Question

answer the 9 question I will mark you as brainiest

Answer 1

2 ,4, 3 are answer
first find the factors of constant and then add to get 9.

Answer 2

Given f(x) = x^3 - 9x^2 + 26x - 24.

Let a, b, c be the zeroes of the polynomial.

= > We know that sum of zeroes = -b/a

a + b + c = -(-9)/1

a + b + c = 9.  ------- (1)

Given that sum of its two zeroes is 5.

5 + c = 9

c = 4.    ---------- (2)


= > We know that product of zeroes = -d/a

abc = -(-24)/1

abc = 24   

ab(4) = 24

ab = 6  -------- (3) 


Substitute (3) in (1)

= > a + b + 4 = 9

= > a + b = 5  ----- (4)


We know that (a + b)^2 = (a - b)^2 + 4ab

= > (5)^2 = (a - b)^2 + 4(6)

= > 25 = (a - b)^2 + 24

= > a - b = 1   ------ (5)


On solving (4) & (5), we get

a + b = 5

a - b = 1

------------------

2a = 6

a = 3.


Substitute a = 3 in (4), we get

= > a + b = 5

= > 3 + b = 5

= > b = 5 - 3

= > b = 2.




Therefore the zeroes of the polynomial are 3,2,4.



Hope this helps!