The polynomial ax3 + bx2 + x - 6 has (x + 2) as a factor and leaves remainder…

For a polynomial P(x), the Factor Theorem states that (x - r) is a factor of P(x) if and only if P(r) = 0. The Remainder Theorem states that when P(x) is divided by (x - r), the remainder equals P(r). Together, these convert a factor/remainder condition directly into a linear equation in the polynomial's unknown coefficients.

Here P(x) = ax3 + bx2 + x - 6. Since (x + 2) is a factor, the Factor Theorem gives P(-2) = 0. Since dividing by (x - 2) leaves remainder 4, the Remainder Theorem gives P(2) = 4. Substituting each value of x produces two simultaneous equations in a and b:

1. P(-2) = a(-2)3 + b(-2)2 + (-2) - 6 = -8a + 4b - 8. Setting this equal to 0 and dividing by 4 gives -2a + b = 2.

2. P(2) = a(2)3 + b(2)2 + 2 - 6 = 8a + 4b - 4. Setting this equal to 4 and dividing by 4 gives 2a + b = 2.

3. Subtracting the first equation from the second eliminates b: (2a + b) - (-2a + b) = 2 - 2, so 4a = 0, giving a = 0.

4. Substituting a = 0 into 2a + b = 2 gives b = 2.

Check: with a = 0 and b = 2, P(x) = 2x2 + x - 6. Then P(-2) = 2(4) - 2 - 6 = 0, confirming (x + 2) is a factor, and P(2) = 2(4) + 2 - 6 = 4, confirming the remainder condition. So a = 0 and b = 2 satisfies both conditions.