Question

What are the zeros of g(x) = x(x2 – 25)(x2 – 3x – 4)?

Answer 1

just more and more of these.

⇒ ANSWER : x = 0, x = 5, x = -5, x = 4 and x = -1

EQUATION GIVEN: g(x) = x(x2 – 25)(x2 – 3x – 4)

1) factorize the equation

g(x)=x(x-5)(x+5)(x²-3x-4)

2) expand the equation

g(x)=x(x-5)(x+5)(x²+x-4x-4)

3) factorize

g(x)=x(x-5)(x+5)(x(x+1)-4(x+1))

4) factor out x+1

g(x)=x(x-5)(x+5)(x-4)(x-1)

5) equate to 0

x(x-5)(x+5)(x-4)(x+1)=0

6) solve for x

x = 0, x = 5, x = -5, x = 4, x = -1

Pain.

Answer 2

The zeros are -5, -1, 0, 4, 5.

Explanation:

x(x2 – 25)(x2 – 3x – 4) = 0

x(x - 5)(x + 5)(x + 1)(x - 4) = 0

so either x or the other 4 factors can be 0 :

x = 0, x - 5 = 0, x + 5 = 0, x + 1 = 0 or x - 4= 0, giving:-

x = 0, 5, -5, -1, 4.