\(x^2\): \(b - a + 1 = -2 \Rightarrow b - 4 + 1 = -2 \Rightarrow b = 1\)
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- Now add \(Ax + B\) (since divisor is quadratic, remainder is linear: \(Ax + B\)):
- x^4 + (a - 1)x^3 + (b - a + 1)x^2 + (a - b + A)x + (b + B)
- Match:
- \(x\): \(a - b + A = 1 \Rightarrow 4 - 1 + A = 1 \Rightarrow A = -2\)
- constant: \(b + B = 5 \Rightarrow 1 + B = 5 \Rightarrow B = 4\)
- So remainder is \(-2x + 4\)? But earlier substitution gave \(3\omega + 4\), contradiction.
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Let the integers be \( n-1 \), \( n \), and \( n+1 \).
Their sum:
n = \frac{96}{3} = 32
The integers are 31, 32, 33. The largest is 33.
#### 33
A line passes through the points (2, 5) and (6, 17). What is the slope of the line?
Slope \( m \) is:
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{17 - 5}{6 - 2} = \frac{12}{4} = 3
If \( x^2 - 5x + 6 = 0 \), what are the values of \( x \)?
Factor the quadratic: