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Here are a few recommended readings before getting started with this lesson.
Polynomial Function | Roots |
---|---|
p(x)=x3−6x2+11x−6 | 1, 2, 3 |
g(x)=x4−4x3−19x2+106x−120 | -5, 2, 3, 4 |
m(x)=x3+6x2−55x−252 | -9, -4, 7 |
LHS−a0=RHS−a0
Factor out xr
For the second property, suppose that the polynomial P(x) has a rational root qp, such that the fraction is written in its simplest form. Again, it is obtained that P(qp)=0.
This time the equation will be modified in a different way.(ba)m=bmam
LHS⋅qn=RHS⋅qn
LHS−a0(qn)=RHS−a0(qn)
Factor out p
The numerator of the root, p, is a factor of a0.
LHS−anpn=RHS−anpn
Factor out q
Equation | Result |
---|---|
6⋅14−59⋅13+94⋅12−49⋅1+8=0 | 0=0✓ |
6⋅24−59⋅23+94⋅22−49⋅2+8=0 | -90=0× |
6⋅44−59⋅43+94⋅42−49⋅4+8=0 | -924=0× |
6⋅84−59⋅83+94⋅82−49⋅8+8=0 | 0=0✓ |
As determined, the integer roots are 1 and 8. The rational roots have a numerator that is a factor of a0 and a denominator that is a factor of an. In this case,an is 6 with factors of 1, 2, 3, and 6. The following table displays the possible fractions that can be formed with the factors. The columns of the table are the factors of 8, and the rows are the factors of 6.
1 | 2 | 4 | 8 | |
---|---|---|---|---|
2 | 21 | 22=1 | 24=2 | 28=4 |
3 | 31 | 32 | 34 | 38 |
6 | 61 | 62=31 | 64=32 | 68=34 |
It is important to understand that the integers do not need to be tested because the integer factors have already been found. Now, each fraction will be tested to see if they are roots of the equation.
Equation | Result |
---|---|
6(21)4−59(21)3+94(21)2−49(21)+8=0 | 0=0✓ |
6(31)4−59(31)3+94(31)2−49(31)+8=0 | 0=0✓ |
6(32)4−59(32)3+94(32)2−49(32)+8=0 | 2722=0× |
6(34)4−59(34)3+94(34)2−49(34)+8=0 | -9100=0× |
6(38)4−59(38)3+94(38)2−49(38)+8=0 | -277280=0× |
6(61)4−59(61)3+94(61)2−49(61)+8=0 | 108235=0× |
Izabella is tutoring some of her friends for an upcoming math exam.
To study, they decided to go over some exercises that the math professor assigned for homework in the past. One polynomial function, in particular, caught Izabella's attention.Equation | Solution |
---|---|
(1)4−(1)3−6(1)2+14(1)−12=0 | -4=0× |
(2)4−(2)3−6(2)2+14(2)−12=0 | 0=0✓ |
(3)4−(3)3−6(3)2+14(3)−12=0 | 30=0× |
(4)4−(4)3−6(4)2+14(4)−12=0 | 140=0× |
(6)4−(6)3−6(6)2+14(6)−12=0 | 936=0× |
(12)4−(12)3−6(12)2+14(12)−12=0 | 18300=0× |
The value of x=2 is a root of the function. Since the polynomial function is of degree 4, it is possible that one of the negative factors is also a root, so these values will be substituted next.
Equation | Solution |
---|---|
(-1)4−(-1)3−6(-1)2+14(-1)−12=0 | -30=0× |
(-2)4−(-2)3−6(-2)2+14(-2)−12=0 | -40=0× |
(-3)4−(-3)3−6(-3)2+14(-3)−12=0 | 0=0✓ |
(-4)4−(-4)3−6(-4)2+14(-4)−12=0 | 156=0× |
(-6)4−(-6)3−6(-6)2+14(-6)−12=0 | 1200=0× |
(-12)4−(-12)3−6(-12)2+14(-12)−12=0 | 21420=0× |
As can be seen in the table, -3 is also a root of p(x). Since every integer root has to be a factor of -12 and every factor was tested, the only integer roots of the given function are -3 and 2.
Bring down 1
2⋅1=2
-1+2=1
2⋅1=2
-6+2=-4
2⋅-4=-8
14−8=6
2⋅6=12
-12+12=0