Question

Solve $(4x^4-5x^3-7x+1)÷(4x-1)$

Answer 1

To find: $(4x^4-5x^3-7x+1)÷(4x-1)$

Procedure:

Step 1:

$\frac{4x^4}{4x}=x^3$

Now, lets do ,$x^3\times (4x-1)=4x^4-x^3$ and then subtract.

Step 2:

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

Now, lets do , $-x^2\times (4x-1)=-4x^3+x^2$ and then subtract.

Step 3:

$\frac{-x^2}{4x}=-\frac{x}{4}$

Now, lets do , $-\frac{x}{4}\times (4x-1)=-x^2+\frac{x}{4}$ and then subtract.

Step 4:

$\frac{-29x}{4x}=-\frac{-29}{16}$

Now, lets do , $-\frac{-29}{16}\times (4x-1)=\frac{-29x}{4}+\frac{29}{16}$ and then subtract.

From the division,

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

Remainder $=-\frac{13}{6}$