Question

Approximate to the nearest tenth the real zeros of f(x)=-5
`x^(3)`+9
`x^(2)`+12x+2

Answer 1

To the nearest tenth the zeroes are:

`x=-0.2, x=-0.7 \:and\: x=2.7`

Step-by-step explanation


`f(x)=-5x^3+9x^2+12x+2`

The constant term is 2, and the coefficient of the highest degree is -5 so test all rational factors of


`-(2)/(5)`

These are;


`(1)/(5),(2)/(5),-(1)/(5), -(2)/(5)`


`f(-(1)/(5))=-5(-(1)/(5))^3++9( -(1)/(5))^2+12( -(1)/(5))+2`


`f(-(1)/(5))=(5)/(125)+(9)/(25)-((12)/(5))+2`


`f(-(1)/(5))=(5+45-300+250)/(125)=(0)/(125)=0`

Good for us the very first one is a root. Therefore,


`x=-(1)/(5)`

is a root. These also means that


`(5x+1)`

is a factor. Therefore we perform long division to find the other factors as follows.

We can now factor the polynomial as,


`f(x)=-(5x+1)(x^2-2x-2)`


`f(x)=-(5x+1)(x^2-2x-2)=0`

This implies that,


`-(5x+1)=0`

Which gives us


`x=-(1)/(5)=-0.2`

Or


`x^2-2x-2=0`


`\Rightarrow x^2-2x=2`

We complete the square to obtain,


`(x-1)^2=2+1`


`\Rightarrow (x-1)^2=3`

Taking square root of both sides gives;


`\Rightarrow x-1=\pm √(3)`

This implies


`x=1-√(3)=-0.7`


`x=1+√(3)=2.7`