Question

e, x² (2x - 3) + 2(2x - 3)=0
g, 3x(x + 5) - 2x -10 =0 

Answer 1

\(a,x^2\left(2x-3\right)+2\left(2x-3\right)=0\\ \Leftrightarrow\left(2x-3\right)\left(x^2+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x^2+2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x^2=-2\left(ktm\right)\end{matrix}\right.\)

\(b,3x\left(x+5\right)-2x-10=0\\ \Leftrightarrow b,3x\left(x+5\right)-2\left(x+5\right)=0\\ \Leftrightarrow\left(x+5\right)\left(3x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+5=0\\3x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{2}{3}\end{matrix}\right.\)

Answer 2

e) \(\left(2x-3\right)\left(x^2+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x^2=-2\left(ktm\right)\end{matrix}\right.\)

g) \(3x\left(x+5\right)-2\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(3x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{2}{3}\end{matrix}\right.\)