Question

using factor theorem factorise the following problem​.....


write answer plz

Answer 1

p(y) = y^3 - 2y^2 - 29y - 42

Take out the L.C.M of the constant

Here you get 42 = 2 × 3 × 7

Thus, we get (y - 2), ( y+2) , ( y - 3), ( y + 3 ) , ( y - 7), (y +7 )

Now , check which of them are factor of p(y)

using (y - 7) Then we get

y - 7 = 0

y = 7

putting value of y in p(y) we get

p(y) = 7^3 - 2(7)^2 - 29 × 7 - 42

= 0

Thus, (y - 7 ) is a factor of p(y )

Now,

p(y) = y^3 - 2y^2 - 29y - 42

Take out the factor from p(y) and then factorise

p( y ) = y^3 - 7y^2 + 5y^2 - 29y - 42

= y^2 ( y - 7) + 5y^2 - 35y + 6y - 42

= y^2 (y - 7) + 5y ( y - 7 ) + 6 ( y - 7 )

= (y - 7) ( y^2 + 5y + 6 )

To get the factor out we convert the numbers in such way that we can get our factor out.

Now by applying Middle Term Factorization we get our answer.

= ( y - 7 ) ( y^2 + 2y + 3y + 6 )

= ( y - 7 ) { y ( y + 2 ) + 3 ( y + 2 )

= (y - 7 ) ( y + 2 ) ( y + 3 ). Answer