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# If the polynomial $f(x)=6{{x}^{4}}+8{{x}^{3}}+17{{x}^{2}}+21x+7$ is divided by another polynomial $g(x)=3{{x}^{2}}+4x+1$ the remainder is (ax+b). Find a and b.

Last updated date: 13th Jun 2024
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Hint: The function given is $f(x)=6{{x}^{4}}+8{{x}^{3}}+17{{x}^{2}}+21x+7$ and it is given that when the function is divided by $g(x)=3{{x}^{2}}+4x+1$ leaves the remainder (ax+b). So, divide the function and get the remainder by long division method and compare with (ax+b) to get the values of a and b.

The given function is $f(x)=6{{x}^{4}}+8{{x}^{3}}+17{{x}^{2}}+21x+7$. So, let us divide the function by $g(x)=3{{x}^{2}}+4x+1$ to get the remainder and quotient. To divide use the following method:
$\begin{matrix} 3{{x}^{2}}+4x+1\overset{2{{x}^{2}}+5}{\overline{\left){6{{x}^{4}}+8{{x}^{3}}+17{{x}^{2}}+21x+7}\right.}} \\ -(6{{x}^{4}}+8{{x}^{3}}+2{{x}^{2}}) \\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ 15{{x}^{2}}+21x+7 \\ -(15{{x}^{2}}+20x+5) \\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ x+2 \\ \end{matrix}$
So, looking at the results of the above division, we can say that the remainder is $\left( x+2 \right)$ . If we compare this with $(ax+b)$, we can say that $a=1$, as the coefficient of x in the remainder is 1 and b=2, as the constant term in the remainder is 2.