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Mathematics 15 Online
OpenStudy (matheducatormcg):

Here's one I came across today: Factor by grouping. x^3+3x^2+15x+125

OpenStudy (matheducatormcg):

\[x^3+3x^2+15x+125\]

OpenStudy (anonymous):

(x^3 + 3x^2) + (15x + 125) Try factoring each one of those separately.

OpenStudy (matheducatormcg):

won't work. you have to use commutative property to get proper grouping.

OpenStudy (anonymous):

Actually, hold on...let me try and solve this one first.

OpenStudy (matheducatormcg):

try not to use a calculator

OpenStudy (matheducatormcg):

my summer school students thought the question was a little unfair how the terms were ordered in question.

OpenStudy (matheducatormcg):

well, it is factorable.

OpenStudy (anonymous):

Ah. Got it. \[x^3 + 125 + 3x^2 + 15x\] \[(x+5)(x+5)(x+5) + 3x(x+5)\] \[(x+5)((x+5)^2 + 3x)\] \[(x+5)(x^2 + 13x + 25)\]

OpenStudy (anonymous):

\[3(x+5)^2(x^2+5x+25)\] I think that looks right

OpenStudy (anonymous):

Ah, perhaps I forgot to factor out the x from the (x+5)... it's late.

OpenStudy (matheducatormcg):

you got the cubic factoring wrong. here is solution: \[x^3+3x^2+15x+125\] \[=(x^3+125)+(3x^2+15x)\] \[=(x+5)(x^2+5x+25)+3x(x+5)\] \[=(x+5)((x^2+5x+25)+3x)\] \[=(x+5)(x^2+8x+25)\] pretty cool. for a factoring.

OpenStudy (anonymous):

Shouldn't that be x^2 + 10x + 25?

OpenStudy (matheducatormcg):

No. 5x +3x is 8x. ?? the cubic is a lurker, thinking that \[(x^3+125)=(x+5)(x+5)(x+5)\] factors this way is common mistake

OpenStudy (anonymous):

Oh. You're right. That's my fault for not putting too much thought into it, heh.

OpenStudy (matheducatormcg):

thanks for checking question out.

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