Question

He polynomial (2x - 1) (x^2 -2) - x (x^2 - x -2)

can be written in the form
ax^3 + bx^2 + cx + d

where a, b, c, and dare constants.
List the values of a, b, c, and d seperated by comas.

Answer 1

a = 1

b = 0

c = -2

d = 2

Explanation:

Given polynomial:

`\sf= (2x-1)(x^2-2)-x(x^2-x-2)\\\\Distributing \\\\= 2x(x^2-2)-1(x^2-2)-x^3+x^2+2x\\\\= 2x^3-4x-x^2+2-x^3+x^2+2x\\\\Combining \ like \ terms\\\\= 2x^3-x^3-x^2+x^2-4x+2x+2\\\\= x^3-2x +2\\\\Comparing \ it \ with \ \bold{ax^3+bx^2+cx+d}, \ we \ get:\\\\a = 1\\\\b = 0\\\\c = -2\\\\d = 2\\\\\rule[225]{225}{2}`

Hope this helped!

~AH1807

Answer 2

`\sf x^3 -2x+2`

a = 1, b = 0, c = -2, d = 2

Explanation:

• 
  `\sf (2x - 1) (x^2 -2) - x (x^2 - x -2)`

using distributive method:

• 
  `\sf 2x^3-4x-x^2+2 - x (x^2 - x -2)`

expand:

• 
  `\sf 2x^3-4x-x^2+2 - x^3 +x^2 +2x`

group terms:

• 
  `\sf 2x^3-x^3 -x^2 +x^2-4x+2x+2`

final form:

• 
  `\sf x^3 -2x+2`

comparing with
`\sf ax^3 + bx^2 + cx + d` || our input:
`\sf x^3 +0x^2 -2x+2`

we can determine that: a = 1, b = 0, c = -2, d = 2