Long division works quite nicely
http://calc101.com/webMathematica/long-divide.jsp#topdoit
Now if a=0, you are dividing
kb^4 by -3b
and if a=1 and b=2 you are dividing
1-7*4+16k by -5 or (-27+16k)/-5 and if that is to come out as an integer, then
-27+16k is a multiple of 5 which means than in 16k has to be ???8 which means k is now a multiple of 3 and ends in 8. Hmmmm. -18 works.
Do the long division.
Hello, this is a question which i don't know how to do.
Find the value of k for which a-3b is a factor of a^4-7a^2b^2 + kb^4
Hence, for this value of k, factorise a^4 - 7a^2b^2 + kb^4 completely.
I tried to do long division for the question, but couldnt. Is there any other methods?
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