Question

How do you divide $\left( {{x}^{3}}-11{{x}^{2}}+22x+40 \right)\div \left( x-5 \right)$ using synthetic division ?

Answer 1

Hint: Synthetic division is a shorthand method of dividing polynomials for the special case of dividing by a linear factor whose leading coefficient is $1$. So the divisor must be of the form $\left( x-k \right)$. Here we can notice that the leading coefficient i.e the coefficient of $x$ is $1$. While solving any polynomial through synthetic division, we only make use of the coefficients of the dividend rather than the entire expression. This reduces the number of steps.

Complete step by step solution:
So now let us see what we need. We need a divisor and a dividend.
Since we are solving through synthetic division, our divisor is not $\left( x-5 \right)$ rather it is the value of $x$ that we get when we equate $\left( x-5 \right)$ to $0$.
$\begin{align}
  & \Rightarrow x-5=0 \\
 & \Rightarrow x=5 \\
\end{align}$
So our divisor is $5$.
Our dividend is not ${{x}^{3}}-11{{x}^{2}}+22x+40$ rather our dividend is only the coefficients of all the terms.
Let us start our division.
\[\begin{align}
  & \Rightarrow 5\left| \!{\underline {\,
  \begin{align}
  & \text{1 -11 22 40} \\
 & \text{ 5 -30 -40} \\
\end{align} \,}} \right. \\
 & \text{ 1 -6 -8 0} \\
\end{align}\]
We are always supposed to write the number present at the start as it is. So $1$ just comes down. So we have to multiply the number that came down , now it is $1$ , and then add it to the next number.
So $5\times 1=5$ . We have to add this to $-11$ . So $-11+5=-6$ . This is the number which comes down. We have to repeat this process until the number that comes down is $0$.
$1,-6,-8$ represent the coefficients of the quadratic equation we get we divide ${{x}^{3}}-11{{x}^{2}}+22x+40$ with $x-5$.
So the quadratic equation is ${{x}^{2}}-6x-8$.
$\therefore $ $\dfrac{{{x}^{3}}-11{{x}^{2}}+22x+40}{x-5}={{x}^{2}}-6x-8$.

Note: The divisor is going to $\left( x-5 \right)$ if we are solving this sum through long division. But we are solving this sum through synthetic division, so the divisor will be only $5$. We have to be very careful while doing the synthetic division as there is a huge scope of making calculations errors. Sometimes, if we are not sure whether the given divisor is a factor of the given dividend but we are asked to divide only through synthetic division, then we quickly have to do long division to just make sure that we are on the right path.