How do you factor completely: 4x^3 + 28x^2 + 7x + 49?

Jul 22, 2015

I found: $\left(x + 7\right) \left(4 {x}^{2} + 7\right)$

Explanation:

I can collect $4 {x}^{2}$ between the first two terms and $7$ between the last two to get:
$4 {x}^{2} \left(x + 7\right) + 7 \left(x + 7\right) =$
collect $\left(x + 7\right)$;
$= \left(x + 7\right) \left(4 {x}^{2} + 7\right)$

Jul 22, 2015

$4 {x}^{3} + 28 {x}^{2} + 7 x + 49 = \left(4 {x}^{2} + 7\right) \left(x + 7\right)$

Explanation:

Notice that
$\textcolor{w h i t e}{\text{XXXX}}$$4 {x}^{3} + 28 {x}^{2} = 4 {x}^{2} \left(x + 7\right)$
and that
$\textcolor{w h i t e}{\text{XXXX}}$$7 x + 49 = 7 \left(x + 7\right)$

So, by grouping we can re-write the expression with a common factor of $\left(x + 7\right)$
$\textcolor{w h i t e}{\text{XXXX}}$$4 {x}^{2} \left(x + 7\right) + 7 \left(x + 7\right)$

and then applying the distributive property:
$\textcolor{w h i t e}{\text{XXXX}}$$\left(4 {x}^{2} + 7\right) \left(x + 7\right)$

Checking the discriminant demonstrates that $\left(4 {x}^{2} + 7\right)$ has no Real roots and the solution is complete.