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.

Big Picture

What this problem is really about

For a monic quadratic, the binomial constants must match both the constant product and the signed linear sum. We’ll use those two conditions to factor, solve every zero-product branch, and check the result through the relationship between the roots’ sum and product and the original coefficients.

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target product=12; target sum=-7; integer pair=correct; factored=(x−3)(x−4); zeros=x=3 only; check=one factor set to zero

target product=12; target sum=-7; integer pair=-3,-4; factored=(x−3)(x−4); zeros=x=3,4; check=sum roots7 gives coefficient-7 and product12

target product=12; target sum=-7; integer pair=correct; factored=(x−3)(x−4); zeros=-3,-4; check=factor constants copied without solving

target product=12; target sum=-7; integer pair=3,4; factored=(x+3)(x+4); zeros=-3,-4; check=product target matches but middle-term sum has opposite sign

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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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