calculusxy
  • calculusxy
factoring quadratics..
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
calculusxy
  • calculusxy
\[5a^2 - 180\]
Zale101
  • Zale101
Hi!
calculusxy
  • calculusxy
Hello!

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Zale101
  • Zale101
Do you have a question? :3
Zale101
  • Zale101
Lol
Zale101
  • Zale101
nvm
calculusxy
  • calculusxy
Yeah i just posted it above but here it is anyway: \[5a^2 - 180\]
Zale101
  • Zale101
It didnt appear for a moment. Alright, let us begin!
calculusxy
  • calculusxy
okay!
Zale101
  • Zale101
Let us first make the equation look clear to us. Let's factor out the 5.
calculusxy
  • calculusxy
5(a^2 - 36)
Zale101
  • Zale101
Awesome!
Zale101
  • 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
  • Zale101
\(a^2-36=a^2+0a+36\) Correct?
Zale101
  • Zale101
Those two equations are technically the same.
calculusxy
  • calculusxy
yes
Zale101
  • 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
  • calculusxy
wait.. i still don't understand that part about 0a
Zale101
  • 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
  • calculusxy
but what two numbers would lead to that 0
calculusxy
  • calculusxy
but when multiplied will give me 36
Zale101
  • Zale101
Correct!
calculusxy
  • calculusxy
ok so what's the next step?
Zale101
  • Zale101
Member, those numbers must equal 36 when multiplied, so it has to be a factor of 36.
Zale101
  • Zale101
Getting those two numbers is the last step
calculusxy
  • calculusxy
ok...
calculusxy
  • calculusxy
1, 2, 3, 4, 9, 12, 18, 36
calculusxy
  • calculusxy
factors of 36
Zale101
  • Zale101
|dw:1444099970567:dw|
calculusxy
  • calculusxy
6 oh!
calculusxy
  • calculusxy
okay so the answer would be : 5(a - 6)(a + 6)
calculusxy
  • calculusxy
would that be correct?
Zale101
  • Zale101
Yes :)
calculusxy
  • calculusxy
thank you! i have another question.
Zale101
  • Zale101
Does it make sense?
calculusxy
  • calculusxy
\[5n^2 + 5n - 360\]
calculusxy
  • calculusxy
and yes
Astrophysics
  • 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
  • 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
  • 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
  • Astrophysics
Haha, I see
Zale101
  • 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
  • calculusxy
i didn't understand
calculusxy
  • calculusxy
@Zale101 Can you please hurry up a bit? I need to go to bed soon..
Zale101
  • Zale101
Are you sure you factored it right?
calculusxy
  • calculusxy
i am not sure.. i desperately need help with this
Zale101
  • Zale101
Try factoring it again.
Zale101
  • Zale101
there was a mistake when you factored 5 out
calculusxy
  • calculusxy
5(n^2 + 1n - 72)
Zale101
  • Zale101
Much better
Zale101
  • Zale101
Now it should not be confusing :)
Zale101
  • Zale101
Can you work that out?
calculusxy
  • calculusxy
5(n+9)(n-8)
Zale101
  • Zale101
Perfect.
calculusxy
  • calculusxy
sorry i just need help with another question
calculusxy
  • calculusxy
9v^2 - 12v + 4
Zale101
  • Zale101
Show me your attempt on ths.
calculusxy
  • 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
  • calculusxy
i don't know what to do
Zale101
  • Zale101
That's because all three terms dont have a common factor. This needs a slightly different approach.
Zale101
  • Zale101
You have your two parenthesis prepared. Are you familiar with group factoring?
calculusxy
  • calculusxy
not that much
calculusxy
  • calculusxy
can you plz help me quickly?
Zale101
  • 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
  • Zale101
@calculusxy how much do you know about group factoring?
Zale101
  • Zale101
Learning how to group factor will save you when factoring these types of problems.
Zale101
  • Zale101
This will help http://www.mesacc.edu/~scotz47781/mat120/notes/factoring/grouping/grouping.html
calculusxy
  • calculusxy
http://www.purplemath.com/modules/factquad2.htm i was looking at this . i don't know if this can be useful
Zale101
  • Zale101
You can try the bar model, it helps too!
calculusxy
  • calculusxy
okay thanks! :)
Zale101
  • Zale101
But after i arranged the equation with -6v-6v replacing -12v, then group factoring would be easier from there.

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