Answer:
The solutions are 0 , 5/2 and -3
Explanation:
The given expression is:
2x³ + x² - 15x
We can note that we can take x as a common factor from all terms. This will give us:
x(2x²+x-15)
Now, the expression 2x² + x - 15 is a second degree polynomial that can be factored using the quadratic formula shown in the attached image.
In the given expression:
a = 2
b = 1
c = -15
Substituting with the values of a, b and c in the formula, we would find that
2x² + x - 15 can be factored as (2x-5)(x+3)
Based on the above, the factored form of the given expression is:
x(2x-5)(x+3)
Now, to find the solutions means to find the values of x that would make the expression equal to zero.
This means that:
x(2x-5)(x+3) = 0
either x = 0
or 2x-5=0 .............> 2x = 5 ...........> x = 5/2
or x+3=0 ............> x = -3
Based on the above, the solutions of the system are:
0 , 5/2 and -3
Hope this helps :)
The solutions are 0 , 5/2 and -3
Explanation:
The given expression is:
2x³ + x² - 15x
We can note that we can take x as a common factor from all terms. This will give us:
x(2x²+x-15)
Now, the expression 2x² + x - 15 is a second degree polynomial that can be factored using the quadratic formula shown in the attached image.
In the given expression:
a = 2
b = 1
c = -15
Substituting with the values of a, b and c in the formula, we would find that
2x² + x - 15 can be factored as (2x-5)(x+3)
Based on the above, the factored form of the given expression is:
x(2x-5)(x+3)
Now, to find the solutions means to find the values of x that would make the expression equal to zero.
This means that:
x(2x-5)(x+3) = 0
either x = 0
or 2x-5=0 .............> 2x = 5 ...........> x = 5/2
or x+3=0 ............> x = -3
Based on the above, the solutions of the system are:
0 , 5/2 and -3
Hope this helps :)
