Question

1. x^3-2x^2+10x+136=0
2. 2x^3+2x^2-19x+20=0

Answer 1

Explanation:

1) x³-2x²+10x+136=0;

the roots are among ±1; ±2; ±4; ±8; ±17... If to substitute x=-4, then 0=0, it means, x= -4 is root of the given equation. Then if to re-write the equation, then (x+4)(x²-6x+34)=0. Finally, the only root, x=-4.

2) 2x³+2x²-19x+20=0;

the roots are among ±0.5; ±1; ±2; ±2.5; ±4; ±5... If to substitute x= -4, then 0=0, it means x=-4 is root of the given equation. Then if to re-write the equation, then (x+4)(x²-3x+2.5)=0. Finally, the only root is x= -4.