Question

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

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

Answer:

A. –2, –1, 0, 2

Step-by-step 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)