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?
To find the value of k for which a-3b is a factor, we can use the Remainder Theorem or the Factor Theorem.
First, let's substitute a = 3b into the expression a^4 - 7a^2b^2 + kb^4:
(3b)^4 - 7(3b)^2b^2 + kb^4
= 81b^4 - 63b^4 + kb^4
= 18b^4 + kb^4
Now, for a-3b to be a factor, the expression 18b^4 + kb^4 should become 0 when we substitute a = 3b. This means:
18b^4 + kb^4 = 0
To satisfy this equation for all values of b, we need the coefficient of b^4 to be 0. Therefore,
k + 18 = 0
k = -18
So, the value of k for which a-3b is a factor is -18.
Now let's factorize a^4 - 7a^2b^2 + kb^4 completely using the value of k = -18:
We can rewrite the expression as follows:
(a^4 - 9a^2b^2) - 16b^4
= (a^2 - 3ab)(a^2 + 3ab) - 16b^4
The expression can be further factorized using the difference of squares:
= (a - √3ab)(a + √3ab)(a^2 + 3ab - 16b^4)
Therefore, for the value of k = -18, the expression a^4 - 7a^2b^2 + kb^4 can be factorized completely as (a - √3ab)(a + √3ab)(a^2 + 3ab - 16b^4).
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.