Question

find the value of k foe which a-3b is a factor of a^4-7a^2b^2+kb^4. hence for the value of k, factorise a^4-7a^2b^2+kb^4 completely. ####plzzzz answer this question# ********15points*******

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

i) Value of k = -18

i) Value of k = -18ii)a⁴-7a²b²-18b⁴

i) Value of k = -18ii)a⁴-7a²b²-18b⁴ =(a²+2b²)(a+3b)(a-3b)

Step-by-step explanation:

Let expression P(a)=a⁴-7a²b²+kb⁴

If (a-3b) is a factor of given expression then p(3b) = 0

=> (3b)⁴-7(3b)²b²+kb⁴=0

=> 81b⁴-7×9b²×b²+kb⁴=0

=> 81b⁴-63b⁴+kb⁴ =0

=> 18b⁴+kb⁴=0

=> b⁴(18+k )=0

=> 18+k = 0

=> k = -18

Now,

substitute k=-18,the

Expression becomes

a⁴-7a²b²-18b⁴

Splitting the middle term,we get

= a⁴ +2a²b²-9a²b²-18b⁴

= a²(a²+2b²)-9b²(a²+2b²)

= (a²+2b²)(a²-9b²)

= (a²+2b²)[a²-(3b)²]

= (a²+2b²)(a+3b)(a-3b)

Therefore,

i) Value of k = -18

ii)a⁴-7a²b²-18b⁴

=(a²+2b²)(a+3b)(a-3b)

•••♪