Question

algebra 2 3.4 solve rational equations
the algebros

pregunta de opción múltiple

factor the trinomial in the denominator.

(x + 3)(x - 4)

(x + 3)(x + 4)

(x - 3)(x + 4)

Explanation:

Step1: Recall trinomial factoring

A trinomial of the form \(x^2 + bx + c\) factors into \((x + m)(x + n)\) where \(m + n = b\) and \(m \times n = c\). For a trinomial like \(x^2 + x - 12\) (assuming the denominator trinomial is \(x^2 + x - 12\), though not shown, we test the options):

• Test \((x + 3)(x - 4)\): Expand \(x^2 - 4x + 3x - 12 = x^2 - x - 12\) (not matching if middle term is \(+x\)). Wait, maybe the trinomial is \(x^2 + x - 12\)? No, wait, let's check each option:
• Option 1: \((x + 3)(x - 4)=x^2 - 4x + 3x - 12 = x^2 - x - 12\)
• Option 2: \((x + 3)(x + 4)=x^2 + 4x + 3x + 12 = x^2 + 7x + 12\)
• Option 3: \((x - 3)(x + 4)=x^2 + 4x - 3x - 12 = x^2 + x - 12\)

Wait, maybe the original trinomial in the denominator is \(x^2 + x - 12\) (common in rational equations). So factoring \(x^2 + x - 12\) gives \((x - 3)(x + 4)\) since \(-3 + 4 = 1\) (middle term coefficient) and \(-3 \times 4 = -12\) (constant term). But wait, maybe I misread. Wait, the problem says "Factor the trinomial in the denominator" (not shown, but typical for rational equations, maybe the trinomial is \(x^2 + x - 12\) or similar). Wait, let's re - check the expansion of each option:

Wait, maybe the trinomial is \(x^2 - x - 12\). Then \((x + 3)(x - 4)=x^2 - 4x + 3x - 12 = x^2 - x - 12\), which would match. But since the problem is about rational equations in Algebra 2, and factoring trinomials, let's assume the trinomial is \(x^2 - x - 12\). Then \((x + 3)(x - 4)\) is the factor. Wait, maybe the original trinomial (not shown) is \(x^2 - x - 12\), so factoring it gives \((x + 3)(x - 4)\).

Step2: Confirm the correct factor

We expand each option:

• For \((x + 3)(x - 4)\): \(x\times x+x\times(-4)+3\times x + 3\times(-4)=x^{2}-4x + 3x-12=x^{2}-x - 12\)
• For \((x + 3)(x + 4)\): \(x\times x+x\times4+3\times x + 3\times4=x^{2}+4x + 3x + 12=x^{2}+7x + 12\)
• For \((x - 3)(x + 4)\): \(x\times x+x\times4-3\times x - 3\times4=x^{2}+4x-3x - 12=x^{2}+x - 12\)

Since the problem is from "Algebra 2 3.4 Solve Rational Equations", and in rational equations, a common denominator trinomial could be \(x^2 - x - 12\) (which factors to \((x + 3)(x - 4)\)) or \(x^2 + x - 12\) (factors to \((x - 3)(x + 4)\)) or \(x^2 + 7x + 12\) (factors to \((x + 3)(x + 4)\)). But since the first option \((x + 3)(x - 4)\) is a valid trinomial factorization (for \(x^2 - x - 12\)), and given the context of rational equations, this is likely the intended answer.

Answer:

\((x + 3)(x - 4)\) (the first option, so the option with \((x + 3)(x - 4)\))