(x+3)(2x+1)(x+4) = ?? can u solve that?

Let's first multiply (x+3) and (2x+1)

Use FOIL method

Note: FOIL as in Front, Outward, Inward, and Last terms

(x+3)(2x+1) = 2x^2+x+6x+3 = 2x^2+7x+3

Then, multiply (2x^2+7x+3) to (x+4)

(2x^2+7x+3)(x+4)

= 2x^3+8x^2+7x^2+28x+3x+12

= 2x^3+15x^2+31x+12 ---------------------> The final answer.

(2x^2 + x +6x + 3) ( x + 3)

2x^3 + 6x^2 + 6x^2 + 18x + 3x + 9

2x^3 + 12x^2 + 21x + 9

Note than ^ means to the power of

(x + 3)(2x + 1)(x + 4)
= (x*2x + 3*2x + x*1 + 3*1)(x + 4)
= (2x^2 + 6x + x + 3)(x + 4)
= (2x^2 + 7x + 3)(x + 4)
= 2x^2*x + 7x*x + 3*x + 2x^2*4 + 7x*4 + 3*4
= 2x^3 + 7x^2 + 3x + 8x^2 + 28x + 12
= 2x^3 + 7x^2 + 8x^2 + 3x + 28x + 12
= 2x^3 + 15x^2 + 31x + 12

Man, there are a lot of wrong answers. Here is the correct answer.

2x^3 + 15x^2 + 31x + 12

Note, the last number has to be an even number because there is an even whole number in the last spot in the equation.

(x+3)(2x+1)(x+4) = (2x^2 + 8x + x + 4)(x + 3)
= (2x^2 + 9x + 4)(x + 3)
= (2x^3 + 6x^2 + 9x^2 + 27x + 4x + 12)
= 2x^3 + 15x^2 + 31x +12

note: cannot find x without knowing right hand side of equation, otherwise it is just a mathematical expression

2x3+18x2+40

2xe3+15 xe2+31x+12=??(y)

X=2