Question

Find all the values of x satisfying the given conditions.

f(x)= 5x^3+x^2-80x-4 and f(x)=12
What is the solution set?

Answer 1

I did 5x^3+x^2-80x-4=12-12
5x^3+x^2-80x-4=0
(5x^3+x^2)-(80x-4)=0

Answer 2

if you subtract 12 you should get \(5x^3+x^2-80x-16=0\)

Answer 3

How is it -16?

Answer 4

you were right o:

Answer 5

@genius12

Answer 6

You know this is a third degree function with a positive leading coefficient. So the graph goes from the third quadrant to the first quadrant.

To find the value of x where f(x) = 12, just equate the two sides.

5x^3 + x^2 - 80x - 4 =12
5x^3 + x^2 - 80x - 16 =0 --> Bring 12 over to the other side.
x^2(5x + 1) - 16(5x + 1) = 0 --> 'Distributive Property'. Factored 'x^2' from first two terms and - 16 from the other two.

(x^2 - 16)(5x + 1) = 0
(x - 4)(x + 4)(5x + 1) = 0 --> x = 4, -4, -1/5

Now our polynomial is completely factored, which means we can find the solutions. Therefore, the solutions are: x = 4, -4, -1/5

@ikoreanx3