Factoring Using the Distributive Property
5a^2-15
Show your steps to find the GCF
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OpenStudy (31356):
@satellite73
OpenStudy (31356):
@jim_thompson5910
jimthompson5910 (jim_thompson5910):
what do 5 and 15 have in common?
jimthompson5910 (jim_thompson5910):
in terms of factors
OpenStudy (31356):
5
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jimthompson5910 (jim_thompson5910):
so factor out the GCF to get ???
jimthompson5910 (jim_thompson5910):
5a^2 - 15 = 5( .... )
what goes in the parenthesis?
OpenStudy (31356):
5?
I don't know
OpenStudy (31356):
I just learned this today
jimthompson5910 (jim_thompson5910):
factor 15 into 2 numbers where 5 is one factor
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OpenStudy (31356):
In class
jimthompson5910 (jim_thompson5910):
15 = 5*____
OpenStudy (31356):
3
jimthompson5910 (jim_thompson5910):
15 = 5*3, good
jimthompson5910 (jim_thompson5910):
so
5a^2 - 15
5a^2 - 5*3
5(a^2 - 3)
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jimthompson5910 (jim_thompson5910):
for that last step, I'm using the distributive property
a*(b+c) = a*b + a*c
OpenStudy (31356):
Thanks!
OpenStudy (31356):
So this is the answer right?
5(a^2 - 3)
jimthompson5910 (jim_thompson5910):
Notice as a check, we can distribute that 5 back through to get
5(a^2 - 3)
5(a^2) + 5(-3)
5a^2 - 15
which is the original expression. So the answer is confirmed
jimthompson5910 (jim_thompson5910):
Yes, it's 5(a^2 - 3)
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