Question 141506
Which of the following is another version of {{{f(x) = 2x^2 - 4x - 7}}} ? 
a. {{{f(x) = 2(x -1)^2 - 9}}}
b. {{{f(x) = (x - 1)^2 - 9}}}
c. {{{f(x) = (x + 1)^2 - 1/2}}}
d. {{{f(x)}}} cannot be factored; prime
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We can eliminate d because none of the choices are
factorizations of f(x) anyway.

The easy way to find which one is correct is to substitute
an arbitrary value for {{{x}}}, ({{{0}}} is a good choice since it's easy)
in the original and in each of the choices.

Substituting x=0 in the original:

{{{f(x) = 2x^2 - 4x - 7}}}
{{{f(0) = 2(0)^2 - 4(0) - 7}}}
{{{f(0) = -7}}}

Now substitute {{{x=0}}} in the answer a.

{{{f(x) = 2(x-1)^2 - 9}}}
{{{f(0) = 2(0-1)^2 - 9}}}
{{{f(0) = 2(-1)^2 - 9}}} 
{{{f(x) = 2(1) - 9}}}
{{{f(x) = 2 - 9}}}
{{{f(x) = -7}}}

So a. is possible.

Now substitute {{{x=0}}} in the answer b.

{{{f(x) = (x-1)^2 - 9}}}
{{{f(0) = (0-1)^2 - 9}}}
{{{f(0) = (-1)^2 - 9}}} 
{{{f(x) = (1) - 9}}}
{{{f(x) =  1- 9}}}
{{{f(x)=-8}}}

So b. is NOT possible.

Now substitute {{{x=0}}} in the choice c.

{{{f(x) = (x + 1)^2 - 1/2}}}
{{{f(x) = (x+1)^2 - 1/2}}}
{{{f(0) = (0+1)^2 - 1/2}}}
{{{f(0) = (1)^2 - 1/2}}} 
{{{f(x) = 2(1) - 1/2}}}
{{{f(x) = 2 - 1/2}}}
{{{f(x) = 4/2-1/2}}}
{{{f(x) = 3/2}}}

So c. is NOT possible.

Thus the only possible choice is a.

--------------

However, that's not what your teacher wants you
to do, although it is one way to get the correct
answer on a multiple choice test. I just thought 
I'd show you the above method of substituting in
each of the answers to rule out the incorrect 
choices. This may save you from having to miss a 
problem which you can't solve on a multiple 
choice test. Of course, this method will not work 
on the kind of test where you have to show your 
work.

-------------

Here is what your teacher wanted you to do:

{{{f(x) = 2x^2 - 4x - 7}}}

Factor 2 out of the first two terms only:

{{{f(x) = 2(x^2-2x) - 7}}}

We will complete the square inside the parentheses:

Multiply the coefficient of x, which is -2, by {{{1/2}}},
getting -1.  Then square -1 getting +1

So add +1 and immediately subtract 1 inside the 
parentheses:

{{{f(x) = 2(x^2-2x+1-1) - 7}}}

Change the parentheses to brackets so they can hold 
parentheses:

{{{f(x) = 2}}}[{{{x^2-2x+1-1}}}]{{{ - 7}}}

Factor only the first three terms inside the brackets:

{{{f(x) = 2}}}[{{{(x-1)(x-1)-1}}}]{{{ - 7}}}

{{{f(x) = 2}}}[{{{(x-1)^2-1}}}]{{{ - 7}}}

Now remove the brackets by distributing the 2, leaving
the {{{(x-1)^2}}} intact:

{{{f(x) = 2(x-1)^2-2 - 7}}}

Combine the {{{-2}}} and the {{{-7}}} and get {{{-9}}}

{{{f(x) = 2(x-1)^2-9}}}

which is choice a.

Edwin</pre>