MATH SOLVE

10 months ago

Q:
# Describe the steps you would use to factor 2x3 + 5x2 – 8x – 20 completely. Then state the factored form.

Accepted Solution

A:

The polynomial [tex]2x^3 + 5x^2-8x-20[/tex] may have solutions which are the divisors of -20, therefore -20 has the following divisors: [tex]\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20[/tex].

If x=1, then [tex]2\cdot1^3 + 5\cdot1^2-8\cdot1-20=-20\neq 0[/tex],

if x=-1, then [tex]2\cdot(-1)^3 + 5\cdot(-1)^2-8\cdot(-1)-20=-9\neq 0[/tex],

if x=2, then [tex]2\cdot2^3 + 5\cdot2^2-8\cdot2-20=0[/tex], then x=2 is a solution and you have the first factor (x-2).

If x=-2, then [tex]2\cdot(-2)^3 + 5\cdot(-2)^2-8\cdot(-2)-20=0[/tex], then x=-2 is a solution, so you have the second factor (x+2).

Since x-2 and x+2 are two factors of [tex]2x^3 + 5x^2-8x-20[/tex] , then the polynomial [tex]x^2-4[/tex] is a divisor of [tex]2x^3 + 5x^2-8x-20[/tex] and dividing the polynomial [tex]2x^3 + 5x^2-8x-20[/tex] by [tex]x^2-4[/tex] you obtain

[tex]2x^3 + 5x^2-8x-20=(x-2)(x+2)(2x+5)[/tex].

If x=1, then [tex]2\cdot1^3 + 5\cdot1^2-8\cdot1-20=-20\neq 0[/tex],

if x=-1, then [tex]2\cdot(-1)^3 + 5\cdot(-1)^2-8\cdot(-1)-20=-9\neq 0[/tex],

if x=2, then [tex]2\cdot2^3 + 5\cdot2^2-8\cdot2-20=0[/tex], then x=2 is a solution and you have the first factor (x-2).

If x=-2, then [tex]2\cdot(-2)^3 + 5\cdot(-2)^2-8\cdot(-2)-20=0[/tex], then x=-2 is a solution, so you have the second factor (x+2).

Since x-2 and x+2 are two factors of [tex]2x^3 + 5x^2-8x-20[/tex] , then the polynomial [tex]x^2-4[/tex] is a divisor of [tex]2x^3 + 5x^2-8x-20[/tex] and dividing the polynomial [tex]2x^3 + 5x^2-8x-20[/tex] by [tex]x^2-4[/tex] you obtain

[tex]2x^3 + 5x^2-8x-20=(x-2)(x+2)(2x+5)[/tex].