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Amal used the tabular method to show her work dividing \(-2x^3 + 11x^2 - 23x + 20\) by \(x^2 - 3x + 4\).

Amal's Work:
[tex]\[
\begin{tabular}{|l|c|c|c|c|}
\hline
& & \multicolumn{2}{|c|}{Quotient} & \\
\hline
& & \(2x\) & \(+5\) & \\
\hline
\multirow{3}{*}{Divisor}
& \(x^2\) & \(-2x^3\) & \(5x^2\) & \(6x^2 + 5x^2 = 11x^2\) \\
\cline{2-5}
& \(-3x\) & \(6x^2\) & \(-15x\) & \(-8x + (-15x) = -23x\) \\
\cline{2-5}
& \(+4\) & \(-8x\) & \(20\) & \\
\hline
\end{tabular}
\][/tex]

Amal's answer:
[tex]\[
\frac{-2x^3 + 11x^2 - 23x + 20}{x^2 - 3x + 4} = 2x + 5
\][/tex]

Which statement about Amal's work is true?

A. Amal's work is correct.

B. Amal's work is incorrect; the wrong divisor is on the right side of the division table.

C. Amal's work is incorrect because it includes a positive two instead of negative two in the answer.

D. Amal's work is incorrect because the like terms in the division table do not simplify to the original polynomial.


Sagot :

Amal's work is correct.

To elaborate, we can break down the polynomial division step-by-step as follows:

1. Step 1: Divide the leading term of the numerator by the leading term of the divisor:
[tex]\[ \frac{-2x^3}{x^2} = -2x \][/tex]

2. Step 2: Multiply the entire divisor by the quotient term and subtract from the original polynomial:
[tex]\[ (-2x^3 + 11x^2 - 23x + 20) - (-2x^3 + 6x^2 - 8x) \][/tex]
This simplifies to:
[tex]\[ (-2x^3 + 11x^2 - 23x + 20) + (2x^3 - 6x^2 + 8x) = 5x^2 - 15x + 20 \][/tex]

3. Step 3: Divide the new polynomial by the leading term of the divisor:
[tex]\[ \frac{5x^2}{x^2} = 5 \][/tex]

4. Step 4: Multiply the entire divisor by the second quotient term and subtract:
[tex]\[ (5x^2 - 15x + 20) - (5x^2 - 15x + 20) = 0 \][/tex]
Since the remainder is zero, the division is complete.

5. Combining the quotients, we get:
[tex]\[ -2x + 5 \][/tex]

Since both intermediate and final steps show the operations were performed correctly, we can conclude:
[tex]\[ \frac{-2 x^3+11 x^2-23 x+20}{x^2-3 x+4} = -2x + 5 \][/tex]

Therefore, Amal's work is indeed correct.