Answered By Jonny C

2y⁴+y³-10y-5

This equation can be solved by a technique called grouping.

Step 1. Group the first two terms and the last two terms. I've bolded the first two terms and the last two terms.

2y⁴+y³-10y-5

Step 2. Factor out any common denominators within the two group terms. In this case the bolded terms will be 2y³ and the italicized terms will be -5. Remember once you factor out a negative, all the of the terms become positive!

y³(2y+1)-5(2y+1)

Step 3. Now you have two common binomials in (2y+1), so they can rearragned as a multiplication espression.

(y³-5)(2y+1)

Now this part most students become lost. How did step 2's equation turn into step 3's equation? Well If you really look at it, step 3's equation is essentially saying the same thing as step 2's equation. If you FOIL this expression it will go back to step 2, so we were basically just working backwards from what you've learned about foiling.

Step 4. Now we solve for the roots of this equation. If either binomial in the equation equals zero, the entire question equals zero. So we just need to solve for the zeros of each individual binomial.

y³-5=0

y³=5

y=1.71 (you'd have to use cuberoot in your calculator)

Now for the second binomial:

2y+1=0

2y=-1

y=-1/2=-0.5

And there's your answer! Remember if you see 4 variables in the equation with higher powers, grouping is your only method of solving it.

Answered By Majid B

 $2y^4+y^3-10y-5=0$2y⁴+y³−10y−5=0        =>       $\left(2y^4-10y\right)+\left(y^3-5\right)=0$(2y⁴−10y)+(y³−5)=0        =>      $2y\left(y^3-5\right)+\left(y^3-5\right)=0$2y(y³−5)+(y³−5)=0       =>

$\left(y^3-5\right)\left(2y+1\right)=0$‍(y³−5)(2y+1)=0           =>        $y^3-5=0$y³−5=0    OR     $2y+1=0$2y+1=0    =>       $y^3=5$y³=5     OR     $2y=-1$2y=−1        =>

 $y=\sqrt[3]{5}$‍y=³√5   OR    $y=-\frac{1}{2}$y=−12      =>        $y=1.71$y=1.71       OR     $y=-0.50$y=−0.50