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

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Answer 1 · 2015-07-22T14:52:14

I found: $(x+7)(4x^2+7)$

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

Answer 2 · 2015-07-22T14:55:50

$4x^3+28x^2+7x+49 = (4x^2+7)(x+7)$

Notice that
$color(white)("XXXX")$ $4x^3+28x^2 = 4x^2(x+7)$
and that
$color(white)("XXXX")$ $7x+49 = 7(x+7)$

So, by grouping we can re-write the expression with a common factor of $(x+7)$
$color(white)("XXXX")$ $4x^2(x+7) + 7(x+7)$

and then applying the distributive property:
$color(white)("XXXX")$ $(4x^2+7)(x+7)$

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

How do you factor completely: 4x^3 + 28x^2 + 7x + 494x^3 + 28x^2 + 7x + 49?