Let a, ß and y be the zeros of polynomial f(x) such that ab = 12

We have, alpha + beta + y = (- b)/a = (- (- 5))/1 = 5

aßy=d= alpha*beta + beta*y + y*alpha = c/a = - 2/1 = - 2 and aßy = d.

-24 = -24

Putting aß = 12 in aßy = -24, we get

12y = - 24

y = -24 = -2

Now,a+B+ y = 5

a+B

a = 7-B

Ask

2=5

⇒a+B=7 ⇒

.:. αβ = 12

⇒ (7-B)B = 12

7beta - beta ^ 2 = 12

B²-7B+12=0 B²-3B-4B + 12 =

⇒ B=4 r*beta = 3

alpha = 3 or alpha = 4

hope this helps ❤️

Note:

If x = A and x = B are the zeros of given quadratic polynomial p(x) ,then p(x) will be given by;

p(x) = x^2 - (A+B)•x + A•B .

Solution:

Here,

The given zeros of required polynomial are

x = -3 and x = 4 .

Thus,

The required polynomial p(x) will be given by;

=> p(x) = x^2 - {(-3) + 4}•x + (-3)•4

=> p(x) = x^2 - x - 12

Hence,

The required polynomial is x^2 - x - 12 .

hope this helps ❤️