Question

Which functions have real zeros at 1 and 4? Check all that apply.

f(x) = x2 + x + 4
f(x) = x2 – 5x + 4
f(x) = x2 + 3x – 4
f(x) = –2x2 + 10x – 8
f(x) = –4x2 – 16x – 1

Answer 1

`f(x)=x^2+x+4=0\ \text{NO SOLUTIONS}\ :(\\\\f(x)=x^2-5x+4=x^2-4x-x+4=x(x-4)-1(x-4)\\=(x-4)(x-1)=0\iff x=4\ \vee\ x=1\ \text{CORRECT}\\\\f(x)=x^2+3x-4=x^2+4x-x-4=x(x+4)-1(x+4)\\=(x+4)(x-1)=0\iff x=-4\ \vee\ x=1\ :(`


`f(x)=-2x^2+10x-8=-2(x^2-5x+4)=-2(x^2-4x-x+4)=\\-2[x(x-4)-1(x-4)]=-2(x-4)(x-1)=0\iff x=4\ \vee\ x=1\ \text{CORRECT}\\\\f(x)=-4x^2-16x-1\to f(1)=-4\cdot1^2-16\cdot1-1=-4-16-1=-21\\eq0\ :(`


Answer:

`\boxed{f(x)=x^2-5x+4\ and\ f(x)=-2x^2+10x-8}`

Answer 2

ED2020 its B and D

Explanation: