Question

Factor \(x^{2}-4 x+3 .\) When testing possible factorizations, why is it unnecessary to test \((x-1)(x+3)\) and \((x+1)(x-3) ?\)

Short answer

The factored form of \(x^{2} - 4x + 3\) is \((x - 3)(x - 1)\). It's unnecessary to test \((x-1)(x+3)\) and \((x+1)(x-3)\) because the sum of these factors together does not equal -4 which is the coefficient of 'x' in the original equation.

Step-by-step solution

Identify the quadratic form

Notice that the given expression (\(x^{2} - 4x + 3\)) is a quadratic equation in the standard form \(ax^{2} + bx + c\). Here, a=1, b=-4 and c=3.

Find two numbers

Find two numbers such that they multiply to give 'ac' (Product of a and c, here a=1, c=3. So ac=3) and add up to give 'b' (Sum is equal to b=-4). The two numbers that satisfy these conditions are -3 and -1 because (-3)*(-1) = 3 and -3 + (-1) = -4.

Factorize the expression

We place these two numbers in the binomials. So, the factored form of \(x^{2} - 4x + 3\) is \((x - 3)(x - 1)\).

Explain unnecessary factorizations

The expressions \((x-1)(x+3)\) and \((x+1)(x-3)\) are unnecessary to test because the sum of the factors (numbers in parentheses) do not equal to -4 (the coefficient of 'x' in the original equation \(x^{2} - 4x + 3\)), which is a necessary condition for correct factorization of quadratic expressions.