Question

${x}^{3} - {4x}^{2} + x + 6$
Find roots with steps

Answer 1

let f(x)=x³-4x²+x+6
for existing zeroes, f(x) = 0
f(3) = 0;
so, (x-3) is a factor.
provided,
x³-4x²+x+6
=x³-3x²-x²+3x-2x+6
=x²(x-3) -x(x-3) -2(x-3)
=(x-3)(x²-x-2)
Now for to find the roots, f(x) has to be equal to 0.
f(x)=0
=>x³-4x²+x+6=0
=>(x-3)(x²-x-2)=0
Either,
x-3=0
=>x=3
Or,
x²-x-2=0
=>x={1±√(1+8)}/2
=>x=2,-1
So, x=-1,2,3