Polynomial Division and Factoring: Solutions

Here are some solutions to the Exercises to accompany the section Polynomial Division and Factoring.  

Polynomial Division

  1. Use polynomial long-division to compute the quotient, q, and remainder, r, after dividing p by d for each of the following pairs of polynomials.  Check your answer by plugging your answers into the equation p = d· q + r and simplifying.
    1. Divide p(x) = x3 - x2 + 3x - 9 by  d(x) = x - 2.
      Solution
      The quotient is q(x) = x2 + x + 5 with remainder r(x) = 1.  
    2. Divide p(x) = 3x4 + 2x2 - 5x + 1 by  d(x) = x2 + 3x - 2.
      Solution
      The quotient is q(x) = 3x2 - 9x + 35 with remainder r(x) = -128x + 71.  
    3. Divide p(x) = 2x2 + 6x - 9 by  d(x) = 3x - 1.
      Solution
      The quotient is q(x) = 2/3x + 20/9 with remainder r(x) = -61/9.  
    4. Pick your own polynomial, p, and a lower degree polynomial, d, and divide d into p to find the quotient, q, and remainder, r.
      Solution
      Multiply out to check your answer. 

      Repeat this Exercise as often as necessary until you are confident in your ability to divide polynomials.

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Likely Factors of Rational Polynomials and Rational Roots

  1. Use the Rational Root Theorem to list all possible rational roots and corresponding integral, linear factors of the following polynomials.  
    1. x3 - 4x2 + 3x - 15
      Solution
      The possible rational roots are x = ±1, x = ±3, x = ±5, and x = ±15, with the corresponding integral, linear factors x ± 1, x ± 3, x ± 5, and x ± 15, corresponding to the factors of 15.
    2. 4x3 - 3x2 + x - 15
      Solution
      The possible rational roots are x = ±1, x = ±3, x = ±5, x = ±15, x = ±1/2, x = ±3/2, x = ±5/2, x = ±15/2, x = ±1/4, x = ±3/4, x = ±5/4, and x = ±15/4, i.e., all the factors of 15 divided by all the factors of 4.  The corresponding integral, linear factors are x ± 1, x ± 3, x ± 5, x ± 15, 2x ± 1, 2x ± 3, 2x ± 5, 2x ± 15, 4x ± 1, 4x ± 3, 4x ± 5, and 4x ± 15.

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  2. Use our strategy for factoring rational polynomials to factor each of the following polynomials as much  as possible (i.e., find as many rational roots as possible).  To help you narrow your search, some values of each polynomial are already given.
    1. Factor p(x) = 2x3 - 3x2 - 9x + 10.  Hint:
      x p(x)
      -9
      -6
      -3
      0
      3
      6
      9      		 -1610
      -476
      -44
      10
      10
      280
      1144
      Solution
      From the table, we begin to look for a root between -3 and 0.  The possible rational roots in that interval are x = -2.5, -2, -1, -0.5.  Plugging in -1 gives 14, so the root must be either -2.5 or -2 (or not rational).  Plugging in -2 gives 0, and dividing out x + 2 gives a quotient of 2x2 - 7x + 5.  The quadratic formula gives the two remaining roots as 1 and 5/2 corresponding to factors of x - 1 and 2x - 5.  This implies that p(x) = (x + 2)(x - 1)(2x - 5).
    2. Factor q(x) = 4x3 + 6x2 + 2x + 3.  Hint:
      x p(x)
      -3
      -1.8
      -0.6
      0.6
      1.8
      3      		 -57
      -4.488
      3.096
      7.224
      49.368
      171
      Solution
      From the table, we begin to look for a root between -1.8 and -0.6.  The possible rational roots in that interval are x = -1.5, -1, -0.75.  Plugging in -1 gives 3, so the root must be -1.5 (or not rational).  Dividing out 2x + 3 gives a quotient of 2x2 + 1.  Since this is always positive, it has no intercepts and does not factor over the rational (or real) numbers any more.  This implies that q(x) = (2x + 3)(2x2 + 1).
    3. Factor r(x) = 60x4 - 40x3 - 5x2 + 5x.  Hint
      x p(x)
      -1
      -0.6
      -0.2
      0.2
      0.6
      1      		 90
      11.616
      -0.784
      0.576
      0.336
      20
      Solution
      We can immediately factor out a common factor of 5 and x, leaving 12x3 - 8x2 - x + 1.  The possible rational roots between -0.6 and -0.2 are x = -0.5, -1/3, -0.25.  Plugging in -1/3 gives 0 and dividing by 3x + 1 gives a quotient of 4x2 - 4x + 1.  The quadratic formula gives only one root of 1/2.  Dividing out 2x - 1 gives a quotient of 2x - 1.  That is, r(x) = 5x(3x + 1)(2x - 1)2; we say that 0.5 is a double root, since the corresponding factor appears twice.

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Factoring Over the Real and Complex Numbers and Prime Polynomials

  1. For each of the following polynomials, factor them as much as possible over each of the following number systems: 
    1. Rational numbers,
    2. Real numbers, and
    3. Complex numbers.
    1. p(x) = -2x3 - 4x2 + 4x
      Solution
      We can immediately factor out a common factor of -2 and x, leaving x2 + 2x - 2.  The only possible rational roots are ±1, ±2, none of which evaluate to 0, so it has no more rational roots, and factors over the rationals as p(x) = -2x(x2 + 2x - 2).  The quadratic formula gives real roots of -1 ± .  This means that over the reals (and complex numbers), this factors further as p(x) = -2x(x - (-1 + ))(x - (-1 - )) = -2x(x  + 1 - )(x + 1 + ).
    2. q(x) = -2x3 - 4x2 - 10x
      Solution
      We can immediately factor out a common factor of -2 and x, leaving x2 + 2x + 5.  The quadratic formula gives complex roots of -1 ± 2i.  This means that over the rationals and reals, this factors as q(x) = -2x(x2 + 2x + 5).  Over the complex numbers, this factors further as q(x) = -2x(x - (-1 + 2i))(x - (-1 - 2i)) = -2x(x  + 1 - 2i)(x + 1 + 2i).
    3. r(x) = x3 - 1
      Solution
      The possible rational roots are ±1, and plugging in shows that 1 is an actual root.  Dividing out x - 1 leaves x2 + x + 1.  We know that this does not factor any more, that is, r(x) = (x - 1)(x2 + x + 1) over the rationals or the reals.  However, it does factor completely over the complex numbers as:

      .  

      Note: Since a root of r(x) = x3 - 1 is a solution of r(x) = x3 - 1 = 0 or x3 = 1, we see that there are three cube roots of 1 (not just the one we are used to); that is, just as we get two different square roots, we get three different cube roots, four different fourth roots, etc., if we work over the complex numbers.
    4. t(x) = x3 - 2
      Solution
      The possible rational roots are ±1, ±2, but plugging in shows that none of these are actual roots.  Therefore, this does not factor at all over the rational numbers.  However, we can see that x =  is a real root.  Dividing out x -  leaves .  From the graph of x3 - 2, we can see that this does not factor any more over the reals, that is,  over the reals.  Since we can simply multiply by the three cube roots of 1 to find the three cube roots of 2, we can conclude that this factors completely over the complex numbers as:

      .  

       

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