Given
[tex]x^3−x^2−4x+4[/tex]EXPLANATION
Part 1: Since the constant in the given equation is 4,the integer root must be a factor of 4. The possible values are
Answer:
[tex]\pm1,\pm2,\pm4[/tex]Part 2: The remainder theorem says when a polynomial a(x) is divided by a linear polynomial b(x) whose zero is x = k, the remainder is given by r = a(k). In this case for one of the possible values to be a root, it must give a zero result when inserted in the original expression.
Therefore,
[tex]\begin{gathered} when\text{ }x=1 \\ f(1)=1^3-(1)^2-4(1)+4=1-1-4+4=0 \\ when\text{ }x=2 \\ f(2)=2^3-(2)^2-4(2)+8=8-4-8+4=0 \end{gathered}[/tex]Answer: Therefore, 1 and 2 are roots of the polynomial
Part 3:
Let the last factor of the polynomial be ax+b
Therefore,
[tex]\begin{gathered} (x-1)(x-2)(ax+b)=x^3-x^2-4x+4 \\ By\text{ comparison of terms} \\ ax^3=x^3 \\ \therefore a=1 \\ Also \\ 2b=4 \\ b=\frac{4}{2}=2 \end{gathered}[/tex]The factors of the equation are therefore;
Answer: (x-1)(x-2)(x+2)