Answer :

(4x^{3} – 2x^{2} + 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,

(4x^{3} – 2x^{2} + 6x + 7) ÷ (2x + 3)

Now we need to divide (4x^{3} – 2x^{2} + 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 4x^{3}.

To get x^{3}, we need to multiply x×x^{2}.

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

Now we carry 6x + 7along with 4x^{2}, as shown below

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

here (2x + 3) × (–4x)

= – 8x^{2} – 12x

here (2x + 3) × 9

= 18x + 27

Therefore, we got the quotient = 2x^{2} – 4x + 9 and

Remainder = –20

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Find the quotientTamilnadu Board Math Term-I

Find the quotientTamilnadu Board Math Term-I

Find the quotientTamilnadu Board Math Term-I

Find the quotientTamilnadu Board Math Term-I

Find the quotientTamilnadu Board Math Term-I