Question

Find the values of a and b such that $ 8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+ax+b $ is exactly divisible by $ 4{{x}^{2}}+3x-2 $ .

Answer 1

Hint: Here, in this question, we can use the fact that the dividend is equal to the product of the divisor and the quotient when added with the remainder, that is,
 $ p\left( x \right)=g\left( x \right)\times q\left( x \right)+r\left( x \right) $ , where p $ \left( x \right) $ , g $ \left( x \right) $ , q $ \left( x \right) $ and r $ \left( x \right) $ are the dividend, divisor, quotient and remainder respectively and do the long division method to get the quotient and remainder.

Complete step-by-step answer:
In this given question, we are asked to find the values of a and b such that $ 8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+ax+b $ is exactly divisible by $ 4{{x}^{2}}+3x-2 $ .
We are going to use the fact that the dividend is equal to the product of the divisor and the quotient when added with the remainder, that is,
 $ p\left( x \right)=g\left( x \right)\times q\left( x \right)+r\left( x \right) $ where p $ \left( x \right) $ , g $ \left( x \right) $ , q $ \left( x \right) $ and r $ \left( x \right) $ are the dividend, divisor, quotient and remainder respectively.
Here, we are, going to perform the long division method to divide $ 8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+ax+b $ by $ 4{{x}^{2}}+3x-2 $ to find our answers as the a and b by equalizing the remainder\[ax+b\] with 0 as the given dividend is exactly divisible by the divisor. In the long division method, the first term of the quotient is chosen such that after multiplying with it the divisor (the polynomial by which the other polynomial is being divided) and then subtracting from the dividend (the polynomial to be divided), the highest power of x from the dividend gets cancelled out. We continue in this way with the resulting remainder as the new dividend until the power of the remainder becomes less than the power of the divisor.
\[4{{x}^{2}}+3x-2\overset{2{{x}^{2}}+2x-1}{\overline{\left){\begin{align}
  & 8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+ax+b \\
 & -\left( 8{{x}^{4}}+6{{x}^{3}}-4{{x}^{2}} \right) \\
 & 8{{x}^{3}}+2{{x}^{2}}+ax+b \\
 & -\left( 8{{x}^{3}}+6{{x}^{2}}-4x \right) \\
 & -4{{x}^{2}}+ax+4x+b \\
 & -\left( -4{{x}^{2}}-3x-2 \right) \\
 & ax+7x+b+2 \\
\end{align}}\right.}}\]
From the above, we get the quotient as \[2{{x}^{2}}+2x-1\] and the remainder as\[ax+7x+b+2=\left( a+7 \right)x+\left( b+2 \right)\] .
So, \[8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+ax+b=\left( 4{{x}^{2}}+3x-2 \right)\times \left( 2{{x}^{2}}+2x-1 \right)+\left( a+7 \right)x+\left( b+2 \right)\].
Now, as the dividend is exactly divisible by the divisor, the remainder must be 0. We know that if a polynomial is equal to 0 then the coefficients of each power of x should be separately equal to 0.
So,
\[\begin{align}
  & \left( a+7 \right)x+\left( b+2 \right)=0 \\
 & \Rightarrow \left( a+7 \right)=0\text{ and }\left( b+2 \right)=0 \\
 & \Rightarrow a=-7\text{ and b=}-2 \\
\end{align}\]
Therefore, the values of a and b are -7 and -2 respectively.

Note: In this sort of question, we must be very careful while doing the long division process about changing the signs of the terms in the quotient. The divisor should be multiplied by the quotient and the resulting whole polynomial should be subtracted from the dividend. If even a single sign is changed in the wrong way, the whole answer could be different and wrong.