Find zeroes of p(x) = abx2 …

We have,
f(x) = abx² + (b² - ac)x - bc
= abx² + b²x - acx - bc
= bx(ax + b ) - c(ax + b)
= (ax + b) (bx - c)
Now r(x)=0 if 
ax+b=0 or bx-c=0
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Thus, the zeroes of f(x) are : {tex}\style{font-family:Arial}{\style{font-size:12px}{\mathrm\alpha}\style{font-size:12px}=\style{font-size:12px}-\frac{\style{font-size:12px}{\mathrm b}}{\style{font-size:12px}{\mathrm a}}\style{font-size:12px}\;\style{font-size:12px}{\mathrm{and}}\style{font-size:12px}\;\style{font-size:12px}{\mathrm\beta}\style{font-size:12px}=\frac{\style{font-size:12px}{\mathrm c}}{\style{font-size:12px}{\mathrm b}}}{/tex}