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Algebra · Quadratic equations

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Solving quadratics by factorising

Why a product being zero forces a factor to be zero, why the equation must equal zero BEFORE factorising helps, and how to factorise with a leading coefficient and the difference of two squares.

Algebra · Quadratic equations

Solving quadratics by factorising

Why a product being zero forces a factor to be zero, why the equation must equal zero BEFORE factorising helps, and how to factorise with a leading coefficient and the difference of two squares.

Why it works

The whole method rests on one special fact about zero. If two numbers multiply to zero, one of them must be zero — no other product forces its factors like that. So from

(x−2)(x+7)=0(x - 2)(x + 7) = 0

we can conclude x−2=0x - 2 = 0 or x+7=0x + 7 = 0, giving x=2x = 2 or x=−7x = -7. Notice the sign flip: the bracket (x+7)(x + 7) dies when xx is −7-7, not 77 — read each root by asking "what kills this bracket?"

This only works when one side IS zero. (x−2)(x+7)=10(x - 2)(x + 7) = 10 tells you almost nothing — 10 factorises as 2×52 \times 5, 1×101 \times 10, −2×−5-2 \times -5, endlessly. Faced with x(x+2)=24x(x + 2) = 24, don't chase factor pairs of 24: expand, drag everything to one side, and then factorise:

x2+2x−24=0⇒(x+6)(x−4)=0⇒x=−6 or 4.x^2 + 2x - 24 = 0 \quad\Rightarrow\quad (x + 6)(x - 4) = 0 \quad\Rightarrow\quad x = -6 \text{ or } 4.

Never divide both sides by the unknown. x2=5xx^2 = 5x invites "divide by xx: so x=5x = 5" — but that throws away the solution x=0x = 0 (you can't divide by zero, and xx might be zero). Factorise instead: x(x−5)=0x(x - 5) = 0, so x=0x = 0 or x=5x = 5. Two roots, both kept.

Factorising is reverse expansion. For x2+bx+cx^2 + bx + c, hunt two numbers that multiply to cc and add to bb — signs included. For x2+3x−28x^2 + 3x - 28: the pair is +7+7 and −4-4 (7×(−4)=−287 \times (-4) = -28, 7+(−4)=37 + (-4) = 3), so (x+7)(x−4)(x + 7)(x - 4). When the x2x^2 has a coefficient, as in 2x2+7x−152x^2 + 7x - 15, the bracket fronts must multiply to 2x22x^2; test (2x  )(x  )(2x \; )(x \; ) pairs and check the cross terms: (2x−3)(x+5)(2x - 3)(x + 5) works because 10x−3x=7x10x - 3x = 7x. Always multiply back out — the check costs seconds.

A difference of two squares has no middle term to hunt. y2−81=(y+9)(y−9)y^2 - 81 = (y + 9)(y - 9) — the ±9x\pm 9x cross terms cancel. Any "something squared minus something squared" splits this way instantly.