Question

50 points..
alpha, beta and gamma are zeroes of cubic polynomial x cube - 12 x square + 44 X + c. if alpha, beta, and gamma are in AP find the value of c.
explanation with steps..

Answer 1

Answer:

As α,β,¥ are in Ap, let then be a-d,a,a+d respectively...(.P)

Comparing with ax3+bx2+cx+d,

a=1 b=-12 c=44 d=c

α+ β +¥ = -b/a

α +β +¥=-(-12)/1=12

From p

a+a+d+a-d =12

3a =12

a=4

But β=a

β=4

As beta is a zero for given eq.

F(β)=0

F(a)=0

F(4)=0

F(x)=x3-12x2+44x+c

F(4)=(4)^3 -12(4)^2+44(4) +c

0=64-192+176+c

C=-48

Answer 2

✌♥✌hey mate here is your answer ✌♥✌

p(X)=x³-12x²+44x+c♥

now alpha=A⛄

beta=B♥

gamma=C⛄

so A,B,C are the roots of p(X)♥

now⛄

as A,B,C are in AP♥

B=A+C/2..........1⛄

now♥

A×B×C=-c......2⛄

A+B+C=12.......3♥

AB+BC+AC=44......4⛄

A³+B³+C³=(A+B+C)(A²+B²+C²-AB-BC-AC)+3ABC..........5♥

A²+B²+C²=(A+B+C)²-2(AB+BC+AC)......6⛄

now...as A,B,C are roots♥

p(A)=A³-12A²+44A+c=0.....7⛄

p(B)=B³-12B²+44B+c=0.......8♥

p(C)=C³-12C²+44C+c=0.......9⛄

add 7,8,9♥

A³+B³+C³-12(A²+ B² +C²)+44(A+B+C)+3c=0⛄

from 5♥

(A+B+C)(A²+B²+C²-AB-BC-AC)+3ABC-12
(A+B+C)(A²+B²+C²-AB-BC-AC)+3ABC+44(A+B+C)+3c=0⛄

now from 6♥

(A+B+C)((A+B+C)²-2(AB+BC+AC)-AB-BC-AC)+3ABC-12(A+B+C)((A+B+C)²-2(AB+BC+AC)-AB-BC-AC)+3ABC+44(A+B+C)+3c=0⛄

now from 12345♥

12((12)²-2(44)-1(44))+3(-c)-12(12)((12)²-2(44)-1(44)+3(-c)+44(12)+3c=0⛄

12(144-88-44)-3c-144(144-88-44)-3c+528+3c=0♥

12(12)-3c-144(12)-3c+528+3c=0⛄

144-1728+528-6c+3c=0♥

-3c=1056⛄

c=-1056/3♥

c=-352⛄

✌♥hope it will help you ♥✌

✌✌✌mark me brainliest ✌✌✌✌