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All courses Algebra I · A-REI.4.b 52 of 76
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 \(x^{2}+7x+12=0\) by factoring. Find two integers whose sum and product match the trinomial, write both linear factors, and use the zero-product property for the complete solution set.
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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

\[\begin{aligned} \text{product}=12;~\text{sum}=7 \\ 3*4=12;~3+4=7 \end{aligned}\]

For x²+7x+12, the two constant terms in the factors must multiply to 12 and add to the linear coefficient 7. The integers are 3 and 4.

02

Factor and verify the trinomial

\[\begin{aligned} x^2+7x+12=(x+3)(x+4) \\ =x^2+4x+3x+12 \end{aligned}\]

Expanding the proposed factors reproduces the quadratic, so the factorization is correct.

03

Apply the zero-product property

\[\begin{aligned} (x+3)(x+4)=0 \\ x+3=0~\text{or}~x+4=0 \end{aligned}\]

A product equals zero exactly when at least one factor equals zero. Both factors must be considered.

04

Solve and check both roots

\[\begin{aligned} x=-3~\text{or}~x=-4 \\ 9-21+12=0;~16-28+12=0 \end{aligned}\]

Solving the linear equations gives -3 and -4. Substitution verifies that each makes the original quadratic zero.

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Another way

  1. Use the quadratic formula with a=1, b=7, c=12 to obtain (-7±1)/2, which gives -3 and -4.

!

Common mistake

Do not report 3 and 4 as roots. They are the constants in x+3 and x+4; setting those factors to zero changes the signs of the solutions.

Solution walkthrough video