Q4. Factorise x3 – 23x2 + 142x – 120
Answers
Answer:
[x-1][x-10][x-12].
Step-by-step explanation:
x3-23x2+142x-120
by trial and error method let us see if [x-1] is a factor of this polynomial
therefore, p[1]= 1³-23×1²+142×1-120
= 1-23+142-120
= 0
so [x-1] is a factor,
now, to divide [x-1] with x3-23x2+142x-120
by doing this we will get x²-22x+120
by doing this we will get x²-22x+120to find the other factors of x3-23x2+142x-120 we can split the middle term
x3-23x2+142x-120
=x²-12x+[-10x] +120
=x[x-12]-10[x-12]
=[x-10][x-12]
therefore, the factors of x3-23x2+142x-120 are:
[x-1][x-10][x-12]
Answer:
Step-by-step explanation:
[x-1][x-10][x-12].
Step-by-step explanation:
x3-23x2+142x-120
by trial and error method let us see if [x-1] is a factor of this polynomial
therefore, p[1]= 1³-23×1²+142×1-120
= 1-23+142-120
= 0
so [x-1] is a factor,
now, to divide [x-1] with x3-23x2+142x-120
by doing this we will get x²-22x+120
by doing this we will get x²-22x+120to find the other factors of x3-23x2+142x-120 we can split the middle term
x3-23x2+142x-120
=x²-12x+[-10x] +120
=x[x-12]-10[x-12]
=[x-10][x-12]
therefore, the factors of x3-23x2+142x-120 are:
[x-1][x-10][x-12]