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Mathematics 12 Online
OpenStudy (anonymous):

Can anyone help me with factoring? I just have some questions about the possess in general.

Parth (parthkohli):

Sure. And welcome to OpenStudy!

OpenStudy (anonymous):

There are a bunch of different ways to factor, how do I know which one to use?

OpenStudy (anonymous):

Is there one way that will always work?

Parth (parthkohli):

Can you give me an example?

Parth (parthkohli):

It works with grouping.

OpenStudy (anonymous):

sure just one min

OpenStudy (anonymous):

k²+5k+6+ (k+2) (k+___) How would you complete this? (its this first thing on faction i found in my text book)

Parth (parthkohli):

Notice that \(6 = 3\times 2\) and \(5 = 3+2\)

OpenStudy (anonymous):

*factoring

OpenStudy (anonymous):

im so sorry i wrote that wrong k²+5k+6= (k+2) (k+___)

Parth (parthkohli):

Do you know the basic trick?

OpenStudy (anonymous):

no

OpenStudy (anonymous):

wait is the trick finding the GCF and braking it down?

Parth (parthkohli):

Let's start with an example.\[x^2 \color{green}{+ 10}x \color{#C00}{+ 25}\]You have to find two numbers that have their sum as the green number and the product as the red one. Here, they are \(5\) and \(5\). You can now break down the \(+10x\) into \(+5x\) and \(+5x\).\[x^2 + 5x + 5x + 25\]Now factor the first two and last two terms.\[x(x + 5) + 5(x + 5)\]Finally\[(x + 5)(x + 5)\]

OpenStudy (anonymous):

ok i think i get it

Parth (parthkohli):

Great!

OpenStudy (anonymous):

Thank you so much

Parth (parthkohli):

No probs!

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