solve cubic equation?

Question

2y^3 + 5y^2 -22y +15 = 0

try to use a short method but give all steps clearly

Answer 1

You can try and factor this and then set each factor equal to zero. Then solve each little equation for an answer. If you use synthetic division trying "1" as a root, this will work. That leaves 2x^2 + 7x - 15 to factor. This factors as (2x -3)(x+5). Setting each of these factors to "0" gives 2x - 3 =0 and x+5 =0. Solving each of these gives the other 2 roots: x=3/2 and x=-5. So the 3 answers will be x=3/2 and x=-5 and x=1.

Answer 2

Suppose r/s is a rational root of 2y^3 + 5y^2 -22y +15 = 0.
Then we know that r divides 15 and s divides 2.
Trying these we see that 1 is a root, so x-1
is a factor of the left-hand side.
Thus 2y^3 + 5y^2 -22y +15 = 0
yields (y-1)(2y²+7y-15) = 0.
Setting 2y²+7y-15 = 0, we get
(2y-3)(y+5) = 0
with solutions y = -5 and y = 3/2.
So the solutions of the original equation are
y = -5, 1 and 3/2.

Answer 3

Find an integer value of y which is a solution.

y = 1 gives 2*1 + 5*1 - 22*1 + 15 which indeed is zero.

Therefore the equation is (y - 1)*(ay^2 + by + c) = 0 for some values of a, b and c. Multiplying it out and equating the coefficients with a, b and c to the actual coefficients 2, 5, -22 and 15, we find a = 2, b = 7, c = -15, therefore :-

(y - 1) (2y^2 + 7y - 15) = 0

Factoring the quadratic by inspection,

(y - 1) (y + 5) (2y - 3) = 0 (multiply the last 2 factors out to check it)

therefore y = 1, y = -5 or y = 3/2.

Answer 4

If y=1, 2y^3+5y^2-22y+15 =0
Therefore (y-1) is a factor
Dividing 2y^3+5y^2-22y+15 by (y-1) gives
(y-1)(2y^2+7y+15)=0
(y-1)(2y-3)(y+5)=0
For this to be true, either (y-1), (2y-3), or(y+5) must be equal to zero
If y-1=0, y=1
If 2y-3=0, 2y=3 and y=3/2
Ify+5=0, y=-5

Solution set is {1, 3/2, -5}

Answer 5

use synthetic division...
then, get the roots...
i can't explain it how i did it but these is my answers...

2y^3 + 5y^2 -22y +15 = 2 ( 2y - 3 ) ( y + 5 ) ( y - 1)
roots: 3/2, -5 and 1

these roots can help... I think.