Given the function:
[tex]f(x)=x^2+5x-36[/tex]Let's find the zeros of the function.
To find the zeros, substitute 0 for f(x) and solve for x.
We have:
[tex]\begin{gathered} 0=x^2+5x-36 \\ \\ x^2+5x-36=0 \end{gathered}[/tex]Factor the expression using the AC method.
Find a pair of intergers whose sum is 5 and whose product is -36.
We have:
-4 and 9
Thus, we have the factors:
[tex](x-4)(x+9)=0[/tex]Now, equate each factor to zero and solve for x:
[tex]\begin{gathered} x-4=0 \\ \text{Add 4 to both sides:} \\ x-4+4=0+4 \\ x=4 \\ \\ \\ x+9=0 \\ Subtract\text{ 9 from both sides:} \\ x+9-9=0-9 \\ x=-9 \end{gathered}[/tex]We have the solutions:
x = 4 and -9
Therefore, the zeros of the function are: 4 and -9
ANSWER:
A) 4 and -9