Question 25
Given that zeroes of cubic polynomial x3−6x2+3x+10 are of the form a,a+b,a+2b for some real numbers a and b, find the values of a and b as well as zeroes of the given polynomial.
Solution:
To find the values of a and b as well as the zeroes of the cubic polynomial x3−6x2+3x+10, we can follow these steps:
Step 1: Identify the polynomial and its coefficients
The given polynomial is:
f(x)=x3−6x2+3x+10
Here, we can identify the coefficients:
a=1
b=−6
c=3
d=10
Step 2: Use the relationship of the roots
We know that the roots of the polynomial are of the form a,a+b,a+2b.
Step 3: Calculate the sum of the roots
According to Vieta’s formulas, the sum of the roots is given by:
Sum of roots=−b/a=−(−6)/1=6
So we have:
a+(a+b)+(a+2b)=6
This simplifies to:
3a+3b=6⇒a+b=2(Equation 1)
Step 4: Calculate the product of the roots taken two at a time
The sum of the products of the roots taken two at a time is given by:
Sum of products of roots=c/a=3/1=3
So we have:
a(a+b)+a(a+2b)+(a+b)(a+2b)=3
Expanding this:
a2+ab+a2+2ab+(a2+3ab+2b2)=3
Combining like terms gives:
3a2+6ab+2b2=3(Equation 2)
Step 5: Substitute b from Equation 1 into Equation 2
From Equation 1, we have:
b=2−a
Substituting b into Equation 2:
3a2+6a(2−a)+2(2−a)2=3
Expanding this:
3a2+12a−6a2+2(4−4a+a2)=3
This simplifies to:
3a2+12a−6a2+8−8a+2a2=3
Combining like terms:
−1a2+4a+5=0
Rearranging gives:
a2−4a−5=0
Step 6: Factor the quadratic equation
Factoring the quadratic:
(a−5)(a+1)=0
Thus, the solutions for a are:
a=5 or a=−1
Step 7: Find corresponding values of b
Using a+b=2:
1. If a=5:
b=2−5=−3
2. If a=−1:
b=2−(−1)=3
Step 8: Find the zeroes of the polynomial
Now we can find the zeroes for both cases:
1. For a=5 and b=−3:
– The roots are 5,5−3=2,5−6=−1.
2. For a=−1 and b=3:
– The roots are −1,−1+3=2,−1+6=5.
Conclusion
In both cases, the zeroes of the polynomial are:
−1,2,5
The values of a and b are:
– a=−1,b=3 or a=5,b=−3.