Find all real zeros for each equation.

Question

1. f(x)=x^3-3x^2-4x

2. f(x)= x^6-x^2

3. f(x)=x^3+3x^2-16x-48

4. f(x)=9x^4-37x^2+4

Answer 1

I will do one, then you try

x (x^2 - 3 x - 4) = 0

x (x-4)(x+1) = 0

so x = 0 , 4 , -1

Answer 2

1. take out common factor of x, the rest factors
2. factor out x^2, then difference of squares.
one of them goes again.

3. x^3 + 3x^2 - 16x - 48
looks like "grouping" works here
= x^2(x+3) - 16(x+3)
= (x+3)(x^2- 16) , and a difference of squares
=(x+3)(x+4)(x-4)

4. sneaky one ...

9x^4-37x^2+4
= (x^2 - 1)(x^2 - 4)
= ..... can you finish it ?

Answer 3

last one should be

4. sneaky one ...

9x^4-37x^2+4
= (9x^2 - 1)(x^2 - 4)
= ..... can you finish it ?