Question

Factorise this! Help me plz

Answer 1

$\huge\purple {♡}\pink { Answer } \purple{♡}$

${2y}^{3} + {y}^{2} - 2y - 1$

${y}^{2} (2y + 1) - (2y + 1)$

$(2y + 1)( {y}^{2} - 1)$

using Identity

${x}^{2} - {y}^{2} = (x - y)(x + y)$

$(2y + 1)(y - 1)(y + 1)$

Answer 2

using hit and trial method,to solve this equation...

we put y=-1

putting this we get

2×(-1)^3+(-1)^2-2×-1-1

-2+1+2-1=0

so,by factor theorem,(y+1),will be the factor of this eqaution....

this means ,(y+1) will give 0 as remainder

dividing 2y^3+y^2-2y-1 with (y+1) we get

f(y)=(y-1)(2y^2+3y+1)+0 (divedend=divisor×quotient +remainder)

f(y)=(y-1){2y^2+2y+y+1}

f(y)=(y-1){2y(y+1)+1(y+1)}

f(y)=(y-1)(2y+1)(y+1)

hence the factors are

hence the factors are(y-1)(2y+1)(y+1)

{hope it helps}