Question

How do you factor `25v ^ { 2} + 29v + 9`?

Answer 1

`25v^2+29v+9=(5v+29/10+(isqrt59)/10)(5v+29/10-(isqrt59)/10)`

Explanation:

`25v^2+29v+9`

= `(5v)^2+2xx5vxx29/10+(29/10)^2-(29/10)^2+9`

= `(5v+29/10)^2-841/100+9`

= `(5v+29/10)^2+59/100`

= `(5v+29/10)^2-(59/100)xxi^2` - as `i^2=-1`

= `(5v+29/10)^2-((isqrt59)/10)^2`

= `(5v+29/10+(isqrt59)/10)(5v+29/10-(isqrt59)/10)`

Answer 2

This quadratic cannot be factored using Real coefficients.

If you allow Complex coefficients then you find:

`(25x^2+29v+9) = (5x+29/10-sqrt(59)/10i)(5x+29/10+sqrt(59)/10i)`

Explanation:

`25v^2+29v+9`

is in the form

`av^2+bv+c`

with `a=25`, `b=29` and `c=9`

This has discriminant `Delta` given by the formula:

`Delta = b^2-4ac = 29^2-4(25)(9) = 841-900 = -59`

Since `Delta < 0` this quadratic cannot be factored using Real coefficients.

If you still want to factor it, it can be done using Complex coefficients.

I will complete the square then use the difference of squares identity:

`A^2-B^2 = (A-B)(A+B)`

using `A=(50x+29)` and `B=sqrt(Delta)=sqrt(-59)=sqrt(59)i`

To avoid some arithmetic requiring fractions, multiply the given quadratic by `100 = 2^2*5^2`, then divide by `100` at the end:

`100(25x^2+29v+9) = 2500x^2+2900v+900`

`color(white)(100(25x^2+29v+9)) = (50x)^2+2(50x)(29)+29^2+59`

`color(white)(100(25x^2+29v+9)) = (50x+29)^2-(sqrt(59)i)^2`

`color(white)(100(25x^2+29v+9)) = ((50x+29)-sqrt(59)i)((50x+29)+sqrt(59)i)`

`color(white)(100(25x^2+29v+9)) = (50x+29-sqrt(59)i)(50x+29+sqrt(59)i)`

So:

`25x^2+29v+9 = 1/100(50x+29-sqrt(59)i)(50x+29+sqrt(59)i)`

`color(white)(25x^2+29v+9) = (5x+29/10-sqrt(59)/10i)(5x+29/10+sqrt(59)/10i)`