Question

fill in the missing values below one at a time to find the quotient when $-x^{3}+20x - 16$ is divided by $x - 4$.

Explanation:

Step1: Find the first term of the quotient

We divide the leading term of the dividend \(-x^{3}\) by the leading term of the divisor \(x\).
\(\frac{-x^{3}}{x}=-x^{2}\)

Step2: Multiply the divisor \((x - 4)\) by \(-x^{2}\)

\(-x^{2}\times x=-x^{3}\) (this fills the first cell in the \(x\) - row)
\(-x^{2}\times(- 4)=4x^{2}\) (this fills the first cell in the \(-4\) - row)

Step3: Consider the next term

Since the dividend is \(-x^{3}+0x^{2}+20x - 16\), after subtracting \(-x^{3}\) (from the first multiplication), we bring down the \(0x^{2}\). Now we divide \(4x^{2}\) (from the multiplication \(-x^{2}(x - 4)\)) by \(x\) (leading term of divisor) to get \(4x\) (second term of the quotient).
\(4x\times x = 4x^{2}\) (second cell in \(x\) - row)
\(4x\times(-4)=-16x\) (second cell in \(-4\) - row)

Step4: Consider the last non - constant term

After subtracting \(4x^{2}\) (from the previous step), we bring down \(20x\). Now we have \(20x-(-16x)=36x\). Divide \(36x\) by \(x\) (leading term of divisor) to get \(4\) (third term of the quotient)
\(4\times x=4x\) (third cell in \(x\) - row)
\(4\times(-4)=-16\) (third cell in \(-4\) - row)

Answer:

The first missing value (top - left cell) is \(-x^{2}\), the second term of the quotient row is \(4x\), the third term of the quotient row is \(4\). The second cell in the \(x\) - row is \(4x^{2}\), the third cell in the \(x\) - row is \(4x\). The first cell in the \(-4\) - row is \(4x^{2}\), the second cell in the \(-4\) - row is \(-16x\), the third cell in the \(-4\) - row is \(-16\). The quotient is \(-x^{2}+4x + 4\)