Question

Find the zero of y=4x^2-12x-16


HELPP

Answer 1

`\large\boxed{x=-1\ or\ x=4}`

Explanation:

`y=4x^2-12x-16\\\\\text{the zeros are for}\ y=0:\\\\4x^2-12x-16=0\qquad\text{divide both sides by 4}\\\\x^2-3x-4=0\\\\x^2+x-4x-4=0\\\\x(x+1)-4(x+1)=0\\\\(x+1)(x-4)=0\iff x+1=0\ \vee\ x-4=0\\\\x+1=0\qquad\text{subtract 1 from both sides}\\\boxed{x=-1}\\\\x-4=0\qquad\text{add 4 to both sides}\\\boxed{x=4}`

Answer 2

-1, 4

Explanation:

factor out common term 4

4(x^2 - 3x - 4) find factors of -4 1 (-4)= -4 (c term) also 1 - 4 = -3 (b term)

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

zeros are -1, 4