The real zeros of the following function [tex]$\quad f(x)=\left(x^4-16\right)\left(x^2+3 x-18\right)$[/tex] are
2,-2,3,-6, and the graph is attached below
This is further explained below.
What are the Zeros?
[tex]$\quad f(x)=\left(x^4-16\right)\left(x^2+3 x-18\right)$[/tex]
as we know the formula a^2-b^2=(a-b)(a+b)
[tex]\begin{aligned}x^4-16 &=\left[\left(x^2\right)^2-(4)^4\right]=\left(x^2-4\right)\left(x^2+4\right) \\&=\left((x)^2-(2)^2\right)\left(x^2+4\right) \\x^4-16 &=(x-2)(x+2)\left(x^2+4\right)\end{aligned}[/tex]
substituting it in equation (1) we get.
[tex]\begin{gathered}f(x)=(x-2)(x+2)\left(x^2+4\right)\left(x^2+3 x-18\right) \\\\\left(x^2+3 x-18\right)=x^2+6 x-3 x-18=x(x+6)-3(x+6) \\\\f(x)=(x-2)(x+2)(x-3)(x+6)\left(x^2+4\right) \\\\x^2+4=(x+2 i)(x-2 i)=\left[(x)^2-(2 i)^2\right] \\\\f(x)=(x-2)(x+2)(x-3)(x+6)(x+2 i)(x-2 i)\end{gathered}[/tex]
Real zuoes =2,-2,3,-6
imaginary zuroes =2 i,-2 i
In conclusion, In the question, we have to answer the real zeroes of the function f(x)
which is 2,-2,3,-6
Read more about zeroes
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