Solving equations by factoring
The zero product property turns a factored equation into simpler ones. It is the first method for solving quadratic and higher-degree equations.
The zero product property
If \(ab = 0\), then \(a = 0\) or \(b = 0\). So if \((x - 4)(x + 7) = 0\), either \(x - 4 = 0\) or \(x + 7 = 0\), giving x = 4 or x = -7. This works only when the product equals zero: \((x - 4)(x + 7) = 8\) cannot be split this way.
Standard steps
To solve by factoring: move every term to one side so the other side is 0, factor completely, set each factor equal to 0, and solve. For \(x^2 = 5x + 14\), write \(x^2 - 5x - 14 = 0\), factor as \((x - 7)(x + 2) = 0\), so x = 7 or x = -2.
Do not divide by the variable
For \(x^2 = 3x\), dividing both sides by x loses a solution. Instead write \(x^2 - 3x = 0\), so \(x(x - 3) = 0\), giving x = 0 or x = 3. Dividing by an expression that could be zero removes solutions.
Higher-degree equations
The same idea solves cubic and higher equations that factor. \(x^3 - 9x = 0\) factors as \(x(x + 3)(x - 3) = 0\), giving three solutions: 0, -3, and 3. A polynomial equation of degree n has at most n real solutions.
Zeros and x-intercepts
The solutions of \(p(x) = 0\) are the zeros of the polynomial and the x-intercepts of the graph of \(y = p(x)\). The graph of \(y = x^2 - x - 6 = (x - 3)(x + 2)\) crosses the x-axis at x = 3 and x = -2, connecting algebra and graphs.
Building a polynomial from zeros
Running the process backwards, a polynomial with zeros 2 and -5 is \((x - 2)(x + 5) = x^2 + 3x - 10\). Any constant multiple, such as \(3(x - 2)(x + 5)\), has the same zeros.
Area and projectile applications
A rectangle's length is 3 m more than its width and its area is 54 m²: \(w(w + 3) = 54\), so \(w^2 + 3w - 54 = 0\), which factors as \((w + 9)(w - 6) = 0\). Width cannot be negative, so the width is 6 m and the length 9 m. Always reject solutions that make no sense in context.
Worked example
A ball is thrown upward so that its height in metres after t seconds is \(h = -5t^2 + 20t\). It hits the ground when \(h = 0\): \(-5t(t - 4) = 0\), so t = 0 (the launch) or t = 4 s. It reaches its highest point halfway, at t = 2 s, where \(h = -20 + 40 = 20\) m.
Key takeaways
The zero product property says a product equals zero only if a factor does, so it applies only to equations set equal to zero. Move all terms to one side, factor completely, set each factor to zero, and solve. Never divide by a variable expression, which can lose solutions. Degree-n polynomial equations have at most n real solutions. Solutions are the zeros of the polynomial and the x-intercepts of its graph, and zeros let you rebuild a polynomial. Reject answers that make no sense in context, such as negative lengths.
Flashcard
If ab = 0, then a = 0 or b = 0
Check your understanding
Solve \(x^2 = 3x\).