Find the quotient the and remainder of the following division.

(4x³ – 2x² + 6x + 7) ÷ (3 + 2x)

(4x³ – 2x² + 6x + 7) ÷ (3 + 2x)

We see that the equation is not arranged in descending order, so we need first arrange it in descending order of the power of x.

Therefore it becomes,

(4x³ – 2x² + 6x + 7) ÷ (2x + 3)

Now we need to divide (4x³ – 2x² + 6x + 7) by (2x + 3)

Now we need to find out by how much should we multiple “x” to get a value as much as 4x³.

To get x³, we need to multiply x×x².

Therefore we need to multiply with 2x² × (2x + 3) and we get (4x³ + 6x²) now subtract (4x³ + 6x²) from 4x³ – 2x² + 6x + 7so we get – 8x².

Now we carry 6x + 7along with 4x², as shown below

So, in same way we have keep dividing till we get rid of x as shown below.

here (2x + 3) × (–4x)

= – 8x² – 12x

here (2x + 3) × 9

= 18x + 27

Therefore, we got the quotient = 2x² – 4x + 9 and

Remainder = –20