Question

Factorise:
$\left(i\right)x^3-2x^2-x+2$

$\left(ii\right)x^3-3x^2-9x-5$

$\left(iii\right)x^3+13x^2+32x+20$

$\left(iv\right)2y^3+y^2-2y-1$

Answer 1

$\left(i\right)Given;x^3-2x^2-x+2$

Factorising
$x^3-x-2x^2+2$

$=x(x^2-1)-2(x^2-1)$
$=(x^2-1)(x-2)$
$=\left[\right(x{)}^2-\left(1{)}^2\right](x-2)$
$=(x-1)(x+1)(x-2)$

$\because \left[\right(a^2-b^2)=(a+b\left)\right(a-b\left)\right]$

Conclusion
$Thus,x^3-2x^2-x+2=(x-1)(x+1)(x-2)$



$\left(ii\right)Given:x^3-3x^2-9x-5$

Factorising
$x^3-3x^2-9x-5$

$=x^3+x^2-4x^2-4x-5x-5$
$=x^2(x+1)-4x(x+1)-5(x+1)$
$=(x+1)(x^2-4x-5)$
$=(x+1)(x^2-5x+x-5)$
$=(x+1)\left[\right(x(x-5)+1(x-5)\left)\right]$
$=(x+1)(x-5)(x+1)$

Conclusion
$Thus,x^3-3x^2-9x-5=(x+1)(x-5)(x+1)$


$\left(iii\right)Given:x^3+13x^2+32x+20$

Factorising
$x^3+13x^2+32x+20$

$=x^3+x^2+12x^2+12x+20x+20$
$=x^2(x+1)+12x(x+1)+20(x+1)$
$=(x+1)(x^2+12x+20)$
$=(x+1)(x^2+2x+10x+20)$
$=(x+1)\left[x\right(x+2)+10(x+2)]$
$=(x+1)(x+2)(x+10)$

Conclusion
$Thus,x^3+13x^2+32x+20=(x+1)(x+2)(x+10)$



$\left(iv\right)Given:2y^3+y^2-2y-1$

Factorising
$2y^3+y^2-2y-1$

$=2y^3-2y^2+3y^2-3y+y-1$
$=2y^2(y-1)+3y(y-1)+1(y-1)$
$=(y-1)(2y^2+3y+1)$
$=(y-1)(2y^2+2y+y+1)$
$=(y-1)\left[\right(2y(y+1)+1(y+1)]$
$=(y-1)(y+1)(2y+1)$

Conclusion:
$Thus,2y^3+y^2-2y-1$
$=(y-1)(y+1)(2y+1)$