Question 1
When dividing 3x4−5x3+2x2−7x+4 by x2−2x+1, what is the remainder?
- −3x+2 (correct answer)
- −5x+6
- 2x−1
- x+3
Explanation: Using polynomial long division: 3x4−5x3+2x2−7x+4=(x2−2x+1)(3x2+x+2)+(−3x+2). The remainder is −3x+2. Choice B results from sign errors during division. Choice C comes from incorrect handling of the constant term. Choice D represents an incomplete calculation stopping too early.
Question 2
If synthetic division of 2x3−7x2+kx−15 by (x−3) results in a remainder of 0, what is the value of k?
- k=2
- k=5
- k=8 (correct answer)
- k=11
Explanation: Since the remainder is 0, (x−3) is a factor, so f(3)=0. Substituting: 2(3)3−7(3)2+k(3)−15=0, which gives 54−63+3k−15=0, so 3k=24 and k=8. Choice A comes from using x=−3 instead of x=3. Choice B results from arithmetic errors in the calculation. Choice D comes from forgetting the negative sign on the constant term.
Question 3
When 6x3−13x2+px+q is divided by (x−2)(x+1), the remainder is 5x−3. What are the values of p and q?
- p=7,q=6
- p=−2,q=9 (correct answer)
- p=4,q=−8
- p=1,q=12
Explanation: Since the divisor has degree 2, we can write 6x3−13x2+px+q=(x−2)(x+1)⋅Q(x)+5x−3. At x=2: 48−52+2p+q=10−3, so 2p+q=11. At x=−1: −6−13−p+q=−5−3, so −p+q=11. Solving: p=−2,q=9. The other choices result from sign errors or incorrect substitution in the system of equations.