Rational root theorem
The rational root theorem is a theorem that you can use to find any rational roots of a polynomial equation.
Rational root theorem
Let anxn + an-1xn-1 + ... + a1x + a0 = 0 be a polynomial equation with integer coefficients.
If p/q is in simplest form and is a rational root of the polynomial equation, then p must be a factor of a0 and q must be a factor of an
Explanation of the rational root theorem
You may now wonder why it makes to use the theorem above. We will now clarify. The reading is much, but in the end you will perfectly understand how the rational root theorem works!
Example #1
First try to write (x + 3)(x - 2)(x - 4) = 0 in standard form
(x + 3)(x2 - 4x - 2x + 8) = 0
(x + 3)(x2 - 6x + 8) = 0
x3 - 6x2 + 8x + 3x2 - 18x + 24 = 0
x3 - 3x2 - 10x + 24 = 0
Suppose you are solving the following polynomial equations
x3 - 3x2 - 10x + 24 = 0
We get (x + 3)(x - 2)(x - 4) = 0 which have -3, 2 and 4 as roots.
Notice that the product of 3, 2 and 4 is equal to 24. in other words, the roots are factors of the constant term.
Let a0 = 24 and an = 1
If we apply the rational root theorem, we can say that p must be a factor of 24 and q must be a factor 1.
24 = 1 x 24, 24 = 2 x 12, 24 = 3 x 8, 24 = 4 x 6
24 = -1 x -24, 24 = -2 x -12, 24 = -3 x -8, 24 = -4 x -6
If p is a factor of 24, then p = {1, 2, 3, 4, 6, 8, 12, 24, -1, -2, -3, -4, -6, -8, -12, -24}
1 = 1 x 1
1 = -1 x -1
If q is a factor of 1, then q = {1, -1}
The p/q in the theorem means that you need to divide each factor of p by each factor of q.
There are a lots of different combinations here as shown below:
1/1, 1/-1, 2/1, 2/-1, 3/1, 3/-1, 4/1, 4/-1, 6/1, 6/-1, 8/1, 8/-1, 12/1, 12/-1, -1/1, -1/-1, -2/1, -2/-1, -3/1, -3/-1, -4/1, -4/-1, -6/1, -6/-1, -8/1, -8/-1, -12/1, -12/-1, -24/1, -24/-1
However, the ones that will give you the answers we already found above are show in red.
2/1 = -2/-1 = 2
-3/1 = 3/-1 = -3
4/1 = -4/-1 = 4.
Example #2
24x3 + 22x2 - 5x - 6 = 0 is equivalent to (x + 2/3)(x - 1/2)(x + 3/4) = 0
To prove that they are equivalent, do the following:
3(x + 2/3)2(x - 1/2)4(x + 3/4) = 3x2x4x0
(3x + 2)(2x - 1)(4x + 3) = 0
I will let you finish it!
The roots are -2/3, 1/2, and -3/4
The numerators 2, 1, and 3 are all factors of the constant term, a0 = -6.
The denominators 3, 2, and 4 are all factors of the leading coefficient, an = 24
We can again apply the rational root theorem in order to see all the rational roots.
We can say that p must be a factor of -6 and q must be a factor 24.
-6 = -1 x 6, -6 = -2 x 3, -6 = 1 x -6, -6 = 2 x -3
If p is a factor of -6, then p = {1, 2, 3, 6, -1, -2, -3, -6}
24 = 1 x 24, 24 = 2 x 12, 24 = 3 x 8, 24 = 4 x 6
24 = -1 x -24, 24 = -2 x -12, 24 = -3 x -8, 24 = -4 x -6
If q is a factor of 24, then q = {1, 2, 3, 4, 6, 8, 12, 24, -1, -2, -3, -4, -6, -8, -12, -24}
The p/q in the theorem means that you need to divide each factor of p by each factor of q. All the different combinations are shown below.
1/1, 1/2, 1/3, 1/4, 1/6, 1/8, 1/12, 1/24, 1/-1, 1/-2, 1/-3, 1/-4, 1/-6, 1/-8, 1/-12, 1/-24, 2/1, 2/2, 2/3, 2/4, 2/6, 2/8, 2/12, 2/24, 2/-1, 2/-2, 2/-3, 2/-4, 2/-6, 2/-8, 2/-12, 2/-24, 3/1, 3/2, 3/3, 3/4, 3/6, 3/8, 3/12, 3/24, 3/-1, 3/-2, 3/-3, 3/-4, 3/-6, 3/-8, 3/-12, 3/-24, 6/1, 6/2, 6/3, 6/4, 6/6, 6/8, 6/12, 6/24, 6/-1, 6/-2, 6/-3, 6/-4, 6/-6, 6/-8, 6/-12, 6/-24
-1/1, -1/2, -1/3, -1/4, -1/6, -1/8, -1/12, -1/24, -1/-1, -1/-2, -1/-3, -1/-4, -1/-6, -1/-8, -1/-12, -1/-24, -2/1, -2/2, -2/3, -2/4, -2/6, -2/8, -2/12, -2/24, -2/-1, -2/-2, -2/-3, -2/-4, -2/-6, -2/-8, -2/-12, -2/-24, -3/1, -3/2, -3/3, -3/4, -3/6, -3/8, -3/12, -3/24, -3/-1, -3/-2, -3/-3, -3/-4, -3/-6, -3/-8, -3/-12, -3/-24, -6/1, -6/2, -6/3, -6/4, -6/6, -6/8, -6/12, -6/24, -6/-1, -6/-2, -6/-3, -6/-4, -6/-6, -6/-8, -6/-12, -6/-24
The ones that will give you the answers we already found above are shown in red. Many answers are repeated here. The lesson about finding rational roots will show you how to sift out unnecessary answers.
100 Tough Algebra Word Problems.
If you can solve these problems with no help, you must be a genius!
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