Question

Y raised to 3 subtracted by y raised to 2 subtracted by 6y subtracted by 4 divided by y+1

Answer 1

Answer:

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