What is the standard form of $y = (x + x^{2}) (6 x - 3) - {(2 x + 2)}^{3}$ $y= (x+x^2)(6x-3) -(2x+2)^3$ ?

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Answer 1 · 2016-01-31T11:54:39

See explanation...

$y = (x + x^{2}) (6 x - 3) - {(2 x + 2)}^{3}$ $y=(x+x^2)(6x-3)-(2x+2)^3$

Multiply $x + x^{2}$ $x+x^2$ and $6 x - 3$ $6x-3$ using Foil method

So,
$(x + x^{2}) (6 x - 3) = 6 x^{2} - 3 x + 6 x^{3} - 3 x^{2} = 3 x^{2} - 3 x + 6 x^{3}$ $(x+x^2)(6x-3)=6x^2-3x+6x^3-3x^2=3x^2-3x+6x^3$

To,simplify ${(2 x + 2)}^{3}$ $(2x+2)^3$ Use the formula(Binomial expansion) $a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$ $a^3+3a^2b+3ab^2+b^3$

${(2 x + 2)}^{3} = 8 x^{3} + 24 x^{2} + 24 x + 8$ $(2x+2)^3=8x^3+24x^2+24x+8$

Watch this video to now about the binomial expansion:

So,

$y = (3 x^{2} - 3 x + 6 x^{3}) - (8 x^{3} + 24 x^{2} + 24 x + 8)$ $y=(3x^2-3x+6x^3)-(8x^3+24x^2+24x+8)$

Change the signs,

$\to y = 3 x^{2} - 3 x + 6 x^{3} - 8 x^{3} - 24 x^{2} - 24 x - 8$ $rarry=3x^2-3x+6x^3-8x^3-24x^2-24x-8$

$\to y = - 21 x^{2} - 3 x + 6 x^{3} - 8 x^{3} - 24 x - 8$ $rarry=-21x^2-3x+6x^3-8x^3-24x-8$

$\to y = - 21 x^{2} - 27 x + 6 x^{3} - 8 x^{3} - 8$ $rarry=-21x^2-27x+6x^3-8x^3-8$

$\to y = - 21 x^{2} - 27 x - 2 x^{3} - 8$ $rarry=-21x^2-27x-2x^3-8$

In Standard form:

$\to y = - 2 x^{3} - 21 x^{2} - 27 x - 8$ $rarry=-2x^3-21x^2-27x-8$

What is the standard form of y= (x+x^2)(6x-3) -(2x+2)^3y=(x+x2)(6x−3)−(2x+2)3y= (x+x^2)(6x-3) -(2x+2)^3?