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Mathematics 14 Online
OpenStudy (anonymous):

Factor a3 - 3 + 3a2 - a. HELP!! A. (a - 1)(a + 1)(a + 3) B. (a2+ 1)(a - 3) C. (a2- 3)(a + 1)

OpenStudy (anonymous):

You should definitely put it in standard form as the first step.

OpenStudy (anonymous):

ok

OpenStudy (whpalmer4):

This can be done by "factoring by grouping". Let's write it out in standard form, as suggested: \[a^3+3a^2-a-3\] Now we'll group it into two chunks: \[(a^3+3a^2)-1(a+3)\] Notice that I pulled out a -1 from the last two terms. The reason why will become apparent shortly.

OpenStudy (whpalmer4):

Now can you factor the first group at all? \((a^3+3a^2)\)

OpenStudy (whpalmer4):

What is the largest common factor of \(a^3+3a^2\)?

OpenStudy (anonymous):

idk

OpenStudy (anonymous):

Find the largest item that is common in both of the items..

OpenStudy (whpalmer4):

Here, let's write it as \[a^3+3a^2 = a*a*a + 3*a*a\]

OpenStudy (anonymous):

^ Exactly

OpenStudy (whpalmer4):

Remember the distributive property? \[x(y+z) = x*y + x*z\]We're trying to do that in reverse.

OpenStudy (whpalmer4):

We've got \(x*y + x*z\) and we want to split it back into \( x (y+z)\)

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