Start by looking for integer zeros. Integer zeros are factors of the constant term.
- 3, 6, \dfrac {1 \pm \sqrt{5}}{2}
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^3-16x^2+21x+18=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 18, the possible integer roots are ± 1, ± 2, ± 3, ± 6, ± 9, and ± 18. Let's check.
x
x^4-4x^3-16x^2+21x+18
P(x)=x^4-4x^3-16x^2+21x+18
1
1^4-4( 1)^3-16( 1)^2+21( 1)+18
20 *
- 1
( - 1)^4-4( - 1)^3-16( - 1)^2+21( - 1)+18
- 14 *
2
2^4-4( 2)^3-16( 2)^2+21( 2)+18
- 20 *
- 2
( - 2)^4-4( - 2)^3-16( - 2)^2+21( - 2)+18
- 40 *
3
3^4-4( 3)^3-16( 3)^2+21( 3)+18
- 90 *
- 3
( - 3)^4-4( - 3)^3-16( - 3)^2+21( - 3)+18
0 ✓
6
6^4-4( 6)^3-16( 6)^2+21( 6)+18
0 ✓
- 6
( - 6)^4-4( - 6)^3-16( - 6)^2+21( - 6)+18
1476 *
9
9^4-4( 9)^3-16( 9)^2+21( 9)+18
2556 *
- 9
( - 9)^4-4( - 9)^3-16( - 9)^2+21( - 9)+18
8010 *
18
18^4-4( 18)^3-16( 18)^2+21( 18)+18
76 860 *
- 18
( - 18)^4-4( - 18)^3-16( - 18)^2+21( - 18)+18
122 760 *
We found that - 3 and 6 are roots for P(x)=0. Therefore, (x+3) and (x-6) are factors of the polynomial. Let's use synthetic division to factor out those factors and thus find the fourth root. We will start by factoring out (x+3).
Using synthetic division to remove the first root, -3, left us with the following polynomial.
x^3-7x^2+5x+6
Following the same procedure, we can factor out (x-6) from x^3-7x^2+5x+6.
rl IR-0.15cm r 6 & |rr 1& -7 & 5 & 6
â–¼
Perform synthetic division
Bring down the first coefficient
rl IR-0.15cm r 6 & |rr 1& -7 & 5 & 6 & & & & c 1& & &
Removing the second root, 6, gave us an even smaller polynomial.
x^2-x-1
Factoring the remaining quadratic factor
We will use the Quadratic Formula to find the remaining factors. To do so, we will need to identify the values of a, b, and c.
x^2-x-1=0 ⇔ 1x^2+( - 1)x+( - 1)=0
We can see above that a= 1, b= -1, and c= -1. Finally, we will substitute these values into the Quadratic Formula.
