very nice video well represented and well explained. Thank you. But this method only works if the constant can be decomposed into thw replacement cube and square term as required,
By inspection x³+x²=12 can be written as x³+x²=2³+2². Hence x=2. To find other root, if any, note that the last equation can be written as x³-2³+x²-2²=0 (x-2)(x²+2x+4)+(x+2)(x-2)=0 (x-2)(x²+3x+4)=0 x=2, x=½[-6±isqrt(3)] For real root x=2 is the only one. Other root is a complex one.
very nice video well represented and well explained. Thank you. But this method only works if the constant can be decomposed into thw replacement cube and square term as required,
Very arreactuve and well explained. Thank you. But this mergid we only work if the constant can be decomposed into the two cubes required
This can be easily solved using synthetic division but you took it to a whole 'nother level. Very nice tips and tricks.
Yes po, you are right. We just use this as an opportunity to share another method they can use in solving this problem
By inspection x³+x²=12 can be written as x³+x²=2³+2². Hence x=2.
To find other root, if any, note that the last equation can be written as
x³-2³+x²-2²=0
(x-2)(x²+2x+4)+(x+2)(x-2)=0
(x-2)(x²+3x+4)=0
x=2, x=½[-6±isqrt(3)]
For real root x=2 is the only one. Other root is a complex one.