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All courses Algebra I · A-SSE.3.a 57 of 76
Factor quadratics to reveal zeros of the function they define.

Factor a monic quadratic to reveal zeros

Problem
Factor \(x^{2}-7x+12\) by finding two integers whose product is \(12\) and whose sum is \(-7\). Then apply the zero-product property and list every zero.
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Hint

List integer factor pairs of c and test their sums.

For x²+bx+c, factor numbers multiply to c and add to b; x−r=0 gives zero r.

Solution walkthrough

01

Set product and sum targets

\[\begin{aligned} x^2-7x+12 \\ \text{product}=12;~\text{sum}=-7 \end{aligned}\]

For a monic quadratic, the two factor constants multiply to the constant 12 and add to the signed linear coefficient -7.

02

Choose the integer pair and factor

\[\begin{aligned} (-3)(-4)=12;~-3+(-4)=-7 \\ (x-3)(x-4) \end{aligned}\]

The integers -3 and -4 meet both targets, so they become the binomial constants.

03

Apply the zero-product property

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

A product equals zero when at least one factor is zero. Solving both branches gives every zero.

04

Check the coefficient relationships

\[\begin{aligned} \text{root}~\text{sum}=3+4=7~->~\text{linear}~\text{coefficient}=-(7)=-7 \\ \text{root}~\text{product}=3*4=12 \end{aligned}\]

For the monic factors (x-r1)(x-r2), expansion gives linear coefficient as the negative root sum and constant as the root product. Both match the original.

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Another way

  1. Expand the factors to verify x squared minus 7x plus 12 before solving.

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Common mistake

The factor constants are -3 and -4, but the zeros are 3 and 4 after solving the factor equations.

Solution walkthrough video