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High school🔤Algebra●●○○○· 4 min
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Factor a trinomial when you know its roots

If $x_1$ and $x_2$ solve $ax^2 + bx + c = 0$, then $ax^2 + bx + c = a(x - x_1)(x - x_2)$. Free factorization.

The formula to remember

If the trinomial ax2+bx+cax^2 + bx + c has two roots x1x_1 and x2x_2, then:

ax2+bx+c=a(xx1)(xx2)ax^2 + bx + c = a(x - x_1)(x - x_2)

Never forget the factor aa in front.

Example

3x212x+93x^2 - 12x + 9:

  • Discriminant: Δ=144108=36>0\Delta = 144 - 108 = 36 > 0.
  • Roots: x1=1266=1x_1 = \dfrac{12 - 6}{6} = 1, x2=12+66=3x_2 = \dfrac{12 + 6}{6} = 3.
  • Factorization: 3(x1)(x3)3(x - 1)(x - 3).

Check by expansion: 3(x1)(x3)=3(x24x+3)=3x212x+93(x-1)(x-3) = 3(x^2 - 4x + 3) = 3x^2 - 12x + 9 ✓.

Case Δ=0\Delta = 0: double root

If the trinomial has a double root x0x_0:

ax2+bx+c=a(xx0)2ax^2 + bx + c = a(x - x_0)^2

Case Δ<0\Delta < 0: impossible

If Δ<0\Delta < 0, the trinomial doesn’t factor over the reals. It stays in expanded form (or factors with complex numbers in advanced classes).

Quick check

After factoring, mentally expand only the constant terms to verify it equals cc.

On the example: 3×(1)×(3)=93 \times (-1) \times (-3) = 9 ✓.

?Your turn

How does x² − 7x + 12 factor?

#factoring#trinomial#roots#quadratic

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