Question

What are the roots of the polynomial equation x4+x3=4x2+4x? Use a graphing calculator and a system of equations.

A.)–2, –1, 0, 2
B.)–2, 0, 1, 2
C.)–1, 0
D.)0, 1

Answer 1

`x^4+x^3=4x^2+4x\\\\x^4+x^3-4x^2-4x=0\\\\x\Big[x^3+x^2-4x-4\Big]=0\\\\ x\Big[x^2\big(x+1\big)-4\big(x+1\big)\Big]=0\\\\x\Big[\big(x+1\big)\big(x^2-4\big)\Big]=0\\\\x\big(x+1\big)\big(x^2-4\big)=0\\\\ x\big(x+1\big)\big(x+2\big)\big(x-2\big)=0\\\\\\x=0\quad\vee\quad x+1=0\quad\vee\quad x+2=0\quad\vee\quad x-2=0\\\\ \boxed{x=0\quad\vee\quad x=-1\quad\vee\quad x=-2\quad\vee\quad x=2}`

Answer A.

Answer 2

A. –2, –1, 0, 2

Explanation:

Given :
`x^(4) +x^(3) = 4x^(2) +4x`

Solution:

`x^(4) +x^(3) -4x^(2) -4x=0`

`x(x^(3) +x^(2) -4x-4)` (taking x common from the equation)

`x[x^(2) (x+1) - 4(x+1)]=0` (taking x+1 common in thye square bracket )

then,
`x[(x+1)(x^(2)-4)] =0`

then ,
`(x)(x+1)(x^(2)-4) =0`

`(x)(x+1)(x+2)(x-2)`

⇒
`x = 0` ,
`x+1 =0` ,
`x+2 =0` ,
`x-2=0`

⇒
`x=0` ,
`x= -1` ,
`x = -2` ,
`x=2`

So, the roots of the polynomial equation
`x^(4) +x^(3) = 4x^(2) +4x` are –2, –1, 0, 2 (option A)