chikooo

24.09.2020 •

Mathematics

A Premier League Soccer field dimensions are 115 yards (length) by 74 yards (Width). Find the ratio of the width to the perimeter of the soccer field in simplest fraction form.

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Ответ:

foziasif724

20.03.2020

Mathematics

The expression is $\frac{y^{3}-y^{2}-6y-4}{y+1}=y^{2}-2y-4$

For $y^{2}-2y-4=0$ Factors are $(1+\sqrt{5})$ , $(1-\sqrt{5})$

Therefore $y^{3}-y^{2}-6y-4y=(y+1)(1+\sqrt{5})(1-\sqrt{5})$

Step-by-step explanation:

Given expression can be written as $\frac{y^{3}-y^{2}-6y-4}{y+1}$

By using synthetic division we can find factors of given expression

1 -1 -6 -4

-1 | 0 -1 2 4

| _____________________

1 0 -4 0

Therefore quadratic equation is y^2-2y-4=0

Here a=1 , b=-2 and c=-4

$y=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

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

$=\frac{2\pm \sqrt{4+16}}{2}$

$=\frac{2\pm \sqrt{20}}{2}$

$=\frac{2\pm \sqrt{4\times 5}}{2}$

$=\frac{2\pm 2\sqrt{5}}{2}$

$=\frac{2(1\pm \sqrt{5})}{2}$

$=1\pm \sqrt{5}$

Therefore $y=1+\sqrt{5}$ and $y=1-\sqrt{5}$

The expression is $\frac{y^{3}-y^{2}-6y-4}{y+1}=y^{2}-2y-4$

For $y^{2}-2y-4=0$ Factors are $(1+\sqrt{5})$ , $(1-\sqrt{5})$

Therefore $y^{3}-y^{2}-6y-4y=(y+1)(1+\sqrt{5})(1-\sqrt{5})$

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