Oh my bad, thanks a lot. Y'all at CrackU are doing a great job. The doubt redressal mechanism and the plethora of video solutions are extremely helpful to me ! Kudos to y'all!

Hi Gautam Ganeriwala,
Thanks for your words. We are happy to hear it form you. We are always there to help you. Feel free to post any doubts.

Hi Gautam Ganeriwala,
You have taken 2r-1 instead of 2r+1 as the $$x^2$$ coefficient. Please correct that, you will get zero.
I hope this helps!!

Hi Arnav Singh,
$$x^{3} + (2r + 1)x^{2} + (4r - 1)x + 2 =0$$, this can be written as $$x^3+(2r+1+2-2)x^2+(4r-1+2-2)x+2=0$$
$$x^3+(2r-2+1+2)x^2+(4r-1+2-2)x+2=0$$
$$x^3+(2r-1+2)x^2+(4r+1-2)x+2=0$$
$$x^3+((2r-1)+2)x^2+((4r+1)-2)x+2=0$$
$$x^3+(2r-1)x^2+2x^2+(4r+1)x-2x+2=0$$
This is the simplification behind that step.
I hope this helps!!

Hey team, this is still pending.

How would we know in an exam scenario where to adjust which numbers?

Hi Arnav Singh,
This comes with practice. Try to practice more questions, then you will get to know how to adjust numbers to get the required answer. 
I hope this helps!!

Hi,
The Steps of Horner's Synthetic Division:
1. Arrange the polynomial in descending order: Write the polynomial in standard form, with the terms arranged in descending order of degree. 
2. Identify the divisor: Identify the linear binomial (x - c) that will be used as the divisor. 
3. Write the coefficients: Write down the coefficients of the polynomial, excluding the variable terms. These coefficients will be used in the synthetic division process. 
4. Set up the synthetic division table: Draw a table with two rows. In the top row, write the coefficients of the polynomial, starting from the highest degree term and moving to the constant term. In the bottom row, write the constant term of the divisor. 
5. Perform the synthetic division: Perform the synthetic division process by following these steps: - Bring down the first coefficient from the top row into the leftmost cell of the bottom row. - Multiply the divisor's constant term by the value in the bottom row and write the result in the cell to the right. - Add the value in the top row with the value in the bottom row and write the sum in the next cell of the bottom row. - Repeat the previous two steps until you reach the end of the row. 
6. Interpret the results: The numbers in the bottom row of the synthetic division table represent the coefficients of the quotient polynomial. The last number in the bottom row is the remainder. 
7. Write the quotient and remainder: Write the coefficients of the quotient polynomial using the numbers in the bottom row. The remainder is the last number in the bottom row.
For example:
Divide the polynomial f(x) = x^4-3x^2+7 by the linear binomial (x + 2).

In this case, first we take the coefficient and then x+2 = 0 => x = -2
Now put the coefficient of the highest power (1) in the bottom row without any operation.
Now start cross multiplying with the divisor coeff (-2)* bottom row number (1), and add the resultant number, which is -2 with the coefficient of the second highest power, which is 0. After we get (-2+0) = -2 put this in the bottom row, and the repeat the process.
I hope this helps!

Hi,
It is given that '-2' is one of the roots of this polynomial. Hence, The polynomial can be written (x+2)(ax^2+bx+c) 
To simplify the calculation, we are writing the same polynomial in a different way so that we can take (x+2) common from the polynomial.
It is given that x^3+(2r+1)x^2+(4r-1)x+2 = 0
=> x^3+2rx^2+x^2+4rx-x+2 = 0
=> x^3+2rx^2+(2x^2-x^2)+4rx-(2x-x)+2 = 0
=> x^3+2x^2+x^2(2r-1)+x(4r-2)+x+2 =0
Now we can take factor of (x+2) from every part of the equation.
=>  x^3+2x^2+x^2(2r-1)+x(4r-2)+x+2 =0
=> x^3+2x^2+x^2(2r-1)+2x(2r-1)+x+2 = 0
=> x^2(x+2)+(2r-1)x(x+2)+(x+2)*1 = 0
=> (x+2)(x^2+(2r-1)x+1) = 0
I hope this helps! Please let us know if the doubt still persists.

Hi Tejas G,

You receive 3 marks for each correct answer, and there are no negative marks for TITA (Type-In-The-Answer) questions. Based on this calculation, you have received the correct marks for the answers you attempted. Please review it and let me know if you have any further questions.
I hope this helps!!