# SOLUTION: A(x-2)^2 + B(x-2) +C &#8801; 3x^2 -13x + 19 What are the values of the constants A, B and C such that the above is a true identity for all values of x.

Algebra ->  Algebra  -> Polynomials-and-rational-expressions -> SOLUTION: A(x-2)^2 + B(x-2) +C &#8801; 3x^2 -13x + 19 What are the values of the constants A, B and C such that the above is a true identity for all values of x.      Log On

 Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

 Algebra: Polynomials, rational expressions and equations Solvers Lessons Answers archive Quiz In Depth

 Question 567617: A(x-2)^2 + B(x-2) +C ≡ 3x^2 -13x + 19 What are the values of the constants A, B and C such that the above is a true identity for all values of x.Answer by Edwin McCravy(8912)   (Show Source): You can put this solution on YOUR website!```Notice that the sign of identity ≡ is different from the sign of equality =. It is stronger than =. When there is an ≡ that means that no matter what value you substitute for x, an equation will always result. There are two ways to do this. I'll show you both ways: 1. By substitution of arbitrary numbers and solving the resulting system of equations. Since the identity: A(x-2)² + B(x-2) + C ≡ 3x² - 13x + 19 contain three constants, A,B, and C. Choose 3 arbitrary numbers. We pick the easy three, 0,1, and 2, but we could pick any three numbers. Substitute 0 for x in A(x-2)² + B(x-2) + C ≡ 3x² - 13x + 19 A(0-2)² + B(0-2) + C = 3(0)² - 13(0) + 19 4A - 2B + C = 19 Substitute 1 for x in A(x-2)² + B(x-2) +C ≡ 3x² - 13x + 19 A(1-2)² + B(1-2) + C = 3(1)² - 13(1) + 19 A - B + C = 9 Substitute 2 for x in A(x-2)² + B(x-2) + C ≡ 3x² - 13x + 19 A(2-2)² + B(2-2) + C = 3(2)² - 13(2) + 19 C = 12 - 26 + 19 C = 5 So we have this system of equations: 4A - 2B + C = 19 A - B + C = 9 C = 5 which is easy to solve and get A=3, B=-1 and C=5 A(x-2)² + B(x-2) + C ≡ 3x² - 13x + 19 becomes 3(x-2)² - 1(x-2) + 5 ≡ 3x² - 13x + 19 ------------------------------------------------- 2. By equating like terms A(x-2)² + B(x-2) + C ≡ 3x² - 13x + 19 First we square out the first term and remove the parentheses: A(x² - 4x + 4) + Bx - 2B + C ≡ 3x² - 13x + 19 Equate the x² terms on both sides: Ax² - 4Ax + 4A + Bx - 2B + C ≡ 3x² - 13x + 19 Ax² ≡ 3x² A = 3 Equate the x terms on both sides: Ax² - 4Ax + 4A + Bx - 2B + C ≡ 3x² - 13x + 19 -4Ax + Bx ≡ -13x -4A + B = -13 and since A = 3 -4(3) + B = -13 -12 + B = -13 B = -1 Equate the constant terms on both sides: Ax² - 4Ax + 4A + Bx - 2B + C ≡ 3x² - 13x + 19 4A - 2B + C = 19 and sinnce A = 3 and B = -1 4(3) - 2(-1) + C = 19 12 + 2 + C = 19 14 + C = 19 C = 5 So we have A=3, B=-1 and C=5 A(x-2)² + B(x-2) + C = 3x² - 13x + 19 becomes 3(x-2)² - 1(x-2) + 5 = 3x² - 13x + 19 Edwin```