Question

Divide and verify the division algorithm of the polynomial: \[{{x}^{2}}+5x+6\] by \[\left( x+3 \right)\]?

Answer 1

Hint: We can solve \[\dfrac{{{x}^{2}}+5x+6}{x+3}\] by simple algebraic division. First, we divide the first term of dividend by first term of divisor and the quotient is equal to the first term of the quotient of the total division. Then we will multiply with all the terms of the divisor. Then we will subtract the result from \[{{x}^{2}}+5x+6\]. Now the result of subtraction is our new dividend and we will continue the same process until the highest power of x in reminder is less than the highest power of x in divisor.

Complete step by step answer:
We have to solve \[\dfrac{{{x}^{2}}+5x+6}{x+3}\]
First the quotient when we divide \[{{x}^{2}}\] by x is equal to x, so the first term of quotient of \[\dfrac{{{x}^{2}}+5x+6}{x+3}\] is equal to 2x
\[x+3\overset{x}{\overline{\left){{{x}^{2}}+5x+6}\right.}}\]
Now multiplying x with \[x+3\] we get \[{{x}^{2}}+3x\]
\[x+3\overset{x}{\overline{\left){\begin{align}
  & {{x}^{2}}+5x+6 \\
 & \underline{{{x}^{2}}+3x} \\
\end{align}}\right.}}\]
Now subtracting \[{{x}^{2}}+3x\] form \[{{x}^{2}}+5x+6\] we get \[2x+6\]
\[x+3\overset{x}{\overline{\left){\begin{align}
  & {{x}^{2}}+5x+6 \\
 & \underline{{{x}^{2}}+3x} \\
 & 0{{x}^{2}}+2x+6 \\
\end{align}}\right.}}\]
Now diving 2x by x we get 2, so 4 is our next term of quotient, the quotient is updated as \[x+4\]. Now multiply 2 with x + 3 we get \[2x+6\]
Subtracting \[2x+6\] from \[2x+6\] we get 0.
\[x+3\overset{x+2}{\overline{\left){\begin{align}
  & {{x}^{2}}+5x+6 \\
 & \underline{{{x}^{2}}+3x} \\
 & 0{{x}^{2}}+2x+6 \\
 & \underline{0{{x}^{2}}+2x+6} \\
 & 0 \\
\end{align}}\right.}}\]
0 is a constant. We can see the highest power of x in 0 is less than in the divisor.
So, 0 is the reminder and \[x+2\] is the quotient.
\[\dfrac{{{x}^{2}}+5x+6}{x+3}=x+2\]

Note:
In the above solution, we can check our answer by factoring the equation \[{{x}^{2}}+5x+6\]. By factoring the equation, we will get \[\left( x+3 \right)\left( x+2 \right)\]. Always remember we should keep the division process continued until the highest power of x in the remainder is less than the highest coefficient of x in the divisor. If we terminate the division before that, then we can still divide the remainder by divisor.