Section 1.6 : Multiplying Polynomials

Multiply each of the following. Show All Solutions Hide All Solutions

1. \(\left( {7x - 4} \right)\left( {7x + 4} \right)\)
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   Most people remember learning the FOIL method of multiplying polynomials from an Algebra class. I’m not very fond of the FOIL method for the simple reason that it only works when you are multiplying two polynomials each of which has exactly two terms (i.e. you’re multiplying two binomials). If you have more than two polynomials or either of them has more, or less than, two terms in it the FOIL method fails.

   The FOIL method has its purpose, but you’ve got to remember that it doesn’t always work. The correct way to think about multiplying polynomials is to remember the rule that every term in the second polynomial gets multiplied by every term in the first polynomial.

   So, in this case we’ve got.

   \[\left( {7x - 4} \right)\left( {7x + 4} \right) = 49{x^2} + 28x - 28x - 16 = 49{x^2} - 16\]

   Always remember to simplify the results if possible and combine like terms.

   This problem was to remind you of the formula

   \[\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}\]
2. \({\left( {2x - 5} \right)^2}\)
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   Remember that \({3^2} = \left( 3 \right)\left( 3 \right)\) and so

   \[{\left( {2x - 5} \right)^2} = \left( {2x - 5} \right)\left( {2x - 5} \right) = 4{x^2} - 20x + 25\]

   This problem is to remind you that

   \[{\left( {a - b} \right)^n} \ne {a^n} - {b^n}\hspace{0.25in}\,\,\,\,\,\,\,{\rm{and}}\hspace{0.5in}{\left( {a + b} \right)^n} \ne {a^n} + {b^n}\]

   so do not make that mistake!

   \[{\left( {2x - 5} \right)^2} \ne 4{x^2} - 25\]

   There are actually a couple of formulas here.

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

   You can memorize these if you’d like, but if you don’t remember them you can always just FOIL out the two polynomials and be done with it…

3. \(2{\left( {x + 3} \right)^2}\)
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   Be careful in dealing with the 2 out in front of everything. Remember that order of operations tells us that we first need to square things out before multiplying the 2 through.

   \[2{\left( {x + 3} \right)^2} = 2\left( {{x^2} + 6x + 9} \right) = 2{x^2} + 12x + 18\]

   Do, do not do the following

   \[2{\left( {x + 3} \right)^2} \ne {\left( {2x + 6} \right)^2} = 4{x^2} + 24x + 36\]

   It is clear that if you multiply the 2 through before squaring the term out you will get very different answers!

   There is a simple rule to remember here. You can only distribute a number through a set of parenthesis if there isn’t any exponent on the term in the parenthesis.

4. \(\left( {2{x^3} - x} \right)\left( {\sqrt x + \frac{2}{x}} \right)\)
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   While the second term is not a polynomial you do the multiplication in exactly same way. The only thing that you’ve got to do is first convert everything to exponents then multiply.

   \[\begin{align*}\left( {2{x^3} - x} \right)\left( {\sqrt x + \frac{2}{x}} \right) & = \left( {2{x^3} - x} \right)\left( {{x^{\frac{1}{2}}} + 2{x^{ - 1}}} \right)\\ & = 4{x^2} + 2{x^{\frac{7}{2}}} - {x^{\frac{3}{2}}} - 2\end{align*}\]
5. \(\left( {3x + 2} \right)\left( {{x^2} - 9x + 12} \right)\)
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   Remember that the FOIL method will not work on this problem. Just multiply every term in the second polynomial by every term in the first polynomial and you’ll be done.

   \[\begin{align*}\left( {3x + 2} \right)\left( {{x^2} - 9x + 12} \right) & = {x^2}\left( {3x + 2} \right) - 9x\left( {3x + 2} \right) + 12\left( {3x + 2} \right)\\ & = 3{x^3} - 25{x^2} + 18x + 24\end{align*}\]