Question

I'm having problem solving these two equations I will include a picture

Answer 1

Given:

1) The equation is,

`4x^3-5x^2-196x+245=0`

To solve this equation,

using synthetic division,

Now solving further,

`\begin{gathered} 4x^3-5x^2-196x+245=(x-7)(4x^2+23x-35) \\ take,\text{ }4x^2+23x-35=0 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x_{}=(-23\pm√(23^2-4\cdot\:4\left(-35\right)))/(2\cdot\:4) \\ x=(-23\pm\:33)/(2\cdot\:4) \\ x_{}=(-23+33)/(2\cdot\:4),\: x_{}=(-23-33)/(2\cdot\:4) \\ x=(5)/(4),\: x=-7 \end{gathered}`

Hence, the solution of given equation is,

`\begin{gathered} 4x^3-5x^2-196x+245=(x-7)(x-(5)/(4))(x+7) \\ \Rightarrow(x-7)(x-(5)/(4))(x+7)=0 \\ \Rightarrow x=\text{ 7,-7,}(5)/(4) \end{gathered}`

2) the equation is,

`9x^3+2x^2+9x+2=0`

Now, factor the equation,

`\begin{gathered} 9x^3+2x^2+9x+2=0 \\ x^2(9x+2)+(9x+2)=0 \\ (9x+2)(x^2+1)=0 \\ \Rightarrow9x+2=0,x^2+1=0 \\ \Rightarrow x=(-2)/(9), \\ \text{and x}^2=-1 \\ x=\pm\sqrt[]{-1} \\ x=\pm i \end{gathered}`

Hence, the solution of above equation is x= -2/9 , i , -i.