Question

Factor completely x^3 + 6x^2 + 8x + 48

Answer 1

Answer:

The factors are (x+6),$x-2i\sqrt{2}$ and $x+2i\sqrt{2}$

Therefore the given polynomial

$x^3+6x^2+8x+48=(x+6)(x-2i\sqrt{2})(x+2i\sqrt{2})$

Step-by-step explanation:

Given polynomial is $x^3+6x^2+8x+48$

To factorise the polynomial equate the given polynomial to zero

$x^3+6x^2+8x+48=0$

By using synthetic division we can find the factors :

$x^3+6x^2+8x+48=0$

_-6|   1       6     8     48

        0     -6     0    -48

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        1       0      8      0

Therefore (x+6) is a factor

The quadratic equation is $x^2+8=0$

To solve it we can use $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Here a=1 b=0 and c=8

$x=\frac{-(0)\pm\sqrt{0^2-4(1)(8)}}{2(1)}$

$=\frac{\pm\sqrt{-32}}{2}$

$=\frac{\pm\sqrt{32i^2}}{2}$ where $i^2=-1$

$=\frac{\pm4i\sqrt{2}}{2}$

$=\pm2i\sqrt{2}$

$x=\pm2i\sqrt{2}$

Therefore $x=2i\sqrt{2}$ and $x=-2i\sqrt{2}$

Therefore the factors are (x+6),$x-2i\sqrt{2}$ and $x+2i\sqrt{2}$

Therefore the given polynomial

$x^3+6x^2+8x+48=(x+6)(x-2i\sqrt{2})(x+2i\sqrt{2})$