Factor x^4 - x^3 - 7x^2 + x + 6.

Factor x^4 - x^3 - 7x^2 + x + 6.

• A. (x - 1)(x + 1)(x - 3)(x + 2)
• B. (x - 1)(x - 1)(x + 3)(x + 2)
• C. (x - 2)(x + 2)(x - 3)(x + 1)
• D. (x + 1)(x + 1)(x - 3)(x - 2)

Answer: (A)

Factoring a quartic like this often comes from writing it as a product of two quadratics and then factoring those quadratics further. Since the constant term is 6, the product of the constant terms of the two quadratics must be 6, so try b and d with bd = 6. If you take the split b = 2 and d = 3, and set up (x^2 + ax + 2)(x^2 + cx + 3), the coefficients give a + c = -1, ac + 2 + 3 = -7, and ad + bc = 1. This leads to a and c being 3 and -4 (in some order), since 3 + (-4) = -1 and 3(-4) + 2 + 3 = -12 + 5 = -7, and 3·3 + 2(-4) = 9 - 8 = 1. Thus you get (x^2 + 3x + 2)(x^2 - 4x + 3). Each quadratic factors further: x^2 + 3x + 2 = (x + 1)(x + 2) and x^2 - 4x + 3 = (x - 3)(x - 1). Put all together, the polynomial factors as (x - 1)(x + 1)(x - 3)(x + 2). This matches the given product, since the order of factors doesn’t matter.

Factoring a quartic like this often comes from writing it as a product of two quadratics and then factoring those quadratics further. Since the constant term is 6, the product of the constant terms of the two quadratics must be 6, so try b and d with bd = 6. If you take the split b = 2 and d = 3, and set up

(x^2 + ax + 2)(x^2 + cx + 3),

the coefficients give a + c = -1, ac + 2 + 3 = -7, and ad + bc = 1. This leads to a and c being 3 and -4 (in some order), since 3 + (-4) = -1 and 3(-4) + 2 + 3 = -12 + 5 = -7, and 3·3 + 2(-4) = 9 - 8 = 1.

Thus you get (x^2 + 3x + 2)(x^2 - 4x + 3). Each quadratic factors further: x^2 + 3x + 2 = (x + 1)(x + 2) and x^2 - 4x + 3 = (x - 3)(x - 1). Put all together, the polynomial factors as (x - 1)(x + 1)(x - 3)(x + 2). This matches the given product, since the order of factors doesn’t matter.