are koi to answer kar do...
lekin correct..
factorise
Answers
Question
Factorise
x3-23x2+142x-120
3x3−x2−3x+1
2y3 + y2 - 2y - 1.
Answer
- Factorise
⇒x3-23x2+142x-120
⇒ by trial and error method let us see if [x-1] is a factor of this polynimial
⇒ 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
⇒ to 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]
- Factorise
⇒3x3−x2−3x+1
=x2(3x−1)−1(3x−1)
=(3x−1)(x2−1)
=(3x−1)(x−1)(x+1)⇒3x3−x2−3x+1
=(3x−1)(x−1)(x+1)3x3-x2-3x+1=x23x-1-13x-1
=3x-1x2-1
=3x-1x-1x+1
⇒3x3-x2-3x+1
=3x-1x-1x+1
therefore the factors of 3x3−x2−3x+1
3x-1x-1x+1
- Factorise
⇒Let f(y) = 2y3 + y2 - 2y - 1.
⇒Substitute y = 1 then f(y) = 0
⇒y -1 is a factor of 2y3 + y2 - 2y - 1.
⇒ y - 1 ) 2y3 + y2 - 2y - 1 ( 2y2 +3y +1
⇒ 2y3 - 2y2 (substract)
⇒3y2 - 2y - 1
⇒ 3y2 - 3y (substract)
⇒ y -1
⇒ y -1 (substract)
0
∴ The quotient is 2y2 +3y +1 = 0
⇒ 2y2 +2y + y +1 + 0
⇒2y(y + 1) +1(y +1) =0
⇒2y + 1 = 0 ,y + 1 = 0
⇒y = -1/2 , -1 .
∴ 2y3 + y2 - 2y - 1 = ( y - 1)( 2y + 1)(y +1).
hope this helps u
no copied content