Factor theorem

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  1. Introduction to Factor Theorem
  1. Using the Factor Theorem to Test For Factors of a Polynomial
    Which of the following binomials are factors of the polynomial
    P(x)=3x3+2x211x10{P}\left( x \right) = 3{x^3} + 2{x^2} - 11x - 10
    (x2)\left( {x - 2} \right)
    (x1)\left( {x - 1} \right)
    (x+1)\left( {x + 1} \right)
    (3x+5)\left( {3x + 5} \right)?
    1. The polynomial P(x)=ax310x23x+bP\left( x \right) = a{x^3} - 10{x^2} - 3x + b
      i) has a factor of (x+5)(x+5)
      ii) has a remainder of -77 when P(x)P(x) is divided by (x2)(x-2) .
      Find the values of aa and bb.
      Topic Notes
      Factor theorem is usually used to factor and find the roots of polynomials. A root or zero is where the polynomial is equal to zero. Therefore, the theorem simply states that when f(k) = 0, then (x – k) is a factor of f(x).

      Factor Theorem

      Before getting into the Factor Theorem and what it means, it is crucial that we fully understand the Remainder Theorem and what factors are first.

      Remainder Theorem:

      The remainder theorem and factor theorem are very closely related concepts in mathematics. One can't exist without the other, so it is very important that we define what the remainder theorem is before getting into the factor theorem.

      According to the Remainder Theorem:

      If we divide a polynomial f(x) by (x - c), the remainder of that division is simply equal to f(c).

      This theorem is especially helpful because it reduces the amount of work we have to do to solve these types of problems. Without this theorem, we would have to go to the trouble of using long division and/or synthetic division to solve for the remainder, which is difficult and time consuming. But, in any case, our lessons on synthetic division and long division of polynomials will help you understand factor theorem better. In addition, a review on the concepts behind the remainder theorem, as well as a visual proof of the theorem, is also helpful.

      What is a Factor?

      Now that we have an understanding of the remainder theorem, the last little thing we must understand before moving onto the factor theorem is what exactly a "factor" is. The understanding of what factors are is crucial to all of mathematics, and it is a term you will hear again and again as you progress with your studies.

      With regards to division, a factor is just a term or expression that, when another term or expression is divided by this factor, the remainder is equal to zero. In its simplest terms, consider the following: 4 is a factor of 20 because, when 20 is divided by 4, we get the whole number 5 and no remainder. 7 is not a factor of 20 because, when 20 is divided by 7, we get 2.86, which is not a whole number (this is the same as saying 2 and a remainder of 0.86).

      What is the Factor Theorem?

      In its most basic terms, the factor theorem really is just a special case of the remainder theorem. Recall, that in the remainder theorem, if we divide a polynomial f(x) by (x-c), the remainder of that division is simply equal to f(c). In the factor theorem, we use this same concept to prove the following:

      According to the Factor Theorem:

      If we divide a polynomial f(x) by (x - c), and (x - c) is a factor of the polynomial f(x), then the remainder of that division is simply equal to 0.

      Thus, according to this theorem, if the remainder of a division like those described above equals zero, (x - c) must be a factor. If the remainder of such a division is not zero, then (x - c) is not a factor.

      How to Use the Factor Theorem:

      The best way to understand the methodology behind factor theorem is to do some factor theorem examples. Note, to factor polynomials, the factors we get are often referred to as "binomials". A binomial factor is simply a factor with two terms, like (x - 4).

      Example 1:

      Using the Factor Theorem, test and see which of the following binomials are factors of the polynomial:

      P(x)=3x3+2x211x10P(x) = 3x^{3} + 2x^{2} - 11x - 10

      REMEMBER: If (x - c) is a factor, then f(c) = 0

      (x2)(x - 2)

      P(2)=3(2)3+2(2)211(2)10P(2) = 3(2)^{3} + 2(2)^{2} - 11(2) - 10

      P(2)=24+82210P(2) = 24 + 8 - 22 - 10

      P(2)=0P(2) = 0

      a) Since P(2)=0,(x2)P(2) = 0, (x - 2) is a factor of the polynomial.

