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            Paul's Online Notes
            Paul's Online Notes
            Home / Algebra Trig Review / Algebra / Factoring
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            Factoring

            Factor each of the following as much as possible. Show All Solutions Hide All Solutions

            1. \(100{x^2} - 81\)
              Show Solution

              We have a difference of squares and remember not to make the following mistake.

              \[100{x^2} - 81 \ne {\left( {10x - 9} \right)^2}\]

              This just simply isn’t correct. To convince yourself of this go back to Problems 1 and 2 in the Multiplying Polynomials section. Here is the correct answer.

              \[100{x^2} - 81 = \left( {10x - 9} \right)\left( {10x + 9} \right)\]
            2. \(100{x^2} + 81\)
              Show Solution

              This is a sum of squares and a sum of squares can’t be factored, except in rare cases, so this is as factored as it will get. As noted there are some rare cases in which a sum of squares can be factored but you will, in all likelihood never run into one of them.

            3. \(3{x^2} + 13x - 10\)
              Show Solution

              Factoring this kind of polynomial is often called trial and error. It will factor as

              \[\left( {ax + b} \right)\left( {cx + d} \right)\]

              where \(ac = 3\) and \(bd = - 10\). So, you find all factors of 3 and all factors of -10 and try them in different combinations until you get one that works. Once you do enough of these you’ll get to the point that you can usually get them correct on the first or second guess. The only way to get good at these is to just do lots of problems.

              Here’s the answer for this one.

              \[3{x^2} + 13x - 10 = \left( {3x - 2} \right)\left( {x + 5} \right)\]
            4. \(25{x^2} + 10x + 1\)
              Show Solution

              There’s not a lot to this problem.

              \[25{x^2} + 10x + 1 = \left( {5x + 1} \right)\left( {5x + 1} \right) = {\left( {5x + 1} \right)^2}\]

              When you run across something that turns out to be a perfect square it’s usually best write it as such.

            5. \(4{x^5} - 8{x^4} - 32{x^3}\)
              Show Solution

              In this case don’t forget to always factor out any common factors first before going any further.

              \[4{x^5} - 8{x^4} - 32{x^3} = 4{x^3}\left( {{x^2} - 2x - 8} \right) = 4{x^3}\left( {x - 4} \right)\left( {x + 2} \right)\]
            6. \(125{x^3} - 8\)
              Show Solution

              Remember the basic formulas for factoring a sum or difference of cubes.

              \[\begin{align*}{a^3} - {b^3} & = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)\\ {a^3} + {b^3} &= \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)\end{align*}\]

              In this case we’ve got

              \[125{x^3} - 8 = \left( {5x - 2} \right)\left( {25{x^2} + 10x + 4} \right)\]
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