How do you divide the expression $\dfrac{3{{x}^{4}}+5{{x}^{3}}+12x+3}{x+3}$?
Answer
595.5k+ views
Hint: According to the above question, we need to divide the given fourth degree polynomial \[3{{x}^{4}}+5{{x}^{3}}+12x+3\] by the linear polynomial \[x+3\]. For this, we have to use the long division method, which is similar to the division of the numbers. We have to start by dividing the first term of the dividend, which is $3{{x}^{4}}$, by the first term of the divisor, which is $x$, which will give us the first term of the quotient as $3{{x}^{3}}$. This term $3{{x}^{3}}$ needs to be multiplied with the divisor \[x+3\] to get $3{{x}^{4}}+9{{x}^{3}}$ which has to be subtracted from the dividend to obtain $-4{{x}^{3}}+12x+3$. This has to be treated as the new dividend and we have to follow the same procedure to complete the required division.
Complete step-by-step solution:
The division given in the question is $\dfrac{3{{x}^{4}}+5{{x}^{3}}+12x+3}{x+3}$. For this, we long divide \[3{{x}^{4}}+5{{x}^{3}}+12x+3\] by \[x+3\] as shown below.
\[x+3\overset{3{{x}^{3}}-4{{x}^{2}}+12x-24}{\overline{\left){\begin{align}
& 3{{x}^{4}}+5{{x}^{3}}+12x+3 \\
& \underline{3{{x}^{4}}+9{{x}^{3}}} \\
& -4{{x}^{3}}+12x+3 \\
& \underline{-4{{x}^{3}}-12{{x}^{2}}} \\
& 12{{x}^{2}}+12x+3 \\
& \underline{12{{x}^{3}}+36x} \\
& -24x+3 \\
& \underline{-24x-72} \\
& \underline{75} \\
\end{align}}\right.}}\]
From the above division, we obtained the quotient equal to \[3{{x}^{3}}-4{{x}^{2}}+12x-24\] and the remainder equal to $75$.
Hence, we have divided $\dfrac{3{{x}^{4}}+5{{x}^{3}}+12x+3}{x+3}$ and obtained the quotient as \[3{{x}^{3}}-4{{x}^{2}}+12x-24\] and the remainder as $75$.
Note: The divisor in the question was given as a linear polynomial of x. Therefore we can also use the method of synthetic division to carry out the same division. We can check the value of the remainder obtained by using the remainder theorem. The remainder theorem states that when a polynomial $p\left( x \right)$ is divided by $\left( x-a \right)$, then the remainder obtained is equal to $p\left( a \right)$. Since the divisor was given as $\left( x+3 \right)$, we need to substitute $x=-3$ in the dividend to directly obtain the value of the remainder.
Complete step-by-step solution:
The division given in the question is $\dfrac{3{{x}^{4}}+5{{x}^{3}}+12x+3}{x+3}$. For this, we long divide \[3{{x}^{4}}+5{{x}^{3}}+12x+3\] by \[x+3\] as shown below.
\[x+3\overset{3{{x}^{3}}-4{{x}^{2}}+12x-24}{\overline{\left){\begin{align}
& 3{{x}^{4}}+5{{x}^{3}}+12x+3 \\
& \underline{3{{x}^{4}}+9{{x}^{3}}} \\
& -4{{x}^{3}}+12x+3 \\
& \underline{-4{{x}^{3}}-12{{x}^{2}}} \\
& 12{{x}^{2}}+12x+3 \\
& \underline{12{{x}^{3}}+36x} \\
& -24x+3 \\
& \underline{-24x-72} \\
& \underline{75} \\
\end{align}}\right.}}\]
From the above division, we obtained the quotient equal to \[3{{x}^{3}}-4{{x}^{2}}+12x-24\] and the remainder equal to $75$.
Hence, we have divided $\dfrac{3{{x}^{4}}+5{{x}^{3}}+12x+3}{x+3}$ and obtained the quotient as \[3{{x}^{3}}-4{{x}^{2}}+12x-24\] and the remainder as $75$.
Note: The divisor in the question was given as a linear polynomial of x. Therefore we can also use the method of synthetic division to carry out the same division. We can check the value of the remainder obtained by using the remainder theorem. The remainder theorem states that when a polynomial $p\left( x \right)$ is divided by $\left( x-a \right)$, then the remainder obtained is equal to $p\left( a \right)$. Since the divisor was given as $\left( x+3 \right)$, we need to substitute $x=-3$ in the dividend to directly obtain the value of the remainder.
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