# F(x)=3x⁴-9x³+x²+15x+k is completely divisible by 3x²-5,find the value of k and hence the other two zeroes of the polynomial.

2
please help me

Log in to add a comment

please help me

Log in to add a comment

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

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