Four equivalent forms of a quadratic function are given. Which form displays the zeros of function h?
A.) h(x) = -4(x2 − 4)
B.) h(x) = -4x2 + 16
C.) h(x) = -4(x − 2)(x + 2)
D.) h(x) = -2(2x2 − 8)

The Answer is : C.) h(x) = -4(x − 2)(x + 2)

The picture in the attached figurewe know thatin the right triangle mkjapplying the pythagorean theoremkj²=km²+mj²> kj²=6²+3²> kj²=36+9kj=√45find cos ∠jcos ∠ j=3/√> equation 1∠j+∠ l=90°> by complementary anglessocos ∠ j=sin ∠ lin the right triangle lkjsin ∠ l=√45/> equation 2equate equation 1 and equation 23/√45=√45/> 3*lj=> lj=45/3lj=15 unitsthe answer is  lj=15 units