Factor x^2 - 7x + 12.

Step-by-step solution

1. 1

   Set up the search. For a monic quadratic $x^2+bx+c$, factoring means finding two numbers whose product is $c=12$ and whose sum is $b=-7$. Then $x^2+bx+c=(x+m)(x+n)$ with those two numbers as $m$ and $n$.

2. 2

   Use the signs to narrow the search before listing anything. The product $+12$ is positive, so the two numbers share a sign; the sum $-7$ is negative, so that shared sign is negative. Only negative pairs need checking — which halves the work.

3. 3

   List the negative factor pairs of 12 and their sums.

   $(-1,-12)\to-13,\qquad(-2,-6)\to-8,\qquad(-3,-4)\to-7\ \checkmark$

   The pair $-3$ and $-4$ works: $(-3)(-4)=12$ and $-3+(-4)=-7$.

4. 4

   Write the factorisation.

   $x^2-7x+12=(x-3)(x-4)$

5. 5

   Check by expanding. $(x-3)(x-4)=x^2-4x-3x+12=x^2-7x+12$ ✓. The middle coefficient is the sum and the constant is the product, exactly as designed.

6. 6

   Read off the roots and verify. The expression is zero at $x=3$ and $x=4$. Comparing the original with the factored form at every integer from $-30$ to $29$ gives exact agreement ✓, and the discriminant $49-48=1$ is a perfect square — the guarantee that integer factors had to exist.