If the zeroes of the polynomial f(x) = x3 12x2 + 39x + a are in ap, find the value of
a.
Answers
We have,
f(x)=x³+12x²+39x+a
Given:
The zeros of the above polynomial are in AP
Implies,the zeros would be of the form p+q,p and p-q
To find the value of a
Recall the polynomial rules for sum of zeros of a cubic polynomial.
If a,b and c are the zeros of a cubic polynomial,then their sum would be:
- x²coefficeint/x³coefficient
Product would be:
- constant term/x³coefficient
Product of two zeros taken once:
x coefficient/x³coefficient
Now,
•Sum of the zeros:
(p+q)+p+(p-q)= -12
→3p= -12
→p= -4
•Product of two zeros taken at once:
p(p+q)+(p+q)(p-q)+p(p-q)=39
→p²+pq+p²-q²+p²-pq=39
Regrouping and cancelling like terms,
→3p²-q²=39
→3(-4)²-39=q²
→q²=9
→q=3
•Product of zeros:
p(p-q)(p+q)= a
→4(4-3)(4+3)= a
→a= 28
For a= 28,the above polynomial has zeros in AP
Answer:
The value of the constant a in the above given polynomial f(x) = x^3 + 12 x^2 + 39 x + a is 28.
Step-by-step explanation:
Let the zeros be u, u+v and u-v as they are arithmetic progression .
Find f(u), f(u+v) and f(u-v) as they are zero.
Solve the equations.
Easy..
See the solution in the image enclosed..
