Question

What are the zeros for the function

Answer 1

The zeroes of the function are $x = -3, \ \ x =2\sqrt{ 3}i, \ \ x =-2\sqrt{ 3}i$.

Solution:

Given function:

$f(x)=x^{3}+3 x^{2}+12 x+36$

To find the zeros of the function:

⇒ f(x) = 0

$\Rightarrow x^{3}+3 x^{2}+12 x+36=0$

$\Rightarrow (x^{3}+3 x^{2})+(12 x+36)=0$

Take x² as common in first bracket and take 12 as common in 2nd bracket.

$\Rightarrow x^2(x+3 )+12( x+3)=0$

Now, take common term (x + 3) outside.

$\Rightarrow (x+3 ) (x^2+12)=0$

Using zero factor principle, If ab = 0 then a = 0 or b = 0.

$x+3=0, \ \ x^2+12=0$

x = –3

$x^2+12=0$

x² = –12

x² = 2² × 3 × –1

Taking square root on both sides, we get

$\sqrt{x^2} =\pm\sqrt{2^2\times 3\times -1}$

$x =\pm2\sqrt{ 3\times -1}$

we know that $\sqrt{-1} =i$.

$x =\pm2\sqrt{ 3}i$

$x =2\sqrt{ 3}i, \ \ x =-2\sqrt{ 3}i$

Hence the zeroes of the function are $x = -3, \ \ x =2\sqrt{ 3}i, \ \ x =-2\sqrt{ 3}i$.