Question

The function below has at least one rational zero.

Answer 1

Answer:

$x=-7,x=-1,x=-0.289898,x=0.689898$

or $x=-7,x=-1,x=\frac{1}{5}-\frac{\sqrt{6}}{5},x=\frac{1}{5}+\frac{\sqrt{6}}{5}$

Step-by-step explanation:

Factor by grouping

$g(x)=5x^4+38x^3+18x^2-22x-7\\\\g(x)=5x^4+33x^3-15x^2-7x+5x^3+33x^2-15x-7\\\\g(x)=x(5x^3+33x^2-15x-7)+1(5x^3+33x^2-15x-7)\\\\g(x)=(x+1)(5x^3+33x^2-15x-7)\\\\g(x)=(x+1)(5x^3+35x^2-2x^2-14x-x-7)\\\\g(x)=(x+1)(5x^2(x+7)-2x(x+7)-1(x+7))\\\\g(x)=(x+1)(x+7)(5x^2-2x-1)\\\\0=(x+1)(x+7)(5x^2-2x-1)$

We can easily see that the first two zeroes are $x=-1$ and $x=-7$, however, the third zero will need the help of the quadratic formula:

$0=5x^2-2x-1\\\\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \\x=\frac{-(-2)\pm\sqrt{(-2)^2-4(5)(-1)}}{2(5)}\\\\x=\frac{2\pm\sqrt{4+20}}{10}\\ \\x=\frac{2\pm\sqrt{24}}{10}\\ \\x=\frac{2\pm2\sqrt{6}}{10}\\ \\x=\frac{2}{10}\pm\frac{2\sqrt{6}}{10}\\ \\x=\frac{1}{5}\pm\frac{\sqrt{6}}{5}\\\\x_1\approx0.689898\\\\x_2\approx-0.289898$

Therefore, the zeroes of the function are $x=-7,x=-1,x=-0.289898,x=0.689898$