When \(16x^4+12x^{3}-10x^{2}+8x+20\) is divided by \(4x-3\), the quotient and the remainder are, respectively
To solve the problem of dividing \(16x^4+12x^3-10x^2+8x+20\) by \(4x-3\), we use polynomial long division. The process involves dividing the terms one by one starting from the highest degree:
Divide: Consider the leading term of the dividend \(16x^4\) and the divisor’s leading term \(4x\). Divide \(16x^4\) by \(4x\), which results in \(4x^3\).
Multiply: Multiply the entire divisor \(4x-3\) by the result \(4x^3\), yielding:
\(4x^3 \times (4x-3) = 16x^4 - 12x^3\).
Subtract: Subtract the result from the original polynomial:
\((16x^4 + 12x^3 - 10x^2 + 8x + 20) - (16x^4 - 12x^3)=24x^3-10x^2+8x+20\).
Repeat: Take the new leading term \(24x^3\) and divide by \(4x\) to get \(6x^2\).
Multiply: Multiply \(6x^2\) by \(4x-3\) to get:
\(6x^2 \times (4x-3) = 24x^3 - 18x^2\).
Subtract: Subtract again:
\((24x^3 - 10x^2 + 8x + 20) - (24x^3 - 18x^2) = 8x^2+8x+20\).
Continue: Divide \(8x^2\) by \(4x\) to get \(2x\), multiply \(2x\) by \(4x-3\):
\(2x \times (4x-3) = 8x^2-6x\).
Subtract: Perform subtraction:
\((8x^2 + 8x + 20) - (8x^2 - 6x) = 14x+20\).
Final step: Divide \(14x\) by \(4x\) to get \(\dfrac{7}{2}\). Multiply \(\dfrac{7}{2}\) by \(4x-3\):
\(\dfrac{7}{2} \times (4x-3) = 14x - \dfrac{21}{2}\).
Subtract: Subtract final time:
\((14x + 20) - (14x - \dfrac{21}{2}) = \dfrac{61}{2}\).
The final quotient is \(4x^3 + 6x^2 + 2x + \dfrac{7}{2}\) and the remainder is \(\dfrac{61}{2}\). Therefore, the correct option is:
A sum was divided among $P$, $Q$ & $R$. $R$ got double than $P$ who got double than $Q$. If the difference between the shares of $Q$ and $R$ is Rs. 3675.00, then the sum in rupees is