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All courses Math II · A-REI.4.b 6 of 73
Solve quadratics by inspection, square roots, completing the square, formula, and factoring; express complex solutions.

Solve a monic quadratic equation by factoring the trinomial and using the zero-product property

Problem
Solve x²+7x+12=0 by factoring the monic trinomial.
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Hint

Look for two numbers that multiply to \(12\) and add to \(7\).

For a monic trinomial \(x^2+bx+c\), factoring means finding two numbers whose product is \(c\) and whose sum is \(b\).

Solution walkthrough

01

Find the factor pair

\[3~\cdot~4~=~12~\text{and}~3+4~=~7\]

The numbers \(3\) and \(4\) match the constant term and the linear coefficient.

02

Factor the trinomial

\[x^2+7x+12~=~(x+3)(x+4)\]

Because the constant is positive and the middle term is positive, both binomial signs are positive.

03

Set each factor equal to zero

\[x+3~=~0~\text{or}~x+4~=~0\]

If a product is zero, at least one of the factors must be zero.

04

Solve for the roots

\[x~=~-3~\text{or}~x~=~-4\]

These are the two solutions to the equation.

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Another way

  1. Check by substitution: both \(x=-3\) and \(x=-4\) make \(x^2+7x+12\) equal \(0\).

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Common mistake

Using \(3\) and \(4\) as the solutions directly. Since the factors are \((x+3)(x+4)\), the solutions are \(-3\) and \(-4\).