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OpenStudy (anonymous):
Take x commonly outside from each term.Try it.
Parth (parthkohli):
Why not take 5x? That'd be much better.
OpenStudy (kaederfds):
5x(x^2-4)
OpenStudy (anonymous):
Yea that will be better.
OpenStudy (anonymous):
so how do i do this??
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OpenStudy (anonymous):
Do you know how to factor out something from an expression?
OpenStudy (anonymous):
if I have -> 4-2
I can take out a factor of 2 to get 2(2-1) yes?
do the same with 5x^3-20x
take out 5x to get...
OpenStudy (anonymous):
so take 5x out from 5x^3-20x
5x(5x^3/5x -20x/5x)
OpenStudy (anonymous):
Well, do you get it or you need help?
OpenStudy (anonymous):
i get it a lil bit!! so would the answer be 4
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OpenStudy (anonymous):
The answer must be 5x(x^2-4) . Actually what we are doing while factoring is we are multiplying and dividing the expression by a term which is present common in the terms in the equation.
So, 5x^3-20x = 5x*x^2 + 5x*4 .
OpenStudy (anonymous):
Since 5x is common in both terms terms,it can be factored out.
OpenStudy (anonymous):
The off course factor further to get 5x(x+2)(x-2) [diff of 2 squares]