Question

If α and β are the zeroes of the quadratic polynomial f(t) = t² - 4t + 3 ,
find the value of a⁴β³ + α³β⁴

Answer 1

l would solve it:-
First of all:
f(t)=t(2sq.)-4t+3.
=t(2sq.)-t-3t+3
=t(t-1)-3(t-1)
=(t-3)(t-1)
So,
The zeros are 3,1.
alpha:3
beta:1
now,
a(4powered.)xB(3powered)+a(3powered)xB(4 powered)
={(3x3x3x3)(1x1x1)}+{(3x3x3)(1x1x1x1)}
=( 81x1) +(27+1)
=81+27
=108
There are some marks you may not understand, if any doubts comment.
Hope it helps!!

Answer 2

f(t)=t²-4t+3

t²-t-3t+3=0

t(t-1)-3(t-1)=0

(t-1)(t-3)=0

t-1=0 | t-3=0
t=1 | t=3

Therefore,the zeroes of the given polynomial are 1 and 3.

Let α=1 and β=3,

a⁴β³ + α³β⁴ = (1)⁴(3)³ + (1)³(3)⁴

= 1(27)+1(81)

=27+81

=108

Hope it helps