Question

C^2 + 5c =3

Tell if this factors or not? And show how it factors or does not factor? And if it factors what are the roots?

Answer 1

The answer to your questions is that it depends on how we view the polynomial. In particular,



`c^2+5c=3\implies c^2+5c-3=0`

If the left hand side were factorizable, then we would be able to write it in the form


`(c-r_1)(c-r_2)=c^2+5c-3=0`

If we expand the leftmost expression, we'd get


`c^2-(r_1+r_2)c+r_1r_2=c^2+5c-3=0`

and so for the two polynomials to be the same, the coefficients must match. In other words, the unknowns
`r_1,r_2` would have to satisfy


`\begin{cases}-(r_1+r_2)=5\\r_1r_2=-3\end{cases}`

Suppose that, moreover, we want integer solutions for
`r_1,r_2`. For this to happen, they must be factor pairs of the constant term.

-3 only has two factor pairs. Either
`r_1=-1` and
`r_2=3`, or
`r_1=1` and
`r_2=-3`. In the first case, we'd get a linear coefficient of
`-(-1+3)=-2\\eq5`, while in the second, we'd get
`-(1-3)=2\\eq5`.

There is no integer solution for this system, so the original quadratic is not factorizable - but only so over the integers.

If we change the scope of the coefficients, i.e. allow for any real numbers/complex numbers to appear in the factorization, then we always factorize a quadratic. The above system is easy to solve.


`r_1r_2=-3\implies r_2=-\frac3{r_1}`

`\implies-\left(r_1-\frac3{r_1}\right)=5`

`\implies{r_1}^2+5r_1-3=0`

`\implies r_1=\frac{-5\pm√(37)}2`


`\implies c^2+5c-3=\left(c+\frac{5-√(37)}2\right)\left(c+\frac{5+√(37)}2\right)`

so the original quadratic is factorizable over the reals.