Introduction to Polynomial Division and Linear Factors
Polynomial division is a fundamental concept in algebra, particularly when determining remainders. This article explores how to find the remainder when a polynomial is divided by linear factors. Specifically, we will delve into a problem where the polynomial ( p(x) ) leaves remainders of -1, 3, and 4 when divided by ( x - 1 ), ( x - 2 ), and ( x 3 ) respectively. This scenario can be solved using the Remainder Theorem and algebraic manipulation techniques.
Understanding the Problem
Given the polynomial ( p(x) ), we know it leaves remainders as follows:
Remainder when ( p(x) ) is divided by ( x - 1 ): -1 Remainder when ( p(x) ) is divided by ( x - 2 ): 3 Remainder when ( p(x) ) is divided by ( x 3 ): 4We need to find the remainder ( r(x) ) when ( p(x) ) is divided by ( (x - 1)(x - 2)(x 3) ). Let's denote the remainder as ( r(x) ).
Solution Outline
To solve this problem, we assume ( r(x) ) is a polynomial of degree at most 2, as the degree of the divisor is 3. Therefore, ( r(x) ax^2 bx c ).
Analyzing the Given Conditions
From the problem, we have the following conditions:
R1: ( r(1) -1 ) R2: ( r(2) 3 ) R3: ( r(-3) 4 )Setting Up Equations
We can set up a system of equations based on the given conditions:
E1: ( a(1)^2 b(1) c -1 ) E2: ( 4a(2)^2 2b(2) c 3 ) E3: ( 9a(-3)^2 - 3b(-3) c 4 )These can be simplified as follows:
E1: ( a b c -1 ) E2: ( 16a 4b c 3 ) E3: ( 81a 9b c 4 )Solving the System of Equations
We can solve these equations to find the values of ( a ), ( b ), and ( c ). Let's start by eliminating ( c ) from equations by subtracting E1 from E2 and E3:
E4: ( 4a 3b 4 ) E5: ( 80a 8b 5 )Solving E4 for ( b ):
[ b frac{4 - 4a}{3} ]
Substituting this into E5:
[ 80a 8 left(frac{4 - 4a}{3}right) 5 ]
Multiplying through by 3 to clear the fraction:
[ 240a 32 - 32a 15 ]
[ 208a -17 ]
[ a frac{-17}{208} frac{21}{20} ]
Substitute ( a frac{21}{20} ) back into the expression for ( b ):
[ b frac{4 - 4left(frac{21}{20}right)}{3} frac{4 - frac{84}{20}}{3} frac{4 - 4.2}{3} frac{-0.2}{3} frac{17}{20} ]
Now substitute ( a ) and ( b ) back into E1 to find ( c ):
[ frac{21}{20} frac{17}{20} c -1 ]
[ frac{38}{20} c -1 ]
[ c -1 - frac{38}{20} -frac{20}{20} - frac{38}{20} -frac{58}{20} -frac{29}{10} ]
So, the remainder polynomial is:
[ r(x) frac{21}{20}x^2 frac{17}{20}x - frac{29}{10} ]
Final Calculation
To find ( r(6) ), we substitute ( x 6 ) into ( r(x) ):
[ r(6) frac{21}{20} cdot 6^2 frac{17}{20} cdot 6 - frac{29}{10} ]
[ r(6) frac{21}{20} cdot 36 frac{17}{20} cdot 6 - frac{29}{10} ]
[ r(6) frac{756}{20} frac{102}{20} - frac{58}{20} ]
[ r(6) frac{756 102 - 58}{20} ]
[ r(6) frac{800}{20} 40 ]
Therefore, the value of ( r(6) ) is:
boxed{40}