To the nearest tenth the zeroes are:
[tex]x=-0.2, x=-0.7 \:and\: x=2.7[/tex]
EXPLANATION
[tex]f(x)=-5x^3+9x^2+12x+2[/tex]
The constant term is 2, and the coefficient of the highest degree is -5 so test all rational factors of
[tex] -\frac{2}{5}[/tex]
These are;
[tex] \frac{1}{5},\frac{2}{5},-\frac{1}{5}, -\frac{2}{5}[/tex]
[tex] f(-\frac{1}{5})=-5(-\frac{1}{5})^3++9( -\frac{1}{5})^2+12( -\frac{1}{5})+2[/tex]
[tex] f(-\frac{1}{5})=\frac{5}{125}+\frac{9}{25}-(\frac{12}{5})+2[/tex]
[tex]f(-\frac{1}{5})=\frac{5+45-300+250}{125}=\frac{0}{125}=0[/tex]
Good for us the very first one is a root. Therefore,
[tex]x=-\frac{1}{5}[/tex]
is a root. These also means that
[tex](5x+1)[/tex]
is a factor. Therefore we perform long division to find the other factors as follows.
We can now factor the polynomial as,
[tex]f(x)=-(5x+1)(x^2-2x-2)[/tex]
[tex]f(x)=-(5x+1)(x^2-2x-2)=0[/tex]
This implies that,
[tex]-(5x+1)=0[/tex]
Which gives us
[tex]x=-\frac{1}{5}=-0.2[/tex]
Or
[tex]x^2-2x-2=0[/tex]
[tex]\Rightarrow x^2-2x=2[/tex]
We complete the square to obtain,
[tex](x-1)^2=2+1[/tex]
[tex]\Rightarrow (x-1)^2=3[/tex]
Taking square root of both sides gives;
[tex]\Rightarrow x-1=\pm \sqrt{3}[/tex]
This implies
[tex]x=1-\sqrt{3}=-0.7[/tex]
[tex]x=1+\sqrt{3}=2.7[/tex]
