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Section 1.5 : Factoring Polynomials
9. Factor the following polynomial.
\[{y^2} + 16y + 60\]Show All Steps Hide All Steps
Start SolutionThe initial form for the factoring will be,
\[\left( {y + \underline {\,\,\,\,\,} } \right)\left( {y + \underline {\,\,\,\,\,} } \right)\]and the factors of 60 are,
\[\begin{array}{c} \left( { - 1} \right)\left( { - 60} \right) & \left( { - 2} \right)\left( { - 30} \right) & \left( { - 3} \right)\left( { - 20} \right) & \left( { - 4} \right)\left( { - 15} \right) & \left( { - 5} \right)\left( { - 12} \right) & \left( { - 6} \right)\left( { - 10} \right)\\ \left( 1 \right)\left( {60} \right) & \left( 2 \right)\left( {30} \right) & \left( 3 \right)\left( {20} \right) & \left( 4 \right)\left( {15} \right) & \left( 5 \right)\left( {12} \right)& \left( 6 \right)\left( {10} \right)\end{array}\]Sometimes there are a lot of factors that we need to deal with. As you get more practice you will start to be able to do most of this in your head and won’t need to actually write all of the factors down.
Show Step 2Now, recalling that we need the pair of factors from the above list that will add to get 16. So, we can see that the correct factoring will then be,
\[{y^2} + 16y + 60 = \require{bbox} \bbox[2pt,border:1px solid black]{{\left( {y + 6} \right)\left( {y + 10} \right)}}\]