Start by looking for integer zeros. Integer zeros are factors of the constant term.
2, ± sqrt(5)
Practice makes perfect
We want to find the zeros of the polynomial function y=P(x). To do so, we need to solve the equation P(x)=0.
x^4-4x^3-x^2+20x-20=0
The degree of P(x) is 4. Thus, by the Fundamental Theorem of Algebra, we know that P(x)=0 has exactly four roots. Let's find them.
Integer roots
By the Rational Root Theorem, we know that integer roots must be factors of the constant term. Since the constant term of P(x) is - 20, the possible integer roots are ± 1, ± 2, ± 4, ± 5, ± 10, and ± 20. Let's check.
x
x^4-4x^3-x^2+20x-20
P(x)=x^4-4x^3-x^2+20x-20
1
1^4-4( 1)^3- 1^2+20( 1)-20
- 4 *
- 1
( - 1)^4-4( - 1)^3-( - 1)^2+20( - 1)-20
- 36 *
2
2^4-4( 2)^3- 2^2+20( 2)-20
0 âś“
- 2
( - 2)^4-4( - 2)^3-( - 2)^2+20( - 2)-20
- 16 *
4
4^4-4( 4)^3- 4^2+20( 4)-20
44 *
- 4
( - 4)^4-4( - 4)^3-( - 4)^2+20( - 4)-20
396 *
5
5^4-4( 5)^3- 5^2+20( 5)-20
180 *
- 5
( - 5)^4-4( - 5)^3-( - 5)^2+20( - 5)-20
980 *
10
10^4-4( 10)^3- 10^2+20( 10)-20
6080 *
- 10
( - 10)^4-4( - 10)^3-( - 10)^2+20( - 10)-20
13 680 *
20
20^4-4( 20)^3- 20^2+20( 20)-20
127 980 *
- 20
( - 20)^4-4( - 20)^3-( - 20)^2+20( - 20)-20
191 180 *
We found that 2 is a root for P(x)=0. Therefore, (x-2) is a factor of the polynomial. Let's use synthetic division to factor out (x-2) and thus find the other roots.
Using synthetic division to remove the first root, 2, left us with the following polynomial.
x^3-2x^2-5x+10
Since there are no other integer roots besides 2, we will use factoring by grouping to factor x^3-2x^2-5x+10 and then apply the Zero Product Property. To start, we will set our equation equal to 0.
We have now found three more roots, 2, sqrt(5) and - sqrt(5). When we solved for integer roots earlier, we also found the root x=2. Therefore, 2 is a root with multiplicity 2. Let's list all of the roots we have found for the given function.
