what are the roots of the polynomial equation $x^{4}+x^{3}=4x^{2}+4x$? use a graphing calculator and a…

Explanation:

Step1: Rearrange the equation

First, rewrite the equation $x^{4}+x^{3}=4x^{2}+4x$ as $x^{4}+x^{3}-4x^{2}-4x = 0$.

Step2: Factor out the common factor

Factor out $x$ from the left - hand side: $x(x^{3}+x^{2}-4x - 4)=0$.

Step3: Group - factor the cubic part

Group the terms in the cubic expression: $x\left[x^{2}(x + 1)-4(x + 1)\right]=0$. Then factor out $(x + 1)$ to get $x(x + 1)(x^{2}-4)=0$.

Step4: Factor the difference of squares

Since $x^{2}-4=(x + 2)(x - 2)$ by the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$ (here $a=x$ and $b = 2$), the factored form of the equation is $x(x + 1)(x + 2)(x - 2)=0$.

Step5: Find the roots

Set each factor equal to zero: If $x=0$, then it is a root. If $x + 1=0$, then $x=-1$. If $x + 2=0$, then $x=-2$. If $x - 2=0$, then $x=2$.

Answer:

-2, -1, 0, 2