Question

Solve: $(4x^4-5x^3-7x+1)÷(4x-1)$.

Answer 1

Divisor $=(4x-1)$

Dividend $=4x^4-5x^3-7x+1$

Quotient $=x^3-x^2-\frac{x}{4}-\frac{29}{16}$

remainder $=\frac{-13}{16}$

We know , dividend $=$ divisor $\times$ quotient $+$ remainde

$4x^4-5x^3-7x+1=(4x-1)(x^3-x^2-\frac{x}{4}-\frac{29}{16})-\frac{13}{16}$