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  • Paul's Online Notes
    Paul's Online Notes
    Home / Algebra Trig Review / Algebra / Factoring
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    I apologize for the outage on the site yesterday and today. Lamar University is in Beaumont Texas and Hurricane Laura came through here and caused a brief power outage at Lamar. Things should be up and running at this point and (hopefully) will stay that way, at least until the next hurricane comes through here which seems to happen about once every 10-15 years. Note that I wouldn't be too suprised if there are brief outages over the next couple of days as they work to get everything back up and running properly. I apologize for the inconvienence.

    August 27, 2020

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    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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