Question

Find the zeros of the polynomial $x^2-3x-m(m+3).$

Answer 1

$x^2-3x-m(m+3)\Rightarrow x^2-3x-m^2-3m$

a = 1

b = -3

c = $-m^2-3m$

Discriminant, D = $b^2-4ac=3^2-4\times 1\times (-m^2-3m)=9+4m^2+12m=4m^2+6m+6m+9=2m(2m+3)+3(2m+3)=(2m+3)(2m+3)=(2m+3{)}^2$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-(-3)\pm \sqrt{(2m+3{)}^2}}{2}=\frac{3\pm (2m+3)}{2}=\frac{3+(2m+3)}{2}or\frac{3-(2m+3)}{2}=\frac{3+2m+3}{2}or\frac{3-2m-3}{2}=\frac{6+2m}{2}or\frac{-2m}{2}=m+3or-m$

The values of zeroes are m+3 and -m