How do you get the real-number solutions of 5b^3 + 15b^2 + 12b = -36?
I get to (b+3)(5b^2 + 12) and I don't know what to do after that. Can someone please show me how to finish this problem?
5b^3 + 15b^2 + 12b = -36
5b^3 + 15b^2 + 12b +36 = 0
you factored it correctly to
(b+3)(5b^2 + 12) = 0
that was the hard part, the rest is easy
so b+3 = 0 or b = -3
or
5b^2 + 12 = 0
b^2 = -12/5 ---> no real solution
so the only real solution is b = -3
Thank you :-)
oops wrong question.
To find the real-number solutions of the equation 5b^3 + 15b^2 + 12b = -36, you correctly factored out the common factor of 3 from the left side of the equation, which gave you (b+3)(5b^2 +12). Now to solve for b, you need to set each factor equal to zero and solve for b separately.
1. Set (b+3) equal to zero:
b + 3 = 0
Subtract 3 from both sides:
b = -3
2. Set (5b^2 + 12) equal to zero:
5b^2 + 12 = 0
To solve this quadratic equation, you can use the quadratic formula:
For a quadratic equation in the form ax^2 + bx + c = 0, the quadratic formula is:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
In the equation 5b^2 + 12 = 0, a = 5, b = 0, and c = 12.
Applying the quadratic formula:
b = (-0 ± sqrt(0^2 - 4(5)(12))) / (2(5))
= (± sqrt(-240)) / 10
Since the square root of a negative number results in complex solutions, there are no real-number solutions for this part of the equation.
3. The real-number solutions for the original equation 5b^3 + 15b^2 + 12b = -36 are b = -3.
So, the equation has one real-number solution, which is b = -3.