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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.

Respuesta :

fieryanswererft

Answer:

[tex]\sf x^3 -2x+2[/tex]

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

Explanation:

  • [tex]\sf (2x - 1) (x^2 -2) - x (x^2 - x -2)[/tex]

using distributive method:

  • [tex]\sf 2x^3-4x-x^2+2 - x (x^2 - x -2)[/tex]

expand:

  • [tex]\sf 2x^3-4x-x^2+2 - x^3 +x^2 +2x[/tex]

group terms:

  • [tex]\sf 2x^3-x^3 -x^2 +x^2-4x+2x+2[/tex]

final form:

  • [tex]\sf x^3 -2x+2[/tex]

comparing with  [tex]\sf ax^3 + bx^2 + cx + d[/tex]  ||   our input: [tex]\sf x^3 +0x^2 -2x+2[/tex]

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

ujalakhan18

Answer:

a = 1

b = 0

c = -2

d = 2

Step-by-step explanation:

Given polynomial:

[tex]\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}[/tex]

Hope this helped!

~AH1807

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