Question

Factor \(x^{2}+2 x-3 .\) When testing possible factorizations, why is it unnecessary to test \((x-1)(x-3)\) and \((x+1)(x+3) ?\)

Short answer

The factorization of \( x^{2}+2 x-3 \) is \( (x-3)(x+1) \). It's unnecessary to test \((x-1)(x-3)\) and \((x+1)(x+3)\) as their multiplication will result in quadratic expressions that don't match the initial one.

Step-by-step solution

Express the middle term

Factorization of quadratic expressions involves finding two numbers that both add up to the coefficient of the middle term (b = 2, in this case) and multiply to give the constant term (c = -3, in this case). So, we need to find two numbers, let's call them m and n, that satisfy these conditions: \( m + n = 2 \) and \( m \times n = -3 \). The two numbers that meet these conditions are 3 and -1.

Factorize the quadratic expression

With the two numbers found in Step 1, we can write the quadratic equation as follows: \( x^{2}+2 x-3 = (x-3)(x+1) \). So, the factorization of the initial quadratic equation is \( (x-3)(x+1) \).

Explain why other binomials are unnecessary to check

The binomials \((x-1)(x-3)\) will result in a quadratic expression \( x^{2} - 4x + 3 \), and \((x+1)(x+3) \) will result in a quadratic expression \( x^{2} + 4x + 3 \) when multiplied out with the FOIL method. Neither of these expressions match the initial quadratic equation \( x^{2}+2 x-3 \), which is why it is unnecessary to test these binomials.