Question
Solve equation by factorization : $21x^2 – 8x – 4 = 0$
✓
Answer
$
\begin{aligned}
& 21 x^2-8 x-4=0 \\
& \Rightarrow 21 x^2-14 x+6 x-4=0 \\
& \Rightarrow 7 x(3 x-2)+2(3 x-2)=0 \\
& \Rightarrow(3 x-2)(7 x+2)=0
\end{aligned}
$
Either $3 x-2=0$,
then $3 x =2$
$
\Rightarrow x =\frac{2}{3}
$
or
$
7 x+2=0 \text {, }
$
then $7 x =-2$
$
\Rightarrow x=\frac{-2}{7}
$
Hence $x=\frac{2}{3}, \frac{-2}{7}$.
\begin{aligned}
& 21 x^2-8 x-4=0 \\
& \Rightarrow 21 x^2-14 x+6 x-4=0 \\
& \Rightarrow 7 x(3 x-2)+2(3 x-2)=0 \\
& \Rightarrow(3 x-2)(7 x+2)=0
\end{aligned}
$
Either $3 x-2=0$,
then $3 x =2$
$
\Rightarrow x =\frac{2}{3}
$
or
$
7 x+2=0 \text {, }
$
then $7 x =-2$
$
\Rightarrow x=\frac{-2}{7}
$
Hence $x=\frac{2}{3}, \frac{-2}{7}$.
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