Ordered pair: (3x³+10x²+x-6, x-1); Answer: No
Ordered pair: (4x² + 13x + 10, x+2); Answer: Yes
Ordered pair: (3x³+10x²+x-6, x+3); Answer: Yes
1. (3x³+10x²+x-6, x-1)
Using the Remainder Theorem, we can determine whether x-1 is a factor of 3x³+10x²+x-6. The theorem states that if f(a) = 0 for some value a, then (x-a) is a factor of f(x).
Step 1: Evaluate f(1):
f(1) = 3(1)³ + 10(1)² + 1(1) - 6 = 16
Since f(1) ≠ 0, then x-1 is not a factor of 3x³+10x²+x-6.
Therefore, the ordered pair is (3x³+10x²+x-6, x-1) and the answer is No.
2. (4x² + 13x + 10, x+2)
Step 1: Evaluate f(-2):
f(-2) = 4(-2)² + 13(-2) + 10 = 0
Since f(-2) = 0, then x+2 is a factor of 4x² + 13x + 10.
Therefore, the ordered pair is (4x² + 13x + 10, x+2) and the answer is Yes.
3. (3x³+10x²+x-6, x+3)
Step 1: Evaluate f(-3):
f(-3) = 3(-3)³ + 10(-3)² + (-3) - 6 = 0
Since f(-3) = 0, then x+3 is a factor of 3x³+10x²+x-6.
Therefore, the ordered pair is (3x³+10x²+x-6, x+3) and the answer is Yes.