How do you multiply (2x^2+8)div(x^4-16)/(x^2+x-6)?

Dec 16, 2017

$\frac{x + 2}{2 \left(x + 3\right)}$

Explanation:

Take into account the formula:
${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$

${x}^{4} - 16 = {x}^{4} - {4}^{2} = \left({x}^{2} - 4\right) \left({x}^{2} + 4\right)$

(2x^2+8)÷(x^4−16)/(x^2+x−6)=2(x^2+4)÷((x^2-4)(x^2+4))/(x^2+x−6)

1/(2cancel((x^2+4)))*((x^2-4)cancel((x^2+4)))/(x^2+x−6)

We can apply this formula one more for $\left({x}^{2} - 4\right)$
We can also do something with (x^2+x−6)
formula for quadratic function: $a {x}^{2} + b x + c \quad \text{or} \quad \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$
where ${x}_{1}$ and ${x}_{2}$ are roots if it's an equation
We can use the following method only when a=1. In this particular example b=1 and c=-6
in order to get solution we have to: $b = {x}_{1} + {x}_{2} \quad \mathmr{and} \quad c = {x}_{1} \cdot {x}_{2}$
The only way how to get 6 is by multiplying 61 or 32 then we have to resolve signs from the second equation

There is no way we can arrange 6 and 1 to get 1. so the solution is 3 and 2 and it's:
(x^2+x−6)=(x+3)(x-2)

$1 = 3 + \left(- 2\right) \quad \mathmr{and} \quad - 6 = 3 \cdot \left(- 2\right)$ it's correct

Now we can rewrite:

1/(2)*((x+2)(x-2))/((x+3)(x−2))=((x+2)cancel((x-2)))/(2(x+3)cancel((x−2)))=(x+2)/(2(x+3))