When x2 – y2 – z2 + 2yz is factored, it can be written in the form (ax + by + cz)(dx + ey + fz), where a, b, c, d, e, and f are all integer constants, and a > 0.
What is the value of a + b + c + d + e + f?
A. -2
B. -1
C. 2
D. 0
E. 1
Correct Answer: C
Explanation/Reference:
Explanation:
Let’s try to factor x2 – y2– z2 + 2yz.
Notice that the last three terms are very close to y2 + z2 – 2yz, which, if we rearranged them, would become y2 – 2yz + z2. We could factor y2 – 2yz + z2 as (y – z)2, using the general rule that p2 – 2pq + q2 = (p – q)2 .
So we want to rearrange the last three terms. Let’s group them together first.
x2 + (-y2 – z2 + 2yz) If we were to factor out a -1 from the last three terms, we would have the following:
x2 – (y2 + z2 – 2yz) Now we can replace y2 + z2 – 2yz with (y – z)2.
x2 -(y – z)2 This expression is actually a difference of squares. In general, we can factor p2 – q2 as (p – q)(p + q). In this case, we can substitute x for p and (y – z) for q.
x2 – (y – z)2 = (x – (y – z))(x + (y – z)) Now, let’s distribute the negative one in the trinomial x -(y – z) (x – (y – z))(x + (y – z)) (x – y + z)(x + y – z) The problem said that factoring x2 – y2 – z2 + 2yz would result in two polynomials in the form (ax + by + cz)(dx + ey + fz), where a, b, c, d, e, and f were all integers, and a > 0.
(x – y + z)(x + y – z) fits this form. This means that a = 1, b = -1, c = 1, d = 1, e = 1, and f = -1. The sum of all of these is 2.
The answer is 2.
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