Factor x² − 4x + 3 and use the factorisation to solve x² − 4x + 3 = 0.

Step-by-step solution

1. 1

   Set up the target form. A monic quadratic that factors over the integers can be written

   $x^2 - 4x + 3 = (x - a)(x - b)$

   Expanding the right side gives $x^2 - (a+b)x + ab$, so the two unknowns are pinned down by two conditions rather than guessed.

2. 2

   Read off the sum and product conditions. Comparing coefficients:

   $a + b = 4, \qquad ab = 3$

   Equivalently, in the form $x^2 + bx + c$ you need two numbers whose product is $c = +3$ and whose sum is $-4$; those numbers are $-1$ and $-3$.

3. 3

   Use the signs to narrow the search fast. The product $+3$ is positive, so the two numbers share a sign; the sum $-4$ is negative, so both must be negative. The only negative factor pair of $3$ is $(-1, -3)$, and indeed $(-1)(-3) = 3$ and $(-1) + (-3) = -4$. Hence

   $x^2 - 4x + 3 = (x - 1)(x - 3)$

4. 4

   Check the factorisation by expanding.

   $(x - 1)(x - 3) = x^2 - 3x - x + 3 = x^2 - 4x + 3 \ \checkmark$

   The middle coefficient comes from the two cross terms $-3x$ and $-x$; this expansion step is what makes the "how did you get that" question answerable.

5. 5

   Apply the zero-product property. A product equals zero only when one of its factors is zero:

   $(x - 1)(x - 3) = 0 \ \Longrightarrow \ x - 1 = 0 \ \text{or} \ x - 3 = 0$

   $x = 1 \quad \text{or} \quad x = 3$

   Substituting back: $1 - 4 + 3 = 0$ and $9 - 12 + 3 = 0$. The quadratic formula gives the same pair, $x = \frac{4 \pm \sqrt{16 - 12}}{2} = \frac{4 \pm 2}{2}$.