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.