one factor of $f(x)=4x^{3}-4x^{2}-16x + 16$ is $(x - 2)$. what are all the roots of the function? use the…

one factor of $f(x)=4x^{3}-4x^{2}-16x + 16$ is $(x - 2)$. what are all the roots of the function? use the remainder theorem.\n$x = 1,x = 2$, or $x = 4$\n$x=-2,x = 1$, or $x = 2$\n$x = 2,x = 4$, or $x = 16$\n$x=-16,x = 2$, or $x = 16$

one factor of $f(x)=4x^{3}-4x^{2}-16x + 16$ is $(x - 2)$. what are all the roots of the function? use the remainder theorem.\n$x = 1,x = 2$, or $x = 4$\n$x=-2,x = 1$, or $x = 2$\n$x = 2,x = 4$, or $x = 16$\n$x=-16,x = 2$, or $x = 16$

Answer

Explanation:

Step1: Use polynomial long - division

Divide (f(x)=4x^{3}-4x^{2}-16x + 16) by ((x - 2)). [ \begin{align*} \frac{4x^{3}-4x^{2}-16x + 16}{x - 2}&=4x^{2}+4x - 8 \end{align*} ]

Step2: Factor the quotient

Factor (4x^{2}+4x - 8). First, factor out the GCF 4: (4(x^{2}+x - 2)). Then factor the quadratic (x^{2}+x - 2=(x + 2)(x - 1)). So (4x^{2}+4x - 8=4(x + 2)(x - 1)).

Step3: Find the roots

Set (f(x)=(x - 2)(4x^{2}+4x - 8)=(x - 2)\times4(x + 2)(x - 1)=0). Using the zero - product property, (x-2 = 0) gives (x = 2), (x+2 = 0) gives (x=-2), and (x - 1=0) gives (x = 1).

Answer:

B. (x=-2,x = 1,\text{ or }x = 2)