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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.

Complete step-by-step answer:
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:
First arrange the term of dividend and the divisor in the decreasing order of their degrees.
To obtain the first term of the quotient divide the highest degree term of the dividend by the highest degree term of the divisor.
To obtain the second term of the quotient, divide the highest degree term of the new dividend obtained as remainder by the highest degree term of the divisor.
Continue this process till the degree of remainder is less than the degree of divisor.
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 \\
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.
Hence, we can conclude that a=1 and b=2.

Note: The possible error that you may encounter could be that the long division was not done properly. So, be careful and don’t miss the step of changing the signs of the terms in the method of long division as students generally make a mistake in that part.