Answers

2015-10-19T10:13:13+05:30
First divide the f(x) by 3x^2-5
at last the remainder is 10+k
Then we know that given f(x) is divisible by 3x^2-5. So 10+k=0 and k=-10
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2015-10-19T15:49:00+05:30
F(x)=3x⁴-9x³+x²+15x+k
g(x) = 3x²-5

If f(x) is completely divisible by g(x), then zeroes of g(x) are zeroes of f(x) too.
find zeroes of g(x).
g(x) = 3x²-5 = 0
⇒ 3x² = 5
⇒ x = +√5/3 and -√5/3

Put any value of x in f(x) and find the remainder.
f(√5/3) = 3(√5/3)⁴-9(√5/3)³+(√5/3)²+15(√5/3)+k = 10+k
Since f(√5/3) = 0
⇒10+k = 0
⇒ k = -10

So f(x) = 3x⁴-9x³+x²+15x-10
divide 3x⁴-9x³+x²+15x-10 by 3x²-5.
You will get x²-3x+2
Now solve x²-3x+2=0
⇒ x²-x-2x+2=0
⇒x(x-1)-2(x-1)=0
⇒ (x-1)(x-2) = 0
⇒ x = 1 and 2

So other two zeroes are 1 and 2
1 5 1