Question

if we divide 3y^4-y^3+12y^2+2 by 3y^2-1 then remainder is​

Answer 1

3y^4-y^3+12y^2+2 divided by 3y^2-1 will be -1/3y+19/3

Answer 2

Remainder is $\bf \red{\frac{ - y}{3} + \frac{19}{3} }$ ,when $3 {y}^{4} - {y}^{3} + 12 {y}^{2} + 2$ is divided by $3 {y}^{2} - 1$

Given:

• A polynomial $3 {y}^{4} - {y}^{3} + 12 {y}^{2} + 2$
• Another polynomial $3 {y}^{2} - 1$

To find:

• Find the remainder after division.

Solution:

Step 1:

Apply long division method.

Let

$p(x) = 3 {y}^{4} - {y}^{3} + 12 {y}^{2} + 2$

and

$g(x) = 3 {y}^{2} - 1$

Step 2:

Multiply a coefficient or expression with g(x) so that it's first term will be equal to first term of p(x).

$\: \:3 {y}^{2} - 1 \: ) \: 3 {y}^{4} - {y}^{3} + 12 {y}^{2} + 2 \: ) {y}^{2} - \frac{y}{3} + \frac{13}{3} \\ 3 {y}^{4} \: \: \: \: \: \: \: \: \: - {y}^{2} \\ ( - ) \: \: \: \: \: \: \: \: \: \: \: ( + ) \\ - - - - - - - \\ - {y}^{3} + 13 {y}^{2} + 2 \\ - {y}^{3} \: \: \: \: \: \: \: \: \: \: \: \: \: + \frac{y}{3} \\ ( + ) \: \: \: \: \: \: \: \: \: \: \: \: \: ( - ) \\ - - - - - - - - \\ 13 {y}^{2} - \frac{y}{3} + 2 \\ 13 {y}^{2} \: \: \: \: \: \: \: - \frac{13}{3} \\ ( - ) \: \: \: \: \: \: \: ( + ) \\ - - - - - - - \\ - \frac{y}{3} + \frac{19}{3} \\ - - - - - - -$

Thus,

Quotient of division is $\bf {y}^{2} - \frac{y}{3} + \frac{13}{3}$

Remainder is $\bf - \frac{y}{3} + \frac{19}{3}$

Thus,

By this way remainder can be calculated and it is (19-y)/3.

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