(x-3)(x-5)(x-7)%2B15-%E5%9B%A0%E6%95%B0%E5%88%86%E8%A7%A3.jpeg "(x-1)(x-3)(x-5)(x-7)+15 因数分解")
Factoring the expression (x-1)(x-3)(x-5)(x-7)+15
This problem involves a clever trick to factor a seemingly complex expression. Let's break it down step-by-step:
1. Recognizing the pattern:
Observe that the first part of the expression consists of four consecutive odd numbers subtracted from x. This suggests a potential pattern we can exploit.
2. Rearranging terms:
Let's rearrange the expression by grouping the first and last terms, and the middle two terms:
[(x-1)(x-7)] * [(x-3)(x-5)] + 15
3. Expanding and simplifying:
Expanding the grouped terms, we get:
(x² - 8x + 7) * (x² - 8x + 15) + 15
Notice that both expressions within the parentheses share the same first two terms (x² - 8x). This is crucial for the next step.
4. Substitution for simplification:
Let's substitute y for (x² - 8x):
(y + 7) * (y + 15) + 15
5. Expanding and factoring:
Expanding the expression:
y² + 22y + 105 + 15 = y² + 22y + 120
Factoring the quadratic:
(y + 10)(y + 12)
6. Substituting back:
Now, we substitute back (x² - 8x) for y:
(x² - 8x + 10)(x² - 8x + 12)
7. Final factorization:
Finally, we factor the remaining quadratics:
(x - 2)(x - 6)(x - 4)(x - 3)
Therefore, the fully factored expression is (x - 2)(x - 6)(x - 4)(x - 3).
This method demonstrates how recognizing patterns and strategic substitutions can lead to a simplified solution for seemingly complicated expressions.