t^3 - 49t =0
The Highest Common Factor of the equation = t
so we have
[tex]t\text{ (}\frac{t^3}{t}-\text{ }\frac{49t}{t})\text{ =0}[/tex]t(t^2 - 49)=0
by factoring out we get
t = 0
and
[tex]t^{2\text{ }}-49\text{ =0}[/tex]by using difference of two squares where
[tex]a^2-b^2=(a-b)(a+b)\text{ }[/tex]t^2 -49= t^2 -7^2
comparing this to difference of two squares.
t^2 + a^2
and b^2 = 7^2
so by expansion
t^2 - 7^2 =(t-7)(t+7)
so (t-7)(t+7) =0
t-7 =0 ; t = 7
t +7 = 0; t = -7
so t = 0,7 and -7
