Question

How do you factor `5x^{2} + 11x + 7`?

Answer 1

Use the discriminant to determine whether or not there are real roots, then use the quadratic formula to solve for those roots

Explanation:

The discriminant is the part of the quadratic formula under the square root

`b^2-4ac`

sub in the terms from the standard form of the equation
ex. `ax^2+bx+c`

so in this case `a=5, b=11, c=7`

If the discriminant is positive there are 2 real roots, if the discriminant equals 0 there is one real root, and if the discriminant is negative there are no real roots

`11^2-4(5)(7)`
`= 121-140`
`=-19`

Since the discriminant is negative, there are no real roots and you cannot factor this equation

Answer 2

`5x^2+11x+7 = 1/20 (10x+11-sqrt(19)i)(10x+11+sqrt(19)i)`

Explanation:

This quadratic has negative discriminant, so can only be factored using Complex coefficients.

`Delta = b^2-4ac = 11^2-4(5)(7) = 121-140 = -19`

Multiply by `20 = 2^2*5`, complete the square, use the difference of squares identity, then divide by `20`...

The difference of squares identity can be written:

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

We use this with `a=10x+11` and `b=sqrt(19)i` as follows:

`20(5x^2+11x+7) = 100x^2+220x+140`

`color(white)(20(5x^2+11x+7)) = (10x)^2+2(11)(10x)+121+19`

`color(white)(20(5x^2+11x+7)) = (10x+11)^2+19`

`color(white)(20(5x^2+11x+7)) = (10x+11)^2-(sqrt(19)i)^2`

`color(white)(20(5x^2+11x+7)) = ((10x+11)-sqrt(19)i)((10x+11)+sqrt(19)i)`

`color(white)(20(5x^2+11x+7)) = (10x+11-sqrt(19)i)(10x+11+sqrt(19)i)`

So:

`5x^2+11x+7 = 1/20 (10x+11-sqrt(19)i)(10x+11+sqrt(19)i)`