SOLUTION: 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.

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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(20056)   (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
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