Can I Solve for c by Replacing b with a Value Greater Than 0?

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mathland
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I replace b with any value greater than 0 and then solve for c. Right?

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What do you get when you expand the LHS and equate the resulting terms with the corresponding terms in the RHS of the given equation
 
Greg said:
What do you get when you expand the LHS and equate the resulting terms with the corresponding terms in the RHS of the given equation

Ok. I will do as you suggested and be back.
 
It's been 7 months now! Are you still working on it?

For those who were wondering, there are two solutions.
(2x- b)(7x+ b)= 14x^2+ 2bx- 7bx- b^2= 14x^3- 5bx- b^2. That is to be equal to 14x^2- cx- 16.

So we must have -5b= -c and -b^2= -16. b^2= 16 so b= 4 or b= -4.
If b= 4, -5b= -20= -c so c= 20.
If b= -4, -5b= 20= -c so c= -20.

Check:
(2x- 4)(7x+ 4)= 14x^2+ 8x- 28x- 16= 14x^2- 20x- 16.
(2x+ 4)(7x- 4)= 14x^2- 8x+ 28x- 16= 14x^2+ 20x- 16.
 
Beer soaked comment follows.
Country Boy said:
It's been 7 months now! Are you still working on it?
...
He's been banned permanently.
 

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