Question

Find the zeros of this polynomial:

p(x)= (x²+4x+3)(x²-4)

Answer 1

Answer: -3, -2, -1, 2

====================================================

Step-by-step explanation:

Set each factor equal to zero and solve for x. I'm using the zero product property which says A*B = 0 leads to either A = 0 or B = 0.

Let's apply that idea to the first factor.

x^2+4x+3 = 0

(x+1)(x+3) = 0

x+1 = 0 or x+3 = 0

x = -1 or x = -3 are two roots

-------------

Do the same for the other factor as well.

x^2 - 4 = 0

(x-2)(x+2) = 0 .... difference of squares rule

x-2 = 0 or x+2 = 0

x = 2 or x = -2 are another two roots

--------------

In total, the four roots are:

• x = -3
• x = -2
• x = -1
• x = 2

Side note: The term "root" is the same as the zeros of a polynomial. Visually, this corresponds to the x intercepts. Plugging any of those four items into p(x) will lead to p(x) = 0.