Question

Shelia is dividing x4 – 4x3 + 4x2 – 6x + 5 by x – 1 using a division table.

Quotient


Divisor
x3
– 3x2
+ x
– 5
x
x4
–3x3
x2
A
– 1
–x3
3x2
B
C
What are the missing values in the table?
A = –5x; B = 1x; C = 5
A = –5x; B = –1x; C = 5
A = 5x; B = –1x; C = 5
A = 5x; B = 1x; C = 5

Answer 1

To determine the values of A, B, and C in the division table while dividing \(x^4 - 4x^3 + 4x^2 - 6x + 5\) by \(x - 1\), we can perform polynomial long division.

1. **First Step**: Divide the leading term of the dividend by the leading term of the divisor.
\[
\frac{x^4}{x} = x^3
\]
Write \(x^3\) in the quotient.

2. **Multiply** \(x^3\) by the divisor \(x - 1\):
\[
x^3(x - 1) = x^4 - x^3
\]
Subtract this from the dividend:
\[
(x^4 - 4x^3 + 4x^2 - 6x + 5) - (x^4 - x^3) = -3x^3 + 4x^2 - 6x + 5
\]

3. **Second Step**: Repeat the division with the new polynomial \(-3x^3 + 4x^2 - 6x + 5\).
\[
\frac{-3x^3}{x} = -3x^2
\]
Write \(-3x^2\) in the quotient.

4. **Multiply** \(-3x^2\) by the divisor:
\[
-3x^2(x - 1) = -3x^3 + 3x^2
\]
Subtract:
\[
(-3x^3 + 4x^2 - 6x + 5) - (-3x^3 + 3x^2) = x^2 - 6x + 5
\]

5. **Third Step**: Divide the new leading term \(x^2\) by \(x\):
\[
\frac{x^2}{x} = x
\]
Write \(x\) in the quotient.

6. **Multiply** \(x\) by the divisor:
\[
x(x - 1) = x^2 - x
\]
Subtract:
\[
(x^2 - 6x + 5) - (x^2 - x) = -5x + 5
\]

7. **Fourth Step**: Divide \(-5x\) by \(x\):
\[
\frac{-5x}{x} = -5
\]
Write \(-5\) in the quotient.

8. **Multiply** \(-5\) by the divisor:
\[
-5(x - 1) = -5x + 5
\]
Subtract:
\[
(-5x + 5) - (-5x + 5) = 0
\]

Now we have the complete quotient:
\[
\text{Quotient} = x^3 - 3x^2 + x - 5
\]

Identifying values:
- From the division table, we can identify:
- \(A = -5\)
- \(B = 1\) (coefficient of \(x\))
- \(C = 5\) (constant term)

Thus, the correct set of values is:
- **A = -5; B = 1; C = 5**

The closest answer matches:
- **A = -5x; B = 1x; C = 5**.