Find the zeros of polynomial f(x) = x^3-12x^2+39x-28, If it is given that the zeros are in A.P.
Answers
Q:p(X)=x³-21x²+39x-28
A:Let the zeroes be:-
a-d,a,a+d
a-d+a+a+d=12.
3a=12
a=4
a(a+d)(a-d)=28
a(a²-d²)=28
4(16-d²)=28
16-d²=7
-d²=-9
d=3
⏺️a-d=1
⏺️a = 4
⏺️a+d=7
ANS:- The ZEROES are:-1,4,7
Answer:
The zeros are given in A.P. are given to be 1, 4 and 7
Step-by-step explanation:
Let the given polynomial be : f(x) = x³- 21x² + 39x - 28
Let the zeroes in the A.P. are : a - d, a, a + d
Now, by comparing the given polynomial we get,
⇒ a - d + a + a + d = 12.
⇒ 3a = 12
⇒ a = 4
Also, a·(a + d)·(a - d) = 28
⇒ a·(a² - d²) = 28
⇒ 4·(16 - d²) = 28
⇒ 16 - d² = 7
⇒ -d² = -9
⇒ d = 3
Thus, The zeros are : a - d = 1
a = 4
a + d = 7
Hence, The zeros are given in A.P. are given to be 1, 4 and 7