If there are any integer roots, 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.
1, - 4, - 1/2
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!
Let Q(x)= a_nx^n+a_(n-1)x^(n-1)+... +a_1x+ a_0 be a polynomial with integer coefficients. 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)= 2x^3+7x^2-5x - 4
We can check for integer and rational roots one at a time.
Checking for Integer Roots
The constant term of this polynomial is - 4. Its factors, and the possible integer roots for P(x)=0, are ± 1, ± 2, and ± 4. Let's check them! Remember, a root will give us a result of 0 after substituted into the polynomial.
x
2x^3+7x^2-5x-4
P(x)=2x^3+7x^2-5x-4
1
2( 1)^3+7( 1)^2-5( 1)-4
0 ✓
- 1
2( - 1)^3+7( - 1)^2-5( - 1)-4
6 *
2
2( 2)^3+7( 2)^2-5( 2)-4
30 *
- 2
2( - 2)^3+7( - 2)^2-5( - 2)-4
18 *
4
2( 4)^3+7( 4)^2-5( 4)-4
216 *
- 4
2( - 4)^3+7( - 4)^2-5( - 4)-4
0 ✓
Looking at the table, we can see that 1 and - 4 are roots for P(x)=0.
Checking for Non-integer Roots
Next, let's try to find non-integer rational roots. The leading coefficient is 2 and the constant term is - 4. Therefore, the possible rational roots are ± 1 2, ± 2 2, and ± 4 2. Note that ± 2 2=± 1 and ± 4 2=± 2 are integer numbers, and they were already considered in the prior table.
x
2x^3+7x^2-5x-4
P(x)=2x^3+7x^2-5x-4
1/2
2( 1/2 )^3+7( 1/2 )^2-5( 1/2 )-4
- 4.5 *
- 1/2
2(- 1/2 )^3+7(- 1/2 )^2-5(- 1/2 )-4
0 ✓
Looking at the table, we can see that - 12 is a root for P(x)=0.
The rational roots of P(x)=0 include 1, - 4, and - 12.
