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If the quotient obtained on dividing $\left( {{x}^{4}}+10{{x}^{3}}+35{{x}^{2}}+50x+29 \right)$ by $\left( x+4 \right)$ is $\left( {{x}^{3}}-a{{x}^{2}}+bx+6 \right)$ then find $a$, $b$ and also the remainder.

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Last updated date: 24th Feb 2024
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IVSAT 2024
Answer
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Hint: In the above question, the dividend is given as $\left( {{x}^{4}}+10{{x}^{3}}+35{{x}^{2}}+50x+29 \right)$, and the divisor is given to be $\left( x+4 \right)$. So we can use the long division method to carry out the required division of the dividend by the divisor. From there, we will get the required value of the remainder. Also, we will get the quotient as a polynomial of third degree, which on comparing with the given quotient $\left( {{x}^{3}}-a{{x}^{2}}+bx+6 \right)$ will yield the required values of the coefficients $a$ and $b$.

Complete step by step solution:
According to the above question, we are given the dividend to be $\left( {{x}^{4}}+10{{x}^{3}}+35{{x}^{2}}+50x+29 \right)$ and the divisor to be $\left( x+4 \right)$. So we can carry out the long division of these to find out the quotient and the remainder as shown below.
\[x+4\overset{{{x}^{3}}+6{{x}^{2}}+11x+6}{\overline{\left){\begin{align}
  & {{x}^{4}}+10{{x}^{3}}+35{{x}^{2}}+50x+29 \\
 & \underline{{{x}^{4}}+4{{x}^{3}}} \\
 & 6{{x}^{3}}+35{{x}^{2}}+50x+29 \\
 & \underline{6{{x}^{3}}+24{{x}^{2}}} \\
 & 11{{x}^{2}}+50x+29 \\
 & \underline{11{{x}^{2}}+44x} \\
 & 6x+29 \\
 & \underline{6x+24} \\
 & \underline{5} \\
\end{align}}\right.}}\]
From the above, we see that the quotient is equal to \[{{x}^{3}}+6{{x}^{2}}+11x+6\]. In the above question, the quotient is given as $\left( {{x}^{3}}-a{{x}^{2}}+bx+6 \right)$. On comparing the coefficients of these two, we get
$\begin{align}
  & \Rightarrow -a=6 \\
 & \Rightarrow a=-6 \\
\end{align}$
And
$\Rightarrow b=11$
Also, from the above division, the remainder is equal to $5$.
Hence, the values of $a$, $b$ and the remainder are $-6$, $11$ and $5$ respectively.

Note: We can also use the remainder theorem to get the value of the remainder directly without carrying out the long division. The remainder theorem states that when a polynomial $p\left( x \right)$ is divided by the divisor $\left( x+a \right)$, then the remainder obtained will be equal to $p\left( -a \right)$. In our case the divisor was given as $\left( x+4 \right)$. Therefore, on substituting $x=-4$ in the given dividend $\left( {{x}^{4}}+10{{x}^{3}}+35{{x}^{2}}+50x+29 \right)$ we will directly obtain the remainder as $5$.
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