Question

Find the quotient and remainder on dividing polynomial 2y³ + 4y² + 3y + 1 by y+1.​

Answer 1

Answer:

Quotient =2y²+2y+1 and remainder =0

Step-by-step explanation:

2y³/y=2y²

2y²×y+1=2y³+2y²

Like this we have to do.

Answer 2

Answer :-

Reminder = 0

Quotient → 2y² + 2y + 1 ( a quadratic equation )

To Find :-

→ Quotient and reminder .

Explanation :-

We have ;-

$\begin{array}{rccccl} y + 1 \bigg) & 2y^3 & + 4y^2 & + 3y & + 1 & \bigg( 2y^2+2y+1 \\ & 2y^3 & 2y^2 &&& \\ \cline{2-5}& & 2y^2 & + 3y & + 1 & \\ & & 2y^2 & + 2y & & \\ \cline{3-5} & & & y & + 1 & \\ & & & y & + 1 & \\ \cline{4-5} & & & & 0\end{array}$

Reminder = 0

Quotient → 2y² + 2y + 1 ( a quadratic equation )

We can solve this quadratic by Shridharacharya principal .

→ If given quadratic is →ax² + bx + c = 0

$\sf{ \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} }$