Asked by Anonymous
                How do you completely factor   64x-x (the -x is to the power of 4)
            
            
        Answers
                    Answered by
            Anonymous
            
    Ok so the problem is 64x - x^-4
The highest common factor of the two terms is x so the answer is:
x(64 - x^3)
Now, 64 - x^3 is the 'difference of two cubes' as 64 = 4^3 so we use the formula:
X^3 - y^3 = (X - y)*(X^2 + X*y + y^2)
In our case the 'X' = '4' and
'y' = 'x'
so:
(4^3 - x^3) = (4 - x)*(4^2 + 4*x + x^2)
= (4 - x)*(16 + 4x +x^2)
So we started with:
64x - x^4
We then took 'x' out
x(64 - x^3) = x(4^3 - x^3)
and we then found out what 4^3 - x^3 was so the final answer is:
x(4 - x)(16 + 4x + x^2)
Ok, Hope that helps!
 
    
The highest common factor of the two terms is x so the answer is:
x(64 - x^3)
Now, 64 - x^3 is the 'difference of two cubes' as 64 = 4^3 so we use the formula:
X^3 - y^3 = (X - y)*(X^2 + X*y + y^2)
In our case the 'X' = '4' and
'y' = 'x'
so:
(4^3 - x^3) = (4 - x)*(4^2 + 4*x + x^2)
= (4 - x)*(16 + 4x +x^2)
So we started with:
64x - x^4
We then took 'x' out
x(64 - x^3) = x(4^3 - x^3)
and we then found out what 4^3 - x^3 was so the final answer is:
x(4 - x)(16 + 4x + x^2)
Ok, Hope that helps!
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