Question

Drag each expression to the correct location on the model. Not all will be used

Answer 1

Answer:
`\frac{x^(2)+2x+1 }{\mathbf {x-1}} \cdot \frac{\mathbf {5x^(2) +15x-20} }{7x^(2) +7x}`

Explanation:

`((5x^(2) +25x+20)/(7x) )/((x^(2) +2x+1)/(7x^(2) +7x) ) =(5x^(2) +25x+20)/(7x) * (7x^(2) +7x)/(x^(2) +2x+1)=(5x^(2) +25x+20)/(7x) * (7x(x +1))/((x+1)^(2) )=\\\\=(5x^(2) +25x+20)/(x+1) =\frac{{5(x+1)(x+4)}}{x+1}=5(x+4)`

`5(x+4)=(5(x-1) \cdot (x+4))/(x-1)=(5x^(2) +15x-20)/(x-1)`

Thus, the expressions will be used: (5x² + 15x - 20) and (x + 4).

Let's check:

`\frac{x^(2)+2x+1 }{\mathbf {x-1}} \cdot \frac{\mathbf {5x^(2) +15x-20} }{7x^(2) +7x} =((x+1)^(2) )/(x-1) \cdot (5(x-1) \cdot (x+4))/(7x(x+1)) =\\\\=((x+1) \cdot 5 \cdot (x+4))/(7x) =(5x^(2) +25x+20)/(7x)`