Question 4
The number of polynomials having zeroes as -2 and 5 is
Solutions:
Let p(x)=ax2+bx+c be the required polynomial whose zeroes are -2 and 5.
∴ Sum of zeroes =−b/a
⇒−b/a=−2+5=3/1=−(3)/1
and product of zeroes =c/a
⇒c/a=−2×5=−10/1
From Eqs. (i) and (ii),
a=1,b=−3 and c=−10
∴p(x)=ax2+bx+c=1x2−3x−10
=x2−3x−10
But we know that, if we multiply/divide any polynomial by any arbitrary constant. Then, the zeroes of polynomial never change.
∴p(x)=kx2−3kx−10k [where, k is a real number]
⇒p(x)=x2/k−(3/k)x−10/k, [where,k is a non-zero real number]
Hence, the required number of polynomials are infinite i.e., more than 3.