Question

are koi to answer kar do...
lekin correct..
factorise​

Answer 1

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

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