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Question 978568: http://www.algebra.com/tutors/students/your-answer.mpl?question=978456
How do u solve this using remainder theorem. Find the value of k for which a-3b is a factor of a^4-7a^2(b^2)+kb^4. Hence, for this value of k, factorise a^4-7a^2(b^2) +kb^4 completely.
Answer by Edwin McCravy(20054) (Show Source):
You can put this solution on YOUR website!
I already solved it for you here:
http://www.algebra.com/tutors/students/your-answer.mpl?question=978038
Maybe you thought that since I used long division instead of synthetic
division that that was not using the remainder theorem. But synthetic
division is only a shorthand form of long division. You can actually use
synthetic division. You'll get the same thing. I thought it might
confuse you since there are two variables. OK, I'll do it with synthetic
division:
a⁴-7a²b²+k⁴
You have to put the b's in for they are coefficients of the a's. Also you
must insert 0 terms for the missing powers of a
3b | 1 0b -7b² 0b³ kb⁴
| 3b 9b² 6b³ 18b⁴
1 3b 2b2 6b3 (kb⁴+18b⁴)
The remainder must = 0 so that a-3b
will be a factor of a⁴-7a²b²+kb⁴.
kb⁴+18b⁴ = 0
(k+18)b⁴ = 0
k+18=0; b⁴=0
k=-18; b=0
There is a trivial case for b=0 and k is any number,
which we ignore.
So k = -18
a⁴-7a²b²-18b⁴
(a²+2b²)(a²-9b²)
(a²+2b²)(a-3b)(a+3b) <--- complete factorization
Edwin
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