Question

Approximate to the nearest tenth the real zeros of f(x)=-5"x^{3}" alt="x^{3}" align="absmiddle" class="latex-formula">+9$x^{2}$+12x+2

Answer 1

To the nearest tenth the zeroes are:
$x=-0.2, x=-0.7 \:and\: x=2.7$

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

$-\frac{2}{5}$

These are;

$\frac{1}{5},\frac{2}{5},-\frac{1}{5}, -\frac{2}{5}$

$f(-\frac{1}{5})=-5(-\frac{1}{5})^3++9( -\frac{1}{5})^2+12( -\frac{1}{5})+2$

$f(-\frac{1}{5})=\frac{5}{125}+\frac{9}{25}-(\frac{12}{5})+2$

$f(-\frac{1}{5})=\frac{5+45-300+250}{125}=\frac{0}{125}=0$

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

$x=-\frac{1}{5}$

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

$x=-\frac{1}{5}=-0.2$

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;

$\Rightarrow x-1=\pm \sqrt{3}$

This implies

$x=1-\sqrt{3}=-0.7$

$x=1+\sqrt{3}=2.7$