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Section 1.5 : Factoring Polynomials
12. Factor the following polynomial.
\[4{z^2} + 19z + 12\]Show All Steps Hide All Steps
Start SolutionThere are two sets of positive factors of 4 and so we will have one of the two following possible initial forms for the factoring.
\[\left( {2z + \underline {\,\,\,\,\,} } \right)\left( {2z + \underline {\,\,\,\,\,} } \right)\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\left( {4z + \underline {\,\,\,\,\,} } \right)\left( {z + \underline {\,\,\,\,\,} } \right)\]and the factors of 12 are,
\[\left( { - 1} \right)\left( { - 12} \right)\,\hspace{0.25in}\left( { - 2} \right)\left( { - 6} \right)\hspace{0.25in}\left( { - 3} \right)\left( { - 4} \right)\hspace{0.25in}\left( 1 \right)\left( {12} \right)\,\hspace{0.25in}\left( 2 \right)\left( 6 \right)\hspace{0.25in}\hspace{0.25in}\left( 3 \right)\left( 4 \right)\] Show Step 2After some trial and error we see that the correct factoring will then be,
\[4{z^2} + 19z + 12 = \require{bbox} \bbox[2pt,border:1px solid black]{{\left( {4z + 3} \right)\left( {z + 4} \right)}}\]