Question

T^3-49t=0 solve by factoring

Answer 1

t^3 - 49t =0

The Highest Common Factor of the equation = t

so we have

`t\text{ (}(t^3)/(t)-\text{ }(49t)/(t))\text{ =0}`

t(t^2 - 49)=0

by factoring out we get

t = 0

and

`t^{2\text{ }}-49\text{ =0}`

by using difference of two squares where

`a^2-b^2=(a-b)(a+b)\text{ }`

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