Question

question: (2 points) which binomial is a factor of x^4 - 4x^2 - 4x + 8? (1) x - 2 (2) x + 2 (3) x - 4 (4) x + 4

Explanation:

Step1: Use the factor - theorem

According to the factor - theorem, if \(x - a\) is a factor of a polynomial \(P(x)\), then \(P(a)=0\). Let \(P(x)=x^{4}-4x^{2}-4x + 8\).

Step2: Test \(x - 2\)

Set \(x = 2\) in \(P(x)\): \(P(2)=2^{4}-4\times2^{2}-4\times2 + 8=16-16 - 8 + 8=0\). Since \(P(2) = 0\), by the factor - theorem, \(x - 2\) is a factor of \(x^{4}-4x^{2}-4x + 8\).

Step3: Test \(x + 2\)

Set \(x=-2\) in \(P(x)\): \(P(-2)=(-2)^{4}-4\times(-2)^{2}-4\times(-2)+8=16 - 16+8 + 8 = 16
eq0\), so \(x + 2\) is not a factor.

Step4: Test \(x - 4\)

Set \(x = 4\) in \(P(x)\): \(P(4)=4^{4}-4\times4^{2}-4\times4 + 8=256-64 - 16+8=184
eq0\), so \(x - 4\) is not a factor.

Step5: Test \(x + 4\)

Set \(x=-4\) in \(P(x)\): \(P(-4)=(-4)^{4}-4\times(-4)^{2}-4\times(-4)+8=256-64 + 16+8=216
eq0\), so \(x + 4\) is not a factor.

Answer:

(1) \(x - 2\)