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Q: Explain how to factor the following trinomials forms: x2 + bx + c and ax2 + bx + c. Is there more than one way to factor this? Give an example for your classmates to factor.
A: Let's factor the trinomial x² + bx + c. We know that if b²-4c >= 0 then this trinomial has zeros and it can be factorized. [ It's zeros are x1 = [ -b + v(b²-4c) ] / 2 and x2 = [ -b - v(b²-4c) ] / 2. So, we can factor it as x² + bx + c = (x-x1)(x-x2). In a case of trinomial ax² + bx + c we can factorize it as ax² + bx + c = a(x² + b/a·x + c/a), where the trinomial x² +
b/a·x + c/a has the same form as x² + bx + c. For example, I will pick 2x + 5 and 6x - 1 as my two linear expressions Multiplying them together (2x + 5)(6x - 1) gives 12x^2 - 2x + 30x - 5 = 12x^2 + 28x - 5, which is a factorable polynomial of the form ax^2 + bx + c ]
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