Problem

Factor \(2x^{2}+5x+2\) using the \(ac\) method. State the product \(ac\), identify the pair that splits \(5x\), factor by grouping, and use the final factors to find every zero.

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What this problem is really about

For a nonmonic quadratic, the ac method turns the middle term into two terms that support grouping. We’ll find a pair with product a c and sum b, split the linear term, factor the resulting groups, solve both final factors carefully, and expand to verify the leading coefficient was preserved.

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ac=4; split pair=2,3 sums to b but product is wrong; split=2x²+2x+3x+2; grouping=binomials do not match; factors=invalid; zeros=invalid; expansion check=fails

ac=4; split pair=4,1; split line uses correct terms; grouping=changes a sign in the second group; factors=common binomials do not match; zeros=not obtained; expansion check=fails

ac=4; split pair=4,1 sum5; split=2x²+4x+x+2; grouping=2x(x+2)+1(x+2); factored=(2x+1)(x+2); zeros=-1/2,-2; check=expands to original

ac=4; split pair=4,1; grouping=correct; factored=(2x+1)(x+2); zeros=-1,-2; expansion check=factors expand to original

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Factor \(2x^{2}+5x+2\) using the \(ac\) method. State the product \(ac\), identify the pair that splits \(5x\), factor by grouping, and use the final factors to find every zero.

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