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Mathematics 18 Online
OpenStudy (anonymous):

How would you factor: 49k^3m+42k^2m^2+9km^3

OpenStudy (anonymous):

I'd start with taking out \(km\) as being common in all of those terms. What would that leave you with? Once you have that, you can factorise the remaining quadratic. Hint: Pay attention to the numbers in front of the \(k^2\), and the \(m^2\).

OpenStudy (anonymous):

7*7*k*k*k*m+7*4*k*k*m*m+3*3*k*m*m*m

OpenStudy (anonymous):

*7*6

OpenStudy (anonymous):

If you take out \(km\) as being common, you'll get \[km(49k^2+42km+9m^2)\] Do you notice anything special about 49, and 9?

OpenStudy (anonymous):

Other than containing exponents of k and m?

OpenStudy (anonymous):

Other than that, yes.

OpenStudy (anonymous):

If I said that other numbers of this type were 1, 4, 16, 25, 36. Would it help?

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