a^{3} b^{3} + 36 a^{5} b^{2}3^{2} a^{4} b^{2} =16 a^{4} b^{3}− 8 a^{3} b^{3} + 36 a^{5} b^{2}9 a^{4} b^{2} =− 2 a+4 a = 2 a
Division of a Polynomial by a Monomial From the properties of fractions we have
Keep in mind that a+bc means (a+b)÷c but
a+ba = aa+ba = 1+ba To divide a polynomial by a monomial, divide every term of the polynomial by the monomial.
EXAMPLE Divide 12 x^{3}6 x^{2}+18 x6 x and simplify. Solution 12 x^{3}6 x^{2}+18 x6 x = 12 x^{3}6 x+− 6 x^{2}6 x+18 x6 x =2 x^{2}x+3
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EXAMPLE Divide 3 a^{3}2 a^{2} ba b^{2}− a b and simplify. Solution 3 a^{3}2 a^{2} ba b^{2}− a b = 3 a^{3}− a b+− 2 a^{2} b− a b+− a b^{2}− a b =− (3 a^{2}b)+2 a+b
EXAMPLE Perform the indicated operations and simplify: 12 a^{4}+4 a^{3}32 a^{2}4 a^{2}  (3 a8) (a+1) Solution 12 a^{4}+4 a^{3}32 a^{2}4 a^{2}  (3 a8) (a+1) =(3 a^{2}+a8)(3 a^{2}5 a8) =3 a^{2}+a83 a^{2}+5 a+8 = 6 a
Division is deﬁned as the inverse operation of multiplication; thus we start with a multiplication problem and then deduce the division operation. (x^{2}+3 x5) (2 x7) = x^{2} (2 x7)+3 x (2 x7)+(− 5) (2 x7) =(2 x^{3}7 x^{2})+(6 x^{2}21 x)+(− 10 x+35) =2 x^{3}x^{2}31 x+35
Hence if (2 x^{3}x^{2}31 x+35) is divided by (2 x7), the result is (x^{2}+3 x5), the first polynomial of multiplication problem. The polynomial (2 x^{3}x^{2}31 x+35) is called the dividend, (2 x7) is called the divisor, and (x^{2}+3 x5) is called the quotient. The first term of the dividend, 2 x^{3}, comes from multiplying the first term of quotient, x^{2}, by the first term of the divisor, 2 x. Thus to obtain the first term of quotient, x^{2}, we divide the first term of the dividend , 2 x^{3} , by the first term of the divisor, 2 x. Multiplying the entire divisor (2 x7) by that ﬁrst term of the quotient, x^{2}, we get 2 x^{3}7 x^{2}. Subtracting 2 x^{3}7 x^{2} from the dividend, (2 x^{3}x^{2}31 x+35)(2 x^{3}7 x^{2}) = 6 x^{2}31 x+35 The quantity 6 x^{2}31 x+35 is our new dividend. The ﬁrst term, 6 x^{2}, of the new dividend comes from multiplying the second term of the quotient, 3 x, by the first term of the divisor, 2 x. Thus to get the second term of the quotient, 3 x, divide the ﬁrst term of the new dividend, 6 x^{2}, by the ﬁrst term of the divisor, 2 x. Multiplying the divisor (2 x7) by the second term of the quotient, 3 x, we get 6 x^{2}21 x. Subtracting 6 x^{2}21 x from the new dividend, (6 x^{2}31 x+35)(6 x^{2}21 x) = − 10 x+35 The quantity − 10 x+35 is the new dividend. Dividing the first term of the new dividend (− 10 x) by the first term of the divisor, 2 x, we get the third term of the quotient (− 5). Multiplying the divisor (2 x7) by the third term of the quotient (− 5), we get − 10 x+35. Subtracting (− 10 x+35) from the dividend (− 10 x+35), We get zero.
Let us start the problem again an arrange it in a manner analogous to that of long division in arithmetic. Hence 2 x^{3}x^{2}31 x+352 x7 = x^{2}+3 x5
DEFINITION The degree of a polynomial in a literal number is the greatest exponent of that literal number in the polynomial. EXAMPLE x^{5} y7 x^{4} y^{2}2 x^{3} y^{3}+9 y^{4} is a polynomial of degree 5 in x and degree 4 in y. To divide two polynomials, we start by arranging the terms of the dividend according to the decreasing exponents of one of the literals, leaving spaces for the missing powers (including terms with zero coefﬁcients for the missing terms). Arrange the terms of the divisor also according to the decreasing exponents of the same literal used in arranging the terms of the dividend. Divide the ﬁrst term of the dividend by the ﬁrst term of the divisor to get the ﬁrst term of the quotient. Multiply the ﬁrst term of the quotient by each term of the divisor and write the product under the like terms in the dividend. Subtract the product from the dividend to arrive at a new dividend.
EXAMPLE Divide (6 x^{3}17 x^{2}+16) by (3 x4). Solution Write the dividend as 6 x^{3}17 x^{2}+0 x+16
EXAMPLE Divide (19 x^{2}10 x^{3}+x^{5}14 x+6) by (x^{2}+12 x). Solution Write the dividend as x^{5}+0 x^{4}10 x^{3}+19 x^{2}14 x+6. Write the divisor as x^{2}2 x+1
Hence x^{5}10 x^{3}+19 x^{2}14 x+6x^{2}2 x+1 =x^{3}+2 x^{2}7 x+3+− x+3x^{2}2 x+1 =x^{3}+2 x^{2}7 x+3x3x^{2}2 x+1 Note This form is similar to the form used in arithmetic when we write 207 = 2+67
EXAMPLE Divide (2 x^{4}3 y^{4}13 x^{2} y^{2}+14 x y^{3}) by (x^{2}+2 x y3 y^{2}) Solution Write the dividend as 2 x^{4}+0 x^{3} y13 x^{2} y^{2}+14 x y^{3}3 y^{4}.
Hence 2 x^{4}13 x^{2} y^{2}+14 x y^{3}3 y^{4}x^{2}+2 x y3 y^{2} = 2 x^{2}4 x y+y^{2}
