Question 28
If x−√5 is a factor of the cubic polynomial x3−3√5x2+13x−3√5, then find all the zeroes of the polynomial.
Solution:
To find all the zeroes of the cubic polynomial P(x)=x3−3√5x2+13x−3√5, given that x−√5 is a factor, we can follow these steps:
Step 1: Identify the known zero
Since x−√5 is a factor, it means that √5 is a zero of the polynomial. Therefore, we can say:
P(√5)=0
Step 2: Use Vieta’s formulas
For a cubic polynomial ax3+bx2+cx+d, the relationships between the coefficients and the roots (zeroes) α,β,γ are given by:
α+β+γ=−b/a
αβ+βγ+γα=c/a
αβγ=−d/a
Here, a=1, b=−3√5, c=13, and d=−3√5.
Step 3: Calculate the sum of the roots
Using Vieta’s first formula:
α+β+γ=−−3√51=3√5
Since we know α=√5, we can substitute:
√5+β+γ=3√5
This simplifies to:
β+γ=3√5−√5=2√5
Step 4: Calculate the product of the roots
Using Vieta’s third formula:
αβγ=−(−3√5)/1=3√5
Substituting α=√5:
√5⋅β⋅γ=3√5
Dividing both sides by √5 (assuming √5≠0):
β⋅γ=3
Step 5: Set up equations for β and γ
Now we have two equations:
1. β+γ=2√5
2. β⋅γ=3
Step 6: Substitute and form a quadratic equation
Let β and γ be the roots of the quadratic equation x2−(sum)x+(product)=0:
x2−(2√5)x+3=0
Step 7: Solve the quadratic equation
Using the quadratic formula:
x=[−b±√(b2−4ac)]/2a
Here, a=1, b=−2√5, and c=3:
x=[2√5±√{(2√5)2−4⋅1⋅3}]/(2⋅1)
Calculating the discriminant:
(2√5)2−12=20−12=8
Thus, we have:
x=(2√5±√8)/2
x=(2√5±2√2)/2
x=√5±√2
Step 8: Find all zeroes
Now we can summarize the zeroes:
α=√5
β=√5+√2
γ=√5−√2
Thus, the zeroes of the polynomial are:
√5+√2,√5−√2,√5