Question

How do you solve \[\dfrac{15{{x}^{4}}-14{{x}^{3}}-7{{x}^{2}}+x+9}{5x-3}\]?

Answer 1

Hint: This question is from the topic of algebra. We are going to solve this question using a long division method which is done for polynomials. We will understand the solution of this question by taking the references from the figures. We will take the term \[15{{x}^{4}}-14{{x}^{3}}-7{{x}^{2}}+x+9\] as dividend and the term \[5x-3\] as divider. From there, we will find the quotient and remainder. That will be our answer.

Complete step by step solution:
Let us solve this question.
In this question, we have asked to solve the term \[\dfrac{15{{x}^{4}}-14{{x}^{3}}-7{{x}^{2}}+x+9}{5x-3}\]
So, we will divide the term \[15{{x}^{4}}-14{{x}^{3}}-7{{x}^{2}}+x+9\] by the term \[5x-3\] using long division method.
\[5x-3\overline{\left){15{{x}^{4}}-14{{x}^{3}}-7{{x}^{2}}+x+9}\right.}\]
From the above figure, we can see that \[15{{x}^{4}}-14{{x}^{3}}-7{{x}^{2}}+x+9\] is dividend and \[5x-3\] is divisor.
Now, we will divide them. The first term of the quotient will be like the first term of dividend divided by the first term of divisor, that will be \[\dfrac{15{{x}^{4}}}{5x}=3{{x}^{3}}\]. So, we can write the above figure as
\[5x-3\overset{3{{x}^{3}}}{\overline{\left){\begin{align}
  & 15{{x}^{4}}-14{{x}^{3}}-7{{x}^{2}}+x+9 \\
 & 15{{x}^{4}}-9{{x}^{3}} \\
\end{align}}\right.}}\]
Here, the two terms in the below have been written as \[5x-3\] multiplied by \[3{{x}^{3}}\]. Now, we will subtract the below two terms from the above terms, the first term will be cancelled out, then we will get
\[5x-3\overset{3{{x}^{3}}}{\overline{\left){\begin{align}
  & 15{{x}^{4}}-14{{x}^{3}}-7{{x}^{2}}+x+9 \\
 & 15{{x}^{4}}-9{{x}^{3}} \\
 & \overline{0-5{{x}^{3}}-7{{x}^{2}}+x+9} \\
\end{align}}\right.}}\]
Now, we have the term \[-5{{x}^{3}}-7{{x}^{2}}+x+9\], so we will divide it by the divisor. For that, we will add a term in quotient so that if we multiply that quotient term to the first term of the divisor, then we will get as \[-5{{x}^{3}}\]. The term we add in the quotient will be \[\dfrac{-5{{x}^{3}}}{5x}=-{{x}^{2}}\]. We can see from the below figure
\[5x-3\overset{3{{x}^{3}}-{{x}^{2}}}{\overline{\left){\begin{align}
  & 15{{x}^{4}}-14{{x}^{3}}-7{{x}^{2}}+x+9 \\
 & 15{{x}^{4}}-9{{x}^{3}} \\
 & \overline{0-5{{x}^{3}}-7{{x}^{2}}+x+9} \\
 & -5{{x}^{3}}+3{{x}^{2}} \\
 & \overline{0-10{{x}^{2}}+x+9} \\
\end{align}}\right.}}\]
In the above figure, we have seen that the term \[-{{x}^{2}}\] is multiplied with the \[5x-3\] and have got the term as \[-5{{x}^{3}}+3{{x}^{2}}\] and is subtracted from the term \[-5{{x}^{3}}-7{{x}^{2}}+x+9\], we have got the subtracted term as \[-10{{x}^{2}}+x+9\].
Now, we will add a term in the quotient which is when multiplied by \[5x-3\], then we will get the first term as the first term of \[-10{{x}^{2}}+x+9\]. So, the term we add in the quotient will be \[\dfrac{-10{{x}^{2}}}{5x}=-2x\].
Now, after adding \[-2x\] in the quotient, we will write
\[5x-3\overset{3{{x}^{3}}-{{x}^{2}}-2x}{\overline{\left){\begin{align}
  & 15{{x}^{4}}-14{{x}^{3}}-7{{x}^{2}}+x+9 \\
 & 15{{x}^{4}}-9{{x}^{3}} \\
 & \overline{0-5{{x}^{3}}-7{{x}^{2}}+x+9} \\
 & -5{{x}^{3}}+3{{x}^{2}} \\
 & \overline{0-10{{x}^{2}}+x+9} \\
 & -10{{x}^{2}}+6x \\
 & \overline{0-5x+9} \\
\end{align}}\right.}}\]
Now, we have written the term after subtracting. We can see from the above figure.
Now, we will add a term -5 to the quotient to do the further process of division.
\[5x-3\overset{3{{x}^{3}}-{{x}^{2}}-2x-1}{\overline{\left){\begin{align}
  & 15{{x}^{4}}-14{{x}^{3}}-7{{x}^{2}}+x+9 \\
 & 15{{x}^{4}}-9{{x}^{3}} \\
 & \overline{\begin{align}
  & 0-5{{x}^{3}}-7{{x}^{2}}+x+9 \\
 & -5{{x}^{3}}+3{{x}^{2}} \\
\end{align}} \\
 & \overline{0-10{{x}^{2}}+x+9} \\
 & -10{{x}^{2}}+6x \\
 & \overline{\begin{align}
  & 0-5x+9 \\
 & -5x+3 \\
 & \overline{0+6} \\
\end{align}} \\
\end{align}}\right.}}\]
We have subtracted the term which is multiplied by -1 to the divisor from the term \[-5x+9\] and get a remainder of (0+6) that is 6.
So, we have divided the term \[15{{x}^{4}}-14{{x}^{3}}-7{{x}^{2}}+x+9\] with the term \[5x-3\], and get the quotient as \[3{{x}^{3}}-{{x}^{2}}-2x-1\] and remainder as 6.
So, we can write the term \[\dfrac{15{{x}^{4}}-14{{x}^{3}}-7{{x}^{2}}+x+9}{5x-3}\] as \[3{{x}^{3}}-{{x}^{2}}-2x-1+\dfrac{6}{5x-3}\]

Note:
For solving this type of question, we should have a better knowledge in dividing the polynomials by long division method. Always remember that the number which we divide is called the dividend, the number by which we divide is called the divisor, the result obtained is called the quotient, and the number left over is called the remainder.
\[5x-3\overset{3{{x}^{3}}}{\overline{\left){\begin{align}
  & 15{{x}^{4}}-14{{x}^{3}}-7{{x}^{2}}+x+9 \\
 & 15{{x}^{4}}-9{{x}^{3}} \\
\end{align}}\right.}}\]
In the above figure, if someone is getting confused that how the numbers are being subtracted then he/she can understand by the following:
We will change the signs of the below term and then we will add the terms. We can take reference for this from the following figure:
\[5x-3\overset{3{{x}^{3}}}{\overline{\left){\begin{align}
  & 15{{x}^{4}}-14{{x}^{3}}-7{{x}^{2}}+x+9 \\
 & \underset{-}{\mathop{+}}\,15{{x}^{4}}\underset{+}{\mathop{-}}\,9{{x}^{3}} \\
 & \overline{0-5{{x}^{3}}-7{{x}^{2}}+x+9} \\
\end{align}}\right.}}\]
We have used the below signs for addition. So, we can use this method too to add the terms.