Question

Factor completely 3x^2+ 12x+ 7
Help please

Answer 1

Single out 3x^2 + 12x and find the gcf

3x( x + 4) + 7
that's the simplest form.

Answer 2

do you know what the determinant is ?
\(b^2-4ac\)
heard of it ?

Answer 3

you mean discriminant? @hartnn

Answer 4

oh, lol yes...discriminant :P

Answer 5

It can't be factored out in (x+ ) (x + )
so the gcf and reduce to simplest form is the best approach i think.

Answer 6

Using the equation \(\left(\dfrac{b}{2} + d\right)\left(\dfrac{b}{2} - d\right)=ac\) we have:
\((6 + d)(6 - d) = 21\)
\(36 - d^2 = 21\)
\(36 - 21 = d^2\)
\(15 = d^2\)
\(\pm\sqrt{15} = d\)\

\((6 + \sqrt{15})(6 - \sqrt{15}) = 21\)
\((6 + \sqrt{15}) + (6 - \sqrt{15}) = 12\)

Now input the last row into the quadratic equation in place of 12:

\(3x^2 + ((6 + \sqrt{15}) + (6 - \sqrt{15}))x + 7\)

\(3x^2 + (6 + \sqrt{15})x + (6 - \sqrt{15})x + 7\)

Factor by grouping:
\(3x^2 + (6 + \sqrt{15})x + (6 - \sqrt{15})x + \dfrac{(6 + \sqrt{15})(6 - \sqrt{15})}{3}\)

\(3x\left(x + \dfrac{6 + \sqrt{15}}{3}\right) + (6 - \sqrt{15})\left(x + \dfrac{6 + \sqrt{15}}{3}\right)\)

\((3x + 6 - \sqrt{15})\left(x + \dfrac{6 + \sqrt{15}}{3}\right)\)

Annoying and abstruse, but it does factor.

Answer 7

i meant it does not factor into integral terms

Answer 8

when b^2-4ac is not a perfect , we cannot factor it as (ax+b)(cx+d)
where, a,b,c,d are integers

Answer 9

It said factor completely so I didn't make any assumptions about factoring limitations. Usually it will say factor if possible.

Answer 10

ohh

Answer 11

That being said, the user probably revised the original question given to him or her.

Answer 12

It is prime