Question

How do you factor `2x^2+x+3`?

Answer 1

`2x^2+x+3 = 1/8(4x+1-sqrt(23)i)(4x+1+sqrt(23)i)`

`color(white)(2x^2+x+3) = 2(x+1/4-sqrt(23)/4 i)(x+1/4+sqrt(23)/4 i)`

Explanation:

Given:

`2x^2+x+3`

Note that this is in standard quadratic form:

`ax^2+bx+c`

with `a=2`, `b=1` and `c=3`.

It has discriminant `Delta` given by the formula:

`Delta = b^2-4ac = color(blue)(1)^2-4(color(blue)(2))(color(blue)(3)) = -23`

Since `Delta < 0` this quadratic has no real zeros and no linear factors with real coefficients.

We can factor it if we use complex coefficients.

We find:

`8(2x^2+x+3) = 16x^2+8x+24`

`color(white)(8(2x^2+x+3)) = (4x)^2+2(4x)+1+23`

`color(white)(8(2x^2+x+3)) = (4x+1)^2+(sqrt(23))^2`

`color(white)(8(2x^2+x+3)) = (4x+1)^2-(sqrt(23)i)^2`

`color(white)(8(2x^2+x+3)) = ((4x+1)-sqrt(23)i)((4x+1)+sqrt(23)i)`

`color(white)(8(2x^2+x+3)) = (4x+1-sqrt(23)i)(4x+1+sqrt(23)i)`

So:

`2x^2+x+3 = 1/8(4x+1-sqrt(23)i)(4x+1+sqrt(23)i)`

`color(white)(2x^2+x+3) = 2(x+1/4-sqrt(23)/4 i)(x+1/4+sqrt(23)/4 i)`

Answer 2

This can't be factored. (Considering you did not learn imaginary numbers) If you did, look at the answer answered by George C.

Explanation:

(Please remember that I am only dealing with real numbers in this explanation. The solution with imaginary and complex numbers is answered by George C.) Before you start factoring, it is important to know whether it is possible or not. Usually, this type of equation is solved by factoring by grouping.

You do this by multiplying `a` by `c`. `2xx3=6.`

Now, we need to find any factors of 6 that add up to 1(the `b`).

Here are the factors: `1xx6=>1+6= 7`, `2xx3=>2+3= 5`, `-1xx-6=>-1+(-6)= -7`, `-2xx-3=>-2+(-3)= -5`.

Notice that none of the sums equal 6.

For this problem, you could have used the quadratic formula, `(-b+-sqrt(b^2-4(a)(c)))/(2(a))`

By plugging in the values, we get `(-1+-sqrt(1^2-4(2)(3)))/(2(2))` or `(-1+-sqrt(-23))/4` which is a complex number.

Since the answer is a complex number, it is likely that this expression was made to throw you off.