Here's one I came across today: Factor by grouping. x^3+3x^2+15x+125
\[x^3+3x^2+15x+125\]
(x^3 + 3x^2) + (15x + 125) Try factoring each one of those separately.
won't work. you have to use commutative property to get proper grouping.
Actually, hold on...let me try and solve this one first.
try not to use a calculator
my summer school students thought the question was a little unfair how the terms were ordered in question.
well, it is factorable.
Ah. Got it. \[x^3 + 125 + 3x^2 + 15x\] \[(x+5)(x+5)(x+5) + 3x(x+5)\] \[(x+5)((x+5)^2 + 3x)\] \[(x+5)(x^2 + 13x + 25)\]
\[3(x+5)^2(x^2+5x+25)\] I think that looks right
Ah, perhaps I forgot to factor out the x from the (x+5)... it's late.
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.
Shouldn't that be x^2 + 10x + 25?
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
Oh. You're right. That's my fault for not putting too much thought into it, heh.
thanks for checking question out.
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