How do you factor $4 x^{2} + 9$ $4x^2 + 9$ ?

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Answer 1 · 2017-01-15T21:33:19

There are no simpler factors with Real coefficients, but:

$4 x^{2} + 9 = (2 x - 3 i) (2 x + 3 i)$ $4x^2+9 = (2x-3i)(2x+3i)$

Note that $x^{2} \geq 0$ $x^2 >= 0$ for any Real value of $x$ $x$ .

Hence $4 x^{2} + 9 \geq 9$ $4x^2+9 >= 9$ for all Real values of $x$ $x$ .

So $4 x^{2} + 9$ $4x^2+9$ has no linear factors with Real coefficients.

It is possible to factor it with Complex coefficients.

The difference of squares identity can be written:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2 = (a-b)(a+b)$

In conjunction with $i^{2} = - 1$ $i^2=-1$ , we find:

$4 x^{2} + 9 = {(2 x)}^{2} - {(3 i)}^{2} = (2 x - 3 i) (2 x + 3 i)$ $4x^2+9 = (2x)^2-(3i)^2 = (2x-3i)(2x+3i)$

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