Question

Find the quotient and remainder on dividing the algebraic equation, p (x) by g (x) in the following case, without actual division.
${\text{p}}\left( {\text{x}} \right) = {{\text{x}}^3} + 4{{\text{x}}^2} - 6{\text{x + 2; g}}\left( {\text{x}} \right) = {\text{x - 3}}$

Answer 1

Hint:In order to find the quotient and remainder of the equation${{\text{x}}^3} + 4{{\text{x}}^2} - 6{\text{x + 2}}$by dividing it with${\text{x - 3}}$, we use the polynomial factorization method, which is similar to the normal factorization method.
We generally end up being left with a constant which is the remainder and the quotient is built in a stepwise manner as we divide the main equation.

Complete step-by-step answer:
Given Data,
${{\text{x}}^3} + 4{{\text{x}}^2} - 6{\text{x + 2}}$is divided by${\text{x - 3}}$without actual division.

In the factorization method we write the equation to be divided (i.e. the dividend) inside the box, the divisor is written to the left of the box.
We perform this in a stepwise manner, we start off by eliminating the term with the highest degree in each step. So we wrote the equation${{\text{x}}^3} + 4{{\text{x}}^2} - 6{\text{x + 2}}$inside the box and (x – 3) to its left hand side outside the box.
Now the term with the highest degree in the equation is${{\text{x}}^3}$, so we have to multiply the divisor with${{\text{x}}^2}$to get a third degree term. The term we multiply the divisor with, becomes the first term of the quotient written above in brackets.
So multiplying${{\text{x}}^2}$and (x-3) we get${{\text{x}}^3} - 3{{\text{x}}^2}$, which is written under the dividend and subtracted from it. The resultant algebraic term is written below. Now this becomes our new dividend.
The same process is performed until we end up with just a constant as a resultant as shown below. And every term we multiply the divisor with, in each step is added to the quotient.
Towards the end, the constant term we are left with is our remainder and the sum of all the terms that are multiplied to the divisor in each step in our quotient.
So performing this polynomial factorization division method using the given equations in the question looks like, as shown below:



$
  {\text{ }}\left( {{{\text{x}}^2} + 7{\text{x + 15}}} \right) \\
  {\text{x - 3}}\left| \!{\underline {\,

  {{\text{x}}^3} + 4{{\text{x}}^2} - 6{\text{x + 2}} \\
  {{\text{x}}^3} - 3{{\text{x}}^2} \\
  {\text{ - + }} \\
\,}} \right. \\
  {\text{ }}\left| \!{\underline {\,

  {\text{ + 7}}{{\text{x}}^2} - 6{\text{x + 2}} \\
  {\text{ 7}}{{\text{x}}^2} - 21{\text{x}} \\
  {\text{ - + }} \\
\,}} \right. \\
  {\text{ }}\left| \!{\underline {\,

  {\text{ + 15x + 2}} \\
  {\text{ + 15x - 45}} \\
  {\text{ - + }} \\
\,}} \right. \\
  {\text{ }}\left| \!{\underline {\,
  {{\text{ + 47}}} \,}} \right. \\
$

Therefore on dividing the algebraic equation p(x) by g(x) we get that the quotient is${{\text{x}}^2} + 7{\text{x + 15}}$and the remainder is 47.

Note:In order to solve this type of questions the key is to know the concept of division by factorization method for polynomials. We can multiply a term of either sign, positive or negative to the divisor to perform our operation on the dividend. It is to be appropriately added in each step to the quotient term.
We can verify the answer we obtain using the formula of a division which is D = (d × q) + R, where D, d, q and R represent dividend, divisor, quotient and remainder respectively.
Therefore, we should prove
$
   \Rightarrow \left( {{{\text{x}}^2} + 7{\text{x + 15}}} \right)\left( {{\text{x - 3}}} \right) + 47 = {{\text{x}}^3} + 4{{\text{x}}^2} - 6{\text{x + 2}} \\
   \Rightarrow {{\text{x}}^3} - 3{{\text{x}}^2} + 7{{\text{x}}^2} - 21{\text{x + 15x - 45 + 47 = }}{{\text{x}}^3} + 4{{\text{x}}^2} - 6{\text{x + 2}} \\
   \Rightarrow {{\text{x}}^3} + 4{{\text{x}}^2} - 6{\text{x + 2}} = {{\text{x}}^3} + 4{{\text{x}}^2} - 6{\text{x + 2}} \\
$
Hence proved.