24y*4+4y*3−8y*2=0 Solve for Y

Question

The numbers after the stars are exponents. Select all.

Question 3 options:

y=−32 

y=−23

y=0 

y=12 

y=23 

y=32

Answer 1

If it looks like a quadratic but the powers are too high, see if you can pull out any factors first.  

4 goes into all terms and y² goes into all terms, so pull that out first.

 

4y² (6y² + y - 2)=0

You can then factor inside the parentheses as you would with a normal quadratic.

 

Look at the first and last numbers 6 and -2.  Multiply together -12. Are there factors of -12 that when combined together would give you the middle number (1).  +4  and -3.  

Using the 6 and -2 or factors of those numbers, set it up where your Outer numbers will multiply to 4 and your Inner numbers will make -3. 

(2y-1)(3y+2)

 

 

Since getting a 0 answer in multiplication means that at least one term has to be equal to 0,  Set each piece = to 0 and solve for y.

2y-1=0

3y+2=0

Don't forget that you have that extra factor you pulled out at the beginning.  Your right answer might come from that factor!

4y²=0

 

 

Answer 2

24y^4+4y^3-8y^2=0

6y^4+y^3-2y^2=0

y^2(6y^2+y-2)=0

y^2(3y+2)(2y-1)=0

y^2=0

y=0

3y+2=0

3y=-2

y=-2/3

2y-1=0

2y=1

y=1/2

y=0 (double root), y=-2/3, y=1/2