anonymous
  • anonymous
Factor completely. \[10a ^{2}+25a-15\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Jhannybean
  • Jhannybean
divide everything out by 5 first
anonymous
  • anonymous
2,5,-3
anonymous
  • anonymous
5(a+3)(2a-1)

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Jhannybean
  • Jhannybean
oh no you're right! there needs to be a 2 there, haha
anonymous
  • anonymous
that was where this one got me confused. Because I kept getting what you got but it didn't add up. @Jhannybean
Jhannybean
  • Jhannybean
\[\large 5(2a^2+5a-3)\] in order to factor this, you can do what @lujanels1 has done,or you can try my method, works everytime. \[\large 5(\color{red}2a^2+5a-\color{red}3)\] multiply them both. \[\large 5(a^2+5a-6)\]find two numbers that multiply to 6 and add to 5 \[\large 5(a-1)(a+6)\]divide by the leading coefficient 2 \[\large (a-\frac{1}{\color{red}2})(a+\frac{6}{\color{red}2})\]simplify again \[\large (2a-1)(a+3)\]
Jhannybean
  • Jhannybean
oh i forgot to multiply the 5 with it too.
Jhannybean
  • Jhannybean
sorry.
Jhannybean
  • Jhannybean
\(\large 5(2a-1)(a+3)\)****
Jhannybean
  • Jhannybean
But i just did that?
anonymous
  • anonymous
It's fine. I just don't understand how to multiply it to check it with it in \[5(a+3)(2a-1)\] form.
Jhannybean
  • Jhannybean
I factorized it though :( It's alittle longer method but it works.
Jhannybean
  • Jhannybean
you use the FOIL method, @Jewels89 \[\large 5[a(2a-1)+3(2a-1)]\] you get \[\large 5[(a)(2a)-(a)(1)+(3)(2a)-(3)(1)]\]
anonymous
  • anonymous
ok. Now I get it. THANK YOU!!! :)
Jhannybean
  • Jhannybean
no problem :)

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