Question

factor ( x^4 + 14x^2 + 49 ) completely. all factors in your answer should have integer coefficients.

Explanation:

Step1: Recognize the perfect square trinomial

Notice that \(x^4=(x^2)^2\) and \(49 = 7^2\), and \(14x^2=2\times x^2\times7\). So the expression \(x^4 + 14x^2+49\) is in the form of a perfect square trinomial \(a^2 + 2ab + b^2=(a + b)^2\), where \(a=x^2\) and \(b = 7\).
\[x^4+14x^2 + 49=(x^2 + 7)^2\]

Step2: Check if further factoring is possible

Now, we check if \(x^2+7\) can be factored over the integers. The discriminant of the quadratic \(x^2+7\) (in the form \(ax^2+bx + c\) with \(a = 1\), \(b=0\), \(c = 7\)) is \(b^2-4ac=0^2-4\times1\times7=- 28<0\). Since the discriminant is negative, \(x^2 + 7\) cannot be factored into linear factors with integer coefficients. So the complete factorization of \(x^4+14x^2 + 49\) is \((x^2 + 7)^2\).

Answer:

\((x^{2}+7)^{2}\)