Question

How do you factor `x^2+9x+28`?

Answer 1

`x^2+9x+28 = (x+9/2-sqrt(31)/2i)(x+9/2+sqrt(31)/2i)`

Explanation:

`x^2+9x+28` is of the form `ax^2+bx+c` with `a=1`, `b=9` and `c=28`

This has discriminant `Delta` given by the formula:

`Delta = b^2-4ac = 9^2-(4*1*28) = 81 - 112 = -31`

Since the discriminant is negative, this quadratic has no Real zeros and no linear factors with Real coefficients.

It can be factored using Complex coefficients derived from the zeros, which we can find using the quadratic formula:

`x = (-b+-sqrt(b^2-4ac))/(2a)`

`=(-b +-sqrt(Delta))/(2a)`

`=(-9+-sqrt(-31))/2`

`=(-9+-sqrt(31)i)/2`

That is `x = -9/2+sqrt(31)/2 i` or `x = -9/2-sqrt(31)/2 i`

These zeros translate into factors of the form `(x-r)` where `r` is a zero:

`x^2+9x+28 = (x+9/2-sqrt(31)/2i)(x+9/2+sqrt(31)/2i)`

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Alternative method

Here's an alternative method involving completing the square and using the difference of squares identity, which may be written:

`a^2-b^2 = (a-b)(a+b)`

We use this with `a=(2x+9)` and `b=sqrt(31)i` as follows:

To cut down on the arithmetic involving fractions, multiply our quadratic by `2^2 = 4`, then divide by `4` at the end:

`4(x^2+9x+28)`

`=4x^2+36x+112`

`=(2x)^2+2(2x)(9)+9^2+31`

`=(2x+9)^2+(sqrt(31))^2`

`=(2x+9)^2-(sqrt(31)i)^2`

`=((2x+9)-sqrt(31)i)((2x+9)+sqrt(31)i)`

`=(2x+9-sqrt(31)i)(2x+9+sqrt(31)i)`

Then divide by `4` to find:

`x^2+9x+28`

`=1/4(2x+9-sqrt(31)i)(2x+9+sqrt(31)i)`

`=(x+9/2-sqrt(31)/2 i)(x+9/2+sqrt(31)/2 i)`