Question

X^3+6x^2+12x+16 (factorise this eq by using the formula (a+b)^3

Answer 1

This can factor if there are rational roots. So we start with looking at what is a possible rational root, which is a factor of the constant term over a constant of the high-order coefficient. Since that is 1, all we have to worry about is the constant of 16. So factors are: 

±1, ±2, ±4, ±8, ±16 

Since everything is addition to other terms, the root would have to be negative, so we'll start testing negative values from the above list to try to find a root. I'll start with -1 and -2: 

x³ + 6x² + 12x + 16 

(-1)³ + 6(-1)² + 12(-1) + 16 and (-2)³ + 6(-2)² + 12(-2) + 16 

-1 + 6(1) - 12 + 16 and -8 + 6(4) - 24 + 16 

-1 + 6 + 4 and -8 + 24 - 8 

9 and 8 

Neither are roots, so I'll test -4 next: 

x³ + 6x² + 12x + 16 

-64 + 6(16) - 48 + 16 

-64 + 96 - 32 

0 

there is our zero. Now that we know the root is -4, the factor is (x + 4). Now that we have a factor, we can divide that by the cubic above to get a resulting quadratic, then it will be easy to find the remaining factor, if one exists: 

. . . . _x²_+_2x_+_4_____ 

x + 4 ) x³ + 6x² + 12x + 16 

. . . . . x³ + 4x² 

. . . . ------------ 

. . . . . . . . 2x² + 12x + 16 

. . . . . . . . 2x² + 8x 

. . . . . . . --------------- 

. . . . . . . . . . . . . . 4x + 16 

. . . . . . . . . . . . . . 4x + 16 

. . . . . . . . . . . . . ------------ 

. . . . . . . . . . . . . . . . . . . 0 

So now we have this factored to be: 

(x + 4)(x² + 2x + 4) 

What's left cannot be factored as the zeroes are non-real. So this is as factored as you can get

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