The zeroes of the function are [tex]x = -3, \ \ x =2\sqrt{ 3}i, \ \ x =-2\sqrt{ 3}i[/tex].
Solution:
Given function:
[tex]f(x)=x^{3}+3 x^{2}+12 x+36[/tex]
To find the zeros of the function:
⇒ f(x) = 0
[tex]\Rightarrow x^{3}+3 x^{2}+12 x+36=0[/tex]
[tex]\Rightarrow (x^{3}+3 x^{2})+(12 x+36)=0[/tex]
Take x² as common in first bracket and take 12 as common in 2nd bracket.
[tex]\Rightarrow x^2(x+3 )+12( x+3)=0[/tex]
Now, take common term (x + 3) outside.
[tex]\Rightarrow (x+3 ) (x^2+12)=0[/tex]
Using zero factor principle, If ab = 0 then a = 0 or b = 0.
[tex]x+3=0, \ \ x^2+12=0[/tex]
x = –3
[tex]x^2+12=0[/tex]
x² = –12
x² = 2² × 3 × –1
Taking square root on both sides, we get
[tex]\sqrt{x^2} =\pm\sqrt{2^2\times 3\times -1}[/tex]
[tex]x =\pm2\sqrt{ 3\times -1}[/tex]
we know that [tex]\sqrt{-1} =i[/tex].
[tex]x =\pm2\sqrt{ 3}i[/tex]
[tex]x =2\sqrt{ 3}i, \ \ x =-2\sqrt{ 3}i[/tex]
Hence the zeroes of the function are [tex]x = -3, \ \ x =2\sqrt{ 3}i, \ \ x =-2\sqrt{ 3}i[/tex].
