What is the standard form of $y = 3 x {(x + 2)}^{2} - x^{3} + {(x + 4)}^{2}$ $y= 3x(x+2)^2-x^3+(x+4)^2$ ?

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Answer 1 · 2018-01-21T10:54:54

$2 x^{3} + 13 x^{2} + 20 x + 16$ $2x^3+13x^2+20x+16$

We need to expand a bunch:

${(x + 2)}^{2} = x^{2} + 4 x + 4$ $(x+2)^2 = x^2 + 4x+4$

${(x + 4)}^{2} = x^{2} + 8 x + 16$ $(x+4)^2=x^2 + 8x + 16$

Substituting these:

$3 x (x^{2} + 4 x + 4) - x^{3} + x^{2} + 8 x + 16$ $3x(x^2 + 4x+4)-x^3+x^2 + 8x + 16$

distributing:

$3 x^{3} + 12 x^{2} + 12 x - x^{3} + x^{2} + 8 x + 16$ $3x^3 + 12x^2+12x-x^3+x^2 + 8x + 16$

collecting like terms:

$2 x^{3} + 13 x^{2} + 20 x + 16$ $2x^3+13x^2+20x+16$

What is the standard form of y= 3x(x+2)^2-x^3+(x+4)^2 y=3x(x+2)2−x3+(x+4)2 y= 3x(x+2)^2-x^3+(x+4)^2 ?