Question

What must be subtracted from $8x^4+14x^3-2x^2+7x-8$ so that the resulting polynomial is exactly divisible by $4x^2+3x-2$ [4 MARKS]

Answer 1

Concept : 1 Mark
Application : 1 Mark
Calculation : 2 Marks

We know that

Dividend = Quotient $\times$ Divisor + Remainder

$\Rightarrow$ Dividend - Remainder = Quotient $\times$ Divisor

Clearly, RHS of the above result is divisible by the divisor.

Therefore, LHS is also divisible by the divisor. Thus, if we subtract remainder from the dividend, then it will be exactly divisible by the divisor.

Dividing $8x^4+14x^3-2x^2+7x-8$ by $4x^2+3x-2$, we get


$\therefore$ Quotient $=2x^2+2x-1$ and Remainder $=14x-10$

Thus, if we subtract the remainder $14x-10$ from $8x^4+14x^3-2x^2+7x-8$, it will be divisible by $4x^2+3x-2$.