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.
- 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! 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)= 1x^3+2x^2+4x + 21
We can check for integer and rational roots one at a time.
Checking for Integer Roots
The constant term of this polynomial is 21. Its factors, and the possible integer roots for P(x)=0, are ± 1, ± 3, ± 7, and ± 21. Let's check them! Remember, a root will give us a result of 0 after substituted into the polynomial.
x
x^3+2x^2+4x+21
P(x)=x^3+2x^2+4x+21
1
1^3+2( 1)^2+4( 1)+21
28 *
- 1
( - 1)^3+2( - 1)^2+4( - 1)+21
18 *
3
3^3+2( 3)^2+4( 3)+21
78 *
- 3
( - 3)^3+2( - 3)^2+4( - 3)+21
0 âś“
7
7^3+2( 7)^2+4( 7)+21
490 *
- 7
( - 7)^3+2( - 7)^2+4( - 7)+21
- 252 *
21
21^3+2( 21)^2+4( 21)+21
10 248 *
- 21
( - 21)^3+2( - 21)^2+4( - 21)+21
- 8442 *
Looking at the table, we can see that - 3 is a root for P(x)=0.
Checking for Non-integer Roots
Next, let's try to find non-integer rational roots. The leading coefficient is 1 and the constant term is 21. Since the leading coefficient is 1, the possible rational roots are all integer numbers, which were already considered in the prior table.
