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​Algebra 1   State Test Practice

Finding a missing constant and zeros of a function

1/14/2016

 
The function f is defined as f(x) = x(x² - 4) - 3x(x - 2).
Part A
An equivalent form of f is given as
f(x) = x(x - 2)(x - a), where a is a constant.
​What is the value of a?
Part B
Which values are zeros of the function, f?
Select all that apply.
A.)  -3     E.)  1
B.)  -2     F.)  2
C.)  -1     G.)  3
D.)  0

Part A
f(x) = x(x² - 4) - 3x(x - 2)
          and
f(x) = x(x - 2)(x - a)
are equivalent.  Find the value of a.


Therefore,

x(x² - 4) - 3x(x - 2) = x(x - 2)(x - a)   {set the functions equal to each other}
x(x + 2)(x - 2) - 3x(x - 2) = x(x - 2)(x - a)   {factored the x² - 4 into (x + 2)(x - 2)}
x + 2 - 3 = x - a   {divided each side by x(x - 2)}
x - 1 = x - a   {combined like terms}
-a = -1   {subtracted x from each side}
a = 1   {divided each side by -1}


Part B
Which values are zeros of the function, f?
f(x) = x(x - 2)(x - 1)   {replaced a with 1 in the equivalent form, because it is already factored}
x(x - 2)(x - 1) = 0   {substituted 0, for f(x), to find the zeros}
x = 0   or   x - 2 = 0   or   x - 1 = 0   {set each factor equal to 0}
x = 0   or   x = 2   or   x = 1   {solved each equation for x}

D.) 0,   E.) 1,  and  F.) 2  are the zeros

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