      (x1)(x - 1)

      P(1)=3(1)3+2(1)211(1)10P(1) = 3(1)^{3} + 2(1)^{2} - 11(1) - 10

      P(1)=3+21110P(1) = 3 + 2 - 11 - 10

      b) Since P(1)P(1) doesn't = 0, (x1)(x - 1) is a not factor of the polynomial.

      (x+1)(x + 1)

      NOTE: According to the Factor Theorem, we must have (x - c). So, we need to change this binomial to suit that form.

      (x+1)=(x(1))(x + 1) = (x - (-1))

      P(1)=3(1)3+2(1)211(1)10P(-1) = 3(-1)^{3} + 2(-1)^{2} - 11(-1) - 10

      P(1)=3+2+1110P(-1) = -3 + 2 + 11 - 10

      P(1)=0P(-1) = 0

      c) Since P(1)=0,(x+1)P(-1) = 0, (x + 1) is a factor of the polynomial.

      (3x+5)(3x + 5)

      NOTE: According to the Factor Theorem, we must have (x - c). So, we need to change this binomial to suit that form.

      3x+5=03x + 5 = 0

      x=53x = - \frac{5}{3}

      P(53)=3(53)3+2(53)211(53)10P(-\frac{5}{3}) = 3(-\frac{5}{3})^{3} +2(-\frac{5}{3})^{2} -11(-\frac{5}{3}) - 10

      P(53)=1259+509+55310P(-\frac{5}{3}) = -\frac{125}{9} + \frac{50}{9} + \frac{55}{3} - 10

      P(53)=0P(-\frac{5}{3}) = 0

      Since P(53)=0P(-\frac{5}{3}) = 0, (3x+5)(3x + 5) is a factor of the polynomial.

      Example 2:

      This example is a little trickier, but with a good understanding of the Factor and Remainder Theorems it should end up being no problem at all!

      Given the following problem statement:

      The polynomial: P(x)=ax310x23x+bP(x) = ax^{3} - 10x^{2} - 3x + b

      1) Has a factor of (x+5)(x + 5)

      Has a remainder of 77-77 when P(x)P(x) is divided by (x2)(x - 2).

      Find the values of a and b.

      Step 1: Use the hint that the polynomial has a factor of (x + 5) with the Factor Theorem

      P(5)=a(5)3+10(5)23(5)+b=0P(-5) = a(-5)^{3} + 10(-5)^{2} - 3(-5) + b = 0

      125a250+15+b=0-125a - 250 +15 + b = 0

      125a+b=235-125a + b = 235

      Step 2: Use the hint that the polynomial has a remainder of 77-77 when P(x)P(x) is divided by (x2)(x - 2) with Remainder Theorem

      P(2)=a(2)310(2)23(2)+b=77P(2) = a(2)^{3} - 10(2)^{2} - 3(2) + b = -77

      8a406+b=778a - 40 - 6 + b = -77

      8a+b=318a + b = -31

      Step 3: Using the two equations we have just developed that include aa and bb, use the substitution method of solving linear systems to find the values for both aa and bb.

      8a+b=318a + b = -31

      b=318ab = -31 - 8a

      Substitute this newly rearranged equation into 125a+b=235-125a + b = 235 to solve for a:

      125a+(318a)=235-125a + (-31 - 8a) = 235

      125a318a=235-125a - 31 - 8a = 235

      133a133=266133\frac{-133a}{-133} = \frac{266}{-133}

      a=2a = -2

      Substitute a=2a = -2 into 8a+b=318a + b = -31 to solve for b:

      8(2)+b=318(-2) + b = -31

      b=15b = -15

      Therefore, a=2a = -2 and b=15b = -15.

      And that's all there is to it! Be sure to practice all of these concepts to master the remainder and factor theorems. Of course, it's also helpful to be able to check your work. So, as a quick way of performing synthetic division, this online tool is very helpful. Lastly, make sure to check out the lesson on how to apply these concepts to integration of rational functions by partial fractions.

      \cdot Factor Theorem: (axb)(ax-b) is a factor of the polynomial P(x)P(x), if and only if P(ba)=0P(\frac{b}{a})=0.
      \cdot Note that the Factor Theorem is simply a result of the Remainder Theorem when the remainder = 0.