Find the zeros of the polynomial P() =
²– 3.
Answers
Answer:
(∝+β)=0 and (∝β)=-3
Step-by-step explanation:
Given polynomial p(x)=x²-3
p(x)=x²-(√3)²
↓
It is in the form of a^{2}-b^{2}a
2
−b
2
and we know that a^{2}-x^{2}=(a+b)(a-b)a
2
−x
2
=(a+b)(a−b)
a^{2}=x^{2}a
2
=x
2
b^{2}=(\sqrt3)^{2}b
2
=(
3
)
2
(x)²-(√3)²=(x+√3)(x-√3)
x+√3=0
x=0-√3
x=-√3
∝=-√3
and,x-√3=0
x=0+√3
x=+√3
β=√3
x²-3⇒x²-0x-3
Sumof zeroes(∝+β)=-b/a
-√3+√3=-0/1
0=0
Product of zeroes(∝β)=c/a
-√3×√3=-3/1
-3=-3