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How would you factor 4x^2 + 9? I think the answer has imaginary numbers in it, but how would you get the solution?
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Let \( 4x^2+9=0\) then \(4x^2=-9\) so that \(x=\pm 3i/2\). Which shows that \((x-3i/2)(x+3i/2)=0\) or \( (2x-3i)(2x+3i)=0\). So, the factors of \(4x^2+9\) are \((2x-3i)(2x+3i)\).
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