Quadratic Help

Question

How would you find the zeros of the following  a ) -3x^2-18x b)-2x^2-12x-16 c) 2x^2-162

Answer 1

(a) - 3x²-18x = -3x(x+6)

      the zeros are x = 0 and - 6

(b)  -2x²-12x-16 = -2(x²+6x+8) = -2(x+2)(x+4)

      the zeros are x = -2 and -4

(c)  2x²-162 = 2(x²-81)

      the zeros are x = +9 and -9

 

Answer 2

The zeros are where f(x) = 0.

a)    -3x² - 18x = 0

        -3x(x + 6) = 0    factor

     -3x = 0,    x+6 = 0     apply the zero product rule

              x = 0, -6

 

b)  -2x² -12x - 16 = 0

       x² + 6x + 8 = 0         divide entire equation by greatest common factor (-2)

    (x = 2)(x + 4) = 0         factor

   x = -2, -4                      zero product rule

 

c)  2x² - 162 = 0

     2x² = 162

      x² = 81

      x = +/- 9