Question

what should be added to 6x^5+4x^4-27x^3-7x^2-27x-6 so that the resulting polynomial is exactly divisible by (2x^2-3)

Answer 1

I hope it helps you

Answer 2

Answer:

$2x^2+54x+\frac{9}{2}$

Step-by-step explanation:

Dividend :$6x^5+4x^4-27x^3-7x^2-27x-6$

Divisor :$2x^2-3$

Since we know that :

$Dividend = (Divisor \times Quotient)+Remainder$

So, $6x^5+4x^4-27x^3-7x^2-27x-6 = (2x^2-3 \times 3x^3+2x^2-9x-\frac{1}{2})+(-54x-\frac{-15}{2})$

The polynomial need to be added to $6x^5+4x^4-27x^3-7x^2-27x-6$ is Divisor -remainder

: $2x^2-3-(-54x-\frac{-15}{2})$

$2x^2+54x+\frac{9}{2}$

Hence $2x^2+54x+\frac{9}{2}$ should be  be added to $6x^5+4x^4-27x^3-7x^2-27x-6$  o that the resulting polynomial is exactly divisible by$2x^2-3$