Question #12158

Explain how to factor the following trinomials forms: x2 + bx + c and ax2 + bx + c. Is there more than one way to factor this? Show your answer using both words and mathematical notation.

Expert's answer

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 + √(b²-4c) ] / 2

and

x2 = [ -b - √(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.

x1 = [ -b + √(b²-4c) ] / 2

and

x2 = [ -b - √(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.

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