Factorize x3−23x2+142x−120
Factorize x3−23x2+142x−120
Let the given polynomial be f(x)=x3−23x2+142x−120
Now, putting x=0 we get,
f(0)=03+23(0)2+142(0)−120
=−120≠0
So, x=0 can't be a root.
Similar way, put x=1, we get
f(1)=13−23(12)+142(1)−120
=1−23+142−120
=143−143
=0
So, x=1 is a root of the given polynomial.
⟹(x−1) is factor of the given polynomial.
To find the other factors, apply long division method :
x2−22x+120x−1)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x3−23x2+142x−120x3−x2(−)−+__________________________________ −22x2+142x−12022x2+22x(−)−−__________________________________120x−120120x−120(−)−+___________________________________0
Therefore, x2−22x+120 
is also a factor of polynomial f(x).
Now, factorize x2−22x+120 as follow:
x2−22x+120=x2−12x−10x+120
=x(x−12)−10(x−12)
∴x2−22x+120=(x−12)(x−10)
Hence, 
(x−1)(x−12)(x−10) are the factors of given polynomial f(x).
Let the given polynomial be f(x)=x3−23x2+142x−120
Now, putting x=0 we get,
f(0)=03+23(0)2+142(0)−120
=−120≠0
So, x=0 can't be a root.
Similar way, put x=1, we get
f(1)=13−23(12)+142(1)−120
=1−23+142−120
=143−143
=0
So, x=1 is a root of the given polynomial.
⟹(x−1) is factor of the given polynomial.
To find the other factors, apply long division method :
x2−22x+120x−1)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x3−23x2+142x−120x3−x2(−)−+__________________________________ −22x2+142x−12022x2+22x(−)−−__________________________________120x−120120x−120(−)−+___________________________________0
Therefore, x2−22x+120 is also a factor of polynomial f(x).
Now, factorize x2−22x+120 as follow:
x2−22x+120=x2−12x−10x+120
=x(x−12)−10(x−12)
∴x2−22x+120=(x−12)(x−10)
Hence, (x−1)(x−12)(x−10) are the factors of given polynomial f(x).
