Factoring and Roots - Precalculus & Trigonometry Step by Step

Factor x² - 5x + 6 into (x-2)(x-3), identify roots 2 and 3, and verify each root satisfies p(r) = 0.

The Factor Theorem states: r is a root of p(x) if and only if (x - r) is a factor of p(x), equivalently p(r) = 0. Example: x² - 5x + 6. Find two numbers that multiply to 6 and add to -5: those are -2 and -3, so roots are x = 2 and x = 3. Factors: (x - 2)(x - 3). Verify: p(2) = 4 - 10 + 6 = 0; p(3) = 9 - 15 + 6 = 0.

Example

Factor the polynomial and read roots from each linear factor.

highlighted = computed this step

Step 1 — Set up

Set up the expression.

x^{2} - 5x + 6

Step 2 — Find root pair

Find numbers: 2 + 3 = 5 and 2 times 3 = 6.

\hl{2} + \hl{3} = 5 \quad 2 \cdot 3 = 6

Step 3 — Factors

Write factors x - 2 and x - 3.

(x- \hl{2} )(x- \hl{3} )

Step 4 — Roots

Set each factor to zero: roots are 2 and 3.

x= \hl{2} ,\quad x= \hl{3}

factor-and-roots To factor a quadratic x² + bx + c, find two numbers that multiply to c and add to b. These numbers are the roots r₁ and r₂, giving the factored form (x - r₁)(x - r₂).