SOLUTION: Factor the following expression completely: {{{ 4a^2c^2-(a^2-b^2+c^2)^2 }}} Since this is an odd number in the textbook, I have the answer. The answer is: {{{ (a+b+c)*(a+b-c)*(a

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Factor the following expression completely: {{{ 4a^2c^2-(a^2-b^2+c^2)^2 }}} Since this is an odd number in the textbook, I have the answer. The answer is: {{{ (a+b+c)*(a+b-c)*(a      Log On


   

Question 1068250: Factor the following expression completely: +4a%5E2c%5E2-%28a%5E2-b%5E2%2Bc%5E2%29%5E2+
Since this is an odd number in the textbook, I have the answer. The answer is: +%28a%2Bb%2Bc%29%2A%28a%2Bb-c%29%2A%28a-b%2Bc%29%2A%28-a%2Bb%2Bc%29+

Even with the answer, I still do not know how to solve the problem.
First, I tried substitution. I made +-b%5E2%2Bc%5E2+=+z.
After substituting, the expression became: +4a%5E2c%5E2-%28a%5E2%2Bz%29%5E2+
I then expanded the binomial which got me the following expression: +4a%5E2c%5E2-%28a%5E4%2B2a%5E2z%2Bz%5E2%29+
I wasn't really sure what to do with +4a%5E2c%5E2-%28a%5E4%2B2a%5E2z%2Bz%5E2%29+, so I just re-expanded the problem replacing z with +-b%5E2%2Bc%5E2+
After messy expansion and simplification, I got: +2a%5E2c%5E2-a%5E4%2B2a%5E2b%5E2-b%5E4%2B2c%5E2b%5E2-c%5E4+
I thought this was a factor by grouping problem, so I separated the polynomial into three sections: +%282a%5E2c%5E2-a%5E4%29%2B%282a%5E2b%5E2-b%5E4%29%2B%282c%5E2b%5E2-c%5E4%29+
After factoring each of the three sections, I got: +a%5E2%282c%5E2-a%5E2%29%2Bb%5E2%282a%5E2-b%5E2%29%2Bc%5E2%282b%5E2-c%5E2%29+
I have no idea what to do next, and I am out of ammunition in solving this problem. Does anyone have any ideas in solving this?
Answer by ikleyn(52794) About Me  (Show Source):
You can put this solution on YOUR website!
.
Factor the following expression completely: +4a%5E2c%5E2-%28a%5E2-b%5E2%2Bc%5E2%29%5E2+
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

  +4a%5E2c%5E2+-+%28a%5E2-b%5E2%2Bc%5E2%29%5E2+ = %282ac%29%5E2+-+%28a%5E2-b%5E2%2Bc%5E2%29%5E2 =     (apply the formula  x%5E2+-+y%5E2 = (x+y)*(x-y) )

= %282ac+%2B+%28a%5E2+-+b%5E2+%2B+c%5E2%29%29%2A%282ac+-+%28a%5E2+-+b%5E2+%2B+c%5E2%29%29 = 

= %28+%28a%5E2+%2B+2ac+%2B+c%5E2%29+-+b%5E2%29%2A%28b%5E2+-+%28a%5E2+-+2ac+%2B+c%5E2%29%29 = 

= %28b%5E2+-+%28a%2Bc%29%5E2%29%2A%28b%5E2+-+%28a-c%29%5E2%29 =                     ( apply again the same formula to EACH of the two factors )

= (b-a-c)*(b+a+c))*(b+a-c)*(b-a+c).


QED.