SOLUTION
Write out the polynomial given
The first group of the expresion is
[tex]\begin{gathered} 3x^3+4x^2 \\ \text{Then the GCE is } \\ x^2(\frac{3x^3}{x^2}+\frac{4x^2}{x^2}) \\ \text{GCE}=x^2 \end{gathered}[/tex]
GCE is x²
For the second group, we have
[tex]\begin{gathered} 75x+100 \\ \text{GCE}=25(\frac{75x}{25}+\frac{100}{25}) \\ \text{GCE}=25 \end{gathered}[/tex]
The GCE for the secod group is 25
To factorise completely, we have
[tex]\begin{gathered} 3x^3+4x^2+75x+100 \\ \\ x^2(\frac{3x^3}{x^2}+\frac{4x^2}{x^2})+25(\frac{75x}{25}+\frac{100}{25}) \end{gathered}[/tex]
Then by simplification, we have
[tex]\begin{gathered} x^2(3x+4)+25(3x+4) \\ \text{Then, we factor completely to get} \\ (3x+4)(x^2+25) \end{gathered}[/tex]
Then factors are (3x +4)(x²+ 25)
To find the real root, we equate each of the factors to zero, hence
[tex]\begin{gathered} (3x+4)(x^2+25)=0 \\ \text{Then} \\ 3x+4=0orx^2+25=0 \\ 3x=-40rx^2=-25 \\ \end{gathered}[/tex]
Thus
[tex]\begin{gathered} \frac{3x}{3}=-\frac{4}{3} \\ x=-\frac{4}{3}\text{ is a real root } \\ or\text{ } \\ x^2=-25 \\ \text{take square root} \\ x=\pm_{}\sqrt[]{-25}\text{ not a real root} \end{gathered}[/tex]
Therefore, since the root of -25 is a complex number,
The only real root is x = -4/3
