Question 2

A quadratic polynomial, whose zeroes are-3 and 4, is

Solution:

Let ax²+bx+c be a required polynomial whose zeroes are -3 and 4,
Then, sum of zeroes =-3+4=1[∵sum of zeroes=-b/a] …(i)
⇒-b/a=1/1⇒-b/a=-(-1)/1
and product of zeroes =-3×4=-12[∵product of zeroes=c/a] …(ii)
⇒c/a=-12/1
From Eqs. (i) and (ii),
a=1,b=-1 and c=-12
=ax²+bx+c
∴ Required polynomial =1.x²-1.x-12
=x²-x-12
=x^2/2-x/2-6
We know that, if we multiply/divide and polynomial by any constant, then the zeroes of polynomial do not change.

Alternate Method
Let the zeroes of a quadric polynomial are α=-3 and β=4.
The, sum of zeroes =α+β=-3+4=1
and product of zeroes =αβ=(-3)(4)=-12
∴ Required polynomial =x²-(sum of zeroes) x+ (product of zeroes)
=x²-(1)x+(-12)=x²-x-12
x²/2-x/2-6