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\[5a^2 - 180\]
Do you have a question? :3
Yeah i just posted it above but here it is anyway: \[5a^2 - 180\]
It didnt appear for a moment. Alright, let us begin!
Let us first make the equation look clear to us. Let's factor out the 5.
5(a^2 - 36)
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.
Those two equations are technically the same.
0a means that there's two numbers that got added and were the same Those two numbers got multiplied and equaled to 36
wait.. i still don't understand that part about 0a
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.
but what two numbers would lead to that 0
but when multiplied will give me 36
ok so what's the next step?
Member, those numbers must equal 36 when multiplied, so it has to be a factor of 36.
Getting those two numbers is the last step
1, 2, 3, 4, 9, 12, 18, 36
factors of 36
okay so the answer would be : 5(a - 6)(a + 6)
would that be correct?
thank you! i have another question.
Does it make sense?
\[5n^2 + 5n - 360\]
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) \[\huge a^2-b^2 = (a-b)(a+b)\]
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
Nice one astro! Someone aleady sent me to try this, but i didn't want to change what i was gearing towards lol
Haha, I see
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 \((a^2−b^2)=(a−b)(a+b)\)
i didn't understand
@Zale101 Can you please hurry up a bit? I need to go to bed soon..
Are you sure you factored it right?
i am not sure.. i desperately need help with this
Try factoring it again.
there was a mistake when you factored 5 out
5(n^2 + 1n - 72)
Now it should not be confusing :)
Can you work that out?
sorry i just need help with another question
9v^2 - 12v + 4
Show me your attempt on ths.
i don't know how to factor the 9 when i have the 12 or 4 (which can't be fully divisible by 9)
i don't know what to do
That's because all three terms dont have a common factor. This needs a slightly different approach.
You have your two parenthesis prepared. Are you familiar with group factoring?
not that much
can you plz help me quickly?
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 \(9v^2 - 12v + 4\) \(9v^2 -6v-6v + 4\)
@calculusxy how much do you know about group factoring?
Learning how to group factor will save you when factoring these types of problems.
This will help http://www.mesacc.edu/~scotz47781/mat120/notes/factoring/grouping/grouping.html
http://www.purplemath.com/modules/factquad2.htm i was looking at this . i don't know if this can be useful
You can try the bar model, it helps too!
okay thanks! :)
But after i arranged the equation with -6v-6v replacing -12v, then group factoring would be easier from there.