Question 1

What is the AC method for factoring trinomials?

Accepted Answer

The AC method factors ax² + bx + c when a ≠ 1. Step 1: Multiply a × c to get the "AC product." Step 2: Find two numbers that multiply to AC and add to b. Step 3: Rewrite the middle term using these two numbers. Step 4: Factor by grouping. Example: 6x² + 11x + 4. AC = 24. Find factors of 24 that add to 11: 3 and 8. Rewrite: 6x² + 3x + 8x + 4 = 3x(2x+1) + 4(2x+1) = (3x+4)(2x+1).

Question 2

How do I factor a perfect square trinomial?

Accepted Answer

A perfect square trinomial has the form a² + 2ab + b² = (a+b)² or a² − 2ab + b² = (a−b)². Recognition: the first and last terms are perfect squares, and the middle term is ±2 times the product of their square roots. Example: x² + 6x + 9 = (x+3)² because √(x²) = x, √9 = 3, and 2·x·3 = 6x. Spotting this pattern saves time over the AC method.

Question 3

How do I factor a difference of squares?

Accepted Answer

Difference of squares: a² − b² = (a+b)(a−b). Both terms must be perfect squares and connected by a minus sign. Example: x² − 25 = (x+5)(x−5). Works for higher powers too: x⁴ − 16 = (x²+4)(x²−4) = (x²+4)(x+2)(x−2). Note: a sum of squares a² + b² does not factor over the real numbers.

Question 4

When is a polynomial not factorable over the integers?

Accepted Answer

A trinomial ax² + bx + c is not factorable over the integers (called "prime" or "irreducible") when the discriminant b² − 4ac is not a perfect square. For example, x² + x + 1 has discriminant −3 (negative), so it has no real roots and cannot be factored over the integers. It can be factored over the complex numbers or the reals, but not using integer coefficients.

Question 5

What is factoring by grouping?

Accepted Answer

Grouping is used for polynomials with 4 or more terms. Split into two pairs, factor each pair separately, then factor out the common binomial. Example: x³ + 2x² + 3x + 6. Group: (x³ + 2x²) + (3x + 6) = x²(x+2) + 3(x+2) = (x²+3)(x+2). Grouping also underlies the AC method: after rewriting the middle term, you factor a 4-term polynomial by grouping.

Question 6

How does factoring relate to finding roots?

Accepted Answer

Factoring and finding roots are two sides of the same coin (Factor Theorem). If r is a root of p(x) (meaning p(r) = 0), then (x − r) is a factor of p(x). Conversely, if (x − r) is a factor, then r is a root. So factoring ax² + bx + c = a(x−r₁)(x−r₂) immediately gives you the roots r₁ and r₂. This connection makes factoring one of the fastest ways to solve polynomial equations when integer or simple fraction roots exist.

Pattern	Formula	Example	Condition
Difference of Squares	a²−b² = (a+b)(a−b)	x²−9 = (x+3)(x−3)	Both terms perfect squares
Perfect Square (+)	a²+2ab+b² = (a+b)²	x²+6x+9 = (x+3)²	Middle = 2√(first·last)
Perfect Square (−)	a²−2ab+b² = (a−b)²	x²−4x+4 = (x−2)²	Discriminant = 0
Simple trinomial	x²+bx+c = (x+p)(x+q)	x²−5x+6 = (x−2)(x−3)	Find p+q=b, pq=c
AC method	ax²+bx+c	6x²+11x+4 = (3x+4)(2x+1)	Find factors of ac summing to b
GCF first	factor out GCF	6x²+12x = 6x(x+2)	Always check GCF first

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