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Find the real zeros and the number of x-intercepts. x^4 + 4x^2 + 4 = f(x)
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\(x^4+4x^2+4\\=x^4+2x^2+2x^2+4\\=x^2(x^2+2)+2(x^2+2)\\=(x^2+2)(x^2+2)\) Now \(f(x) = 0\) means \((x^2+2)(x^2+2)=0\) means \(x^2=-2\) which has no real solutions.
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