If there are any integerroots, they must be factors of the constant term. If there are any rational roots, they have the form ± pq, where p is a factor of the constant term and q a factor of the leading coefficient.
5/2, - 1/3
Practice makes perfect
One way of finding the roots for P(x)=0 is to guess and check. This is inefficient unless there is a way to minimize the number of possible roots. The Rational Root Theorem helps us with this! Consider the following general polynomial with integer coefficients.
Q(x)= a_nx^n+a_(n-1)x^(n-1)+... +a_1x+ a_0
There are a limited number of possible roots for Q(x)=0.
Rational roots must reduce to be p q, where p is an integer factor of a_0, and q is an integer factor of a_n.Now let's consider the given polynomial.
P(x)= 6x^4-13x^3+13x^2-39x - 15
We can check for integer and rational roots one at a time.
Checking for Integer Roots
The constant term of this polynomial is - 15. Its factors, and the possible integer roots for P(x)=0, are ± 1, ± 3, ± 5, and ± 15. Let's check them! Remember, a root will give us a result of 0 after substituted into the polynomial.
x
6x^4-13x^3+13x^2-39x-15
P(x)=6x^4-13x^3+13x^2-39x-15
1
6( 1)^4-13( 1)^3+13( 1)^2-39( 1)-15
- 48 *
- 1
6( - 1)^4-13( - 1)^3+13( - 1)^2-39( - 1)-15
56 *
3
6( 3)^4-13( 3)^3+13( 3)^2-39( 3)-15
120 *
- 3
6( - 3)^4-13( - 3)^3+13( - 3)^2-39( - 3)-15
1056 *
5
6( 5)^4-13( 5)^3+13( 5)^2-39( 5)-15
2240 *
- 5
6( - 5)^4-13( - 5)^3+13( - 5)^2-39( - 5)-15
5880 *
15
6( 15)^4-13( 15)^3+13( 15)^2-39( 15)-15
262 200 *
- 15
6( - 15)^4-13( - 15)^3+13( - 15)^2-39( - 15)-15
351 120 *
There are no integer roots for P(x)=0.
Checking for Rational Roots
Next, let's try to find rational roots. The leading coefficient is 6 and the constant term is - 15. Therefore, the possible rational roots are ± 1 2, ± 3 2, ± 5 2, ± 15 2, ± 1 3, ± 3 3, ± 5 3, and ± 15 3. Note that ± 3 3=± 1 and ± 15 3=± 5 are integer numbers, and they were already considered in the prior table.
