The roots of a function, are the zeros of the function.
The roots of the equation are: -7, 2 and 4
The function is given as:
[tex]\mathbf{f(x) = 5x^3 + 5x^2 - 170x + 280}[/tex]
If one of the roots is (x + 7), then it means (x + 7) can divide f(x).
So, we have:
[tex]\mathbf{\frac{f(x)}{x + 7} = \frac{5x^3 + 5x^2 - 170x + 280}{x +7}}[/tex]
Factorize the numerator
[tex]\mathbf{\frac{f(x)}{x + 7} = \frac{(5x^2 - 30x + 40)(x + 7)}{x +7}}[/tex]
Cancel out the common terms
[tex]\mathbf{\frac{f(x)}{x + 7} = 5x^2 - 30x + 40}[/tex]
Factor out 5
[tex]\mathbf{\frac{f(x)}{x + 7} = 5(x^2 - 6x + 8)}[/tex]
Expand
[tex]\mathbf{\frac{f(x)}{x + 7} = 5(x^2 - 2x - 4x + 8)}[/tex]
Factorize
[tex]\mathbf{\frac{f(x)}{x + 7} = 5(x(x - 2) - 4(x - 2))}[/tex]
Factor out x - 2
[tex]\mathbf{\frac{f(x)}{x + 7} = 5(x - 4)(x - 2)}[/tex]
Multiply both sides by x + 7
[tex]\mathbf{f(x) = 5(x - 4)(x - 2)(x + 7)}[/tex]
Set f(x) to 0
[tex]\mathbf{5(x - 4)(x - 2)(x + 7) = 0}[/tex]
Divide through by 5
[tex]\mathbf{(x - 4)(x - 2)(x + 7) = 0}[/tex]
Split
[tex]\mathbf{(x - 4)= 0 \,or (x - 2)= 0\ or (x + 7) = 0}[/tex]
Solve for x
[tex]\mathbf{x= 4 \,or\ x= 2\ or\ x = -7}[/tex]
Hence, the roots of the equation are: -7, 2 and 4
Read more about roots of functions at:
https://brainly.com/question/16684818
