Which long division problem can be used to prove the formula for factoring the difference of two perfect cubes?
2 answers:
You might be interested in
Answer:
D
Step-by-step explanation:
If the polynomial p(x) is divided by a polynomial of the form x+k (which accounts for all of the possible answer choices in this question), the result can be written as
p(x)/x+k = q(x) + r/x+k
where q(x) is a polynomial and r is the remainder. Since x+k is a degree-1 polynomial (meaning it only includes and no higher exponents), the remainder is a real number.
Therefore, p(x) can be rewritten as p(x) = (x+k)q(x) + r, where r is a real number.
The question states that p(3) = −2, so it must be true that
−2 = p(3) = (3+k)q(3) + r
Now we can plug in all the possible answers. If the answer is A, B, or C, r will be 0, while if the answer is D, r will be −2.
A. −2=p(3)=(3+(−5))q(3)+0
−2=(3−5)q(3)
−2=(−2)q(3)
This could be true, but only if q(3)=1
B. −2=p(3)=(3+(−2))q(3)+0
−2=(3−2)q(3)
−2=(−1)q(3)
This could be true, but only if q(3)=2
C. −2=p(3)=(3+2)q(3)+0
−2=(5)q(3)
This could be true, but only if q(3) = −2/5
D. −2=p(3)=(3+(−3))q(3)+(−2)
−2=(3−3)q(3)+(−2)
−2=(0)q(3)+(−2)
This will always be true no matter what q(3) is.
Of the answer choices, the only one that must be true about p(x) is D, that the remainder when p(x) is divided by x−3 is -2.
The final answer is D.