factoring quadratics..

- calculusxy

factoring quadratics..

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- calculusxy

\[5a^2 - 180\]

- Zale101

Hi!

- calculusxy

Hello!

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- Zale101

Do you have a question? :3

- Zale101

Lol

- Zale101

nvm

- calculusxy

Yeah i just posted it above but here it is anyway:
\[5a^2 - 180\]

- Zale101

It didnt appear for a moment. Alright, let us begin!

- calculusxy

okay!

- Zale101

Let us first make the equation look clear to us. Let's factor out the 5.

- calculusxy

5(a^2 - 36)

- Zale101

Awesome!

- Zale101

What can you say about x^2-36?
Well, we know that there are 2 x's multiplied together. (x )(x )
We need to find the missing numbers.
To find the missing numbers, we need to know the factors of 36 that when added together is zero and multiplied together is 36.

- Zale101

\(a^2-36=a^2+0a+36\)
Correct?

- Zale101

Those two equations are technically the same.

- calculusxy

yes

- Zale101

0a means that there's two numbers that got added and were the same
Those two numbers got multiplied and equaled to 36

- calculusxy

wait.. i still don't understand that part about 0a

- Zale101

Okay, here's a different way of explaining it
a^2 means x*x
0a means number + number =0
36 means number * number =36
But, those numbers that got added and multiplied together have to be the same two numbers that got added and multiplied.

- calculusxy

but what two numbers would lead to that 0

- calculusxy

but when multiplied will give me 36

- Zale101

Correct!

- calculusxy

ok so what's the next step?

- Zale101

Member, those numbers must equal 36 when multiplied, so it has to be a factor of 36.

- Zale101

Getting those two numbers is the last step

- calculusxy

ok...

- calculusxy

1, 2, 3, 4, 9, 12, 18, 36

- calculusxy

factors of 36

- Zale101

|dw:1444099970567:dw|

- calculusxy

6 oh!

- calculusxy

okay so the answer would be :
5(a - 6)(a + 6)

- calculusxy

would that be correct?

- Zale101

Yes :)

- calculusxy

thank you! i have another question.

- Zale101

Does it make sense?

- calculusxy

\[5n^2 + 5n - 360\]

- calculusxy

and yes

- Astrophysics

Nice one @Zale101
There is a trick to avoid all of this for the future, it's called complete the square (of course if you understand it that is) \[\huge a^2-b^2 = (a-b)(a+b)\]

- calculusxy

5(n^2 + n - 75)
factors of 75: 1,3,5,15,25,75
i can't find a number whose sum will be n or 1

- Zale101

Nice one astro! Someone aleady sent me to try this, but i didn't want to change what i was gearing towards lol

- Astrophysics

Haha, I see

- Zale101

Dont look at the factors, sorry if i included this term. look at the 2 numbers that can be multiplied to get the last term but that's of course after factoring. Always look for the difference in squares. When we had 5(x^2-36)
We can always use the rule astro given to us \((a^2−b^2)=(a−b)(a+b)\)

- calculusxy

i didn't understand

- calculusxy

@Zale101 Can you please hurry up a bit? I need to go to bed soon..

- Zale101

Are you sure you factored it right?

- calculusxy

i am not sure.. i desperately need help with this

- Zale101

Try factoring it again.

- Zale101

there was a mistake when you factored 5 out

- calculusxy

5(n^2 + 1n - 72)

- Zale101

Much better

- Zale101

Now it should not be confusing :)

- Zale101

Can you work that out?

- calculusxy

5(n+9)(n-8)

- Zale101

Perfect.

- calculusxy

sorry i just need help with another question

- calculusxy

9v^2 - 12v + 4

- Zale101

Show me your attempt on ths.

- calculusxy

i don't know how to factor the 9 when i have the 12 or 4 (which can't be fully divisible by 9)

- calculusxy

i don't know what to do

- Zale101

That's because all three terms dont have a common factor. This needs a slightly different approach.

- Zale101

You have your two parenthesis prepared. Are you familiar with group factoring?

- calculusxy

not that much

- calculusxy

can you plz help me quickly?

- Zale101

So, before i group factor i'll try to use a different approach and this approach is examining what got added that made 12. Okay, -6v-6v=-12 v
we can choose to replace -12v with -6v-6v
\(9v^2 - 12v + 4\)
\(9v^2 -6v-6v + 4\)

- Zale101

@calculusxy how much do you know about group factoring?

- Zale101

Learning how to group factor will save you when factoring these types of problems.

- Zale101

This will help http://www.mesacc.edu/~scotz47781/mat120/notes/factoring/grouping/grouping.html

- calculusxy

http://www.purplemath.com/modules/factquad2.htm i was looking at this . i don't know if this can be useful

- Zale101

You can try the bar model, it helps too!

- calculusxy

okay thanks! :)

- Zale101

But after i arranged the equation with -6v-6v replacing -12v, then group factoring would be easier from there.

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