# Find all Real solutions to this equation: (x - 2)(9x^2 - 8x)=(x - 2)(16x - 16)

Find all Real solutions to this equation:
$\left(x-2\right)\left(9×2-8x\right)=\left(x-2\right)\left(16x-16\right)$
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$\left(x-2\right)\left(9×2-8x\right)=\left(x-2\right)\left(16x-16\right)\phantom{\rule{0ex}{0ex}}9{x}^{3}-26{x}^{2}+16x=16{x}^{2}-48x+32$
Subtract $16{x}^{2}-48x+32$ from both sides
$9{x}^{3}-26{x}^{2}+16x-16{x}^{2}+48x-32=0\phantom{\rule{0ex}{0ex}}9{x}^{3}-42{x}^{2}+64x-32=0$
One factor of this polynomial is x-2
Divide the polynomial $9{x}^{3}-42{x}^{2}+64x-32$ by x-2. So we get:
$\left(9{x}^{3}-42{x}^{2}+64x-32\right)/\left(x-2\right)=9{x}^{2}-24x+16$
Our polynomial becomes:
$\left(9{x}^{2}-24x+16\right)=\left(3x-4{\right)}^{2}\phantom{\rule{0ex}{0ex}}\left(x-2\right)\left(3x-4{\right)}^{2}=0$
see factors equal to 0
x-2=0 or 3x-4=0
x=2 or x=4/3
Real solutions are x=2, 4/3