Solution/s:
1.
[tex] \displaystyle x^3 + 13x^2 + 30x = ? [/tex]
[tex] \displaystyle \textsf{Factor out x:} [/tex]
[tex] \displaystyle = x(x^2 + 13x + 30) [/tex]
[tex] \displaystyle \textsf{Factor quadratic trinomial:} [/tex]
[tex] \displaystyle = \boxed{\green{ x(x+10)(x+3) }} [/tex]
[tex] \displaystyle \textsf{(Optional)Let us solve for x:} [/tex]
[tex] \displaystyle \textsf{Equate to 0:} [/tex]
[tex] \displaystyle = x=0,(x+10=0),(x+3=0) [/tex]
[tex] \displaystyle = x=0,(x=-10),(x=-3) [/tex]
[tex] \displaystyle = \boxed{ x=0, -10, -3} [/tex]
[tex] \displaystyle x^3 + 13x^2 + 30x = x(x+10)(x+3) [/tex]
2.
[tex] \displaystyle x^3 + 4x^2 + 5x + 2 = ? [/tex]
[tex] \displaystyle \textsf{Factor out x:} [/tex]
[tex] \displaystyle x^3 + 4x^2 + 5x + 2 [/tex]
[tex] \displaystyle \textsf{Let us find the possible roots:} [/tex]
[tex] \displaystyle \pm 2 \textsf{\ and\ } \pm 1 [/tex]
[tex] \displaystyle \textsf{Let us try: \ } \pm 1 [/tex]
[tex] \displaystyle 1^3 + 4(1)^2 + 5(1) + 2 = 0 \quad ? [/tex]
[tex] \displaystyle 1 + 4 + 5 + 2 = 0 \quad ? [/tex]
[tex]\displaystyle 12 \ne 0 \quad \times \[/tex]
[tex] \displaystyle (-1)^3 + 4(-1)^2 + 5(-1) + 2 = 0 \quad ? [/tex]
[tex] \displaystyle -1 + 4 + -5 + 2 = 0 \quad ? [/tex]
[tex]\displaystyle 0 = 0 \quad \checkmark \[/tex]
[tex]\displaystyle \textsf{-1 is a solution, therefore x+1 is a factor:}[/tex]
[tex] \displaystyle = (x+1)(x^2+3x+2) [/tex]
[tex] \displaystyle \textsf{Now let us factor the quadratic trinomial:} [/tex]
[tex] \displaystyle = \boxed{\green{ (x+1)(x+1)(x+2) }} [/tex]
[tex] \displaystyle x^3 + 4x^2 + 5x + 2 = (x+1)(x+1)(x+2) [/tex]
