Question

a function f(x) has x-intercepts of -3 and -5. what is the constant term in the function? f(x) = x² + 8x + \boxed{1}

Explanation:

Step1: Recall root - factor relationship

If a function \( f(x) \) has \( x \)-intercepts (roots) at \( x = a \) and \( x = b \), then the function can be written in factored form as \( f(x)=(x - a)(x - b) \). Here, the roots are \( x=-3 \) and \( x = - 5 \), so \( a=-3 \) and \( b=-5 \). Then the factored form is \( f(x)=(x-(-3))(x - (-5))=(x + 3)(x + 5) \).

Step2: Expand the factored form

We use the FOIL method to expand \( (x + 3)(x + 5) \).
First: \( x\times x=x^{2} \)
Outer: \( x\times5 = 5x \)
Inner: \( 3\times x=3x \)
Last: \( 3\times5 = 15 \)
Then combine like terms: \( x^{2}+5x+3x + 15=x^{2}+8x + 15 \)

Answer:

15