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Mathematics 8 Online
OpenStudy (j2lie):

x^3 - 125. How do I factor it? @jim_thompson5910

jimthompson5910 (jim_thompson5910):

have you learned the difference of cubes rule?

OpenStudy (j2lie):

The difference is when you subtract a cube

jimthompson5910 (jim_thompson5910):

have you learned this formula? \[\Large a^3 - b^3 = (a-b)*(a^2+ab+b^2)\]

OpenStudy (j2lie):

yes. I just learned it today.

jimthompson5910 (jim_thompson5910):

ok you'll use this formula to factor

jimthompson5910 (jim_thompson5910):

in this case, a = x and b = 5 since 5^3 = 125

OpenStudy (j2lie):

ok. I got x(x^2-5)(x+25) as the answer.

jimthompson5910 (jim_thompson5910):

\[\Large a^3 - b^3 = (a-b)*(a^2+ab+b^2)\] \[\Large x^3 - 5^3 = (x-5)*(x^2+x*5+5^2)\] \[\Large x^3 - 5^3 = (x-5)*(x^2+5x+25)\] \[\Large x^3 - 125 = (x-5)*(x^2+5x+25)\]

OpenStudy (j2lie):

I think you have to simplify after you fill in the pattern.

jimthompson5910 (jim_thompson5910):

that's as simplified as it gets

jimthompson5910 (jim_thompson5910):

we cannot factor `x^2 + 5x + 25`

OpenStudy (j2lie):

Never mind, because in some problems you have to simplify.

OpenStudy (j2lie):

if b was 10x we can factor.

jimthompson5910 (jim_thompson5910):

I'm not sure what you mean

OpenStudy (j2lie):

x^2 + 10x + 25 = (x+5)(x+5). it only works if there was a ten.

jimthompson5910 (jim_thompson5910):

oh I see, well if b = 10, then b^2 would be 100 so that 25 would change to 100

OpenStudy (j2lie):

The problem is not fair.

